# 3.9. Real Valued Functions#

In this section, we discuss results related to real valued functions on metric spaces. It is suggested to review the material from Real Valued Functions on arbitrary sets. We assume $$(X,d)$$ to be an arbitrary metric space in this section. Unless otherwise specified, $$f : X \to \RR$$ is a partial real valued function from $$X$$ to $$\RR$$ with $$\dom f \subseteq X$$.

When the codomain of a function is $$\RR$$, it provides an additional structure of total order on the range of possible values of $$f$$.

1. We can introduce the notion of local and global maximum or minimum values (local and global extrema).

2. We can construct the epigraphs, hypographs, sublevel sets, superlevel sets and contours of a function. This allows us to think about the properties of these sets. Of particular interest are functions whose epigraphs are closed or all sublevel sets are closed.

3. When we discuss limits at some point $$a \in \dom f$$, we can think in terms of whether the nearby values are above or below $$f(a)$$. For each deleted neighborhood of $$a$$, we can find out the largest (supremum) or the smallest (infimum) values. This enables us to introduce the notions of limit superior and limit inferior. Naturally, the limit exists when the limit superior and limit inferior agree.

4. Similarly, the idea of continuity can be split into continuity from above or below. Accordingly, the functions can be classified into lower and upper continuous functions. Continuous functions are both lower and upper continuous.

5. All of these notions easily carry over to extended valued functions (with signatures $$f: X \to \ERL$$).

This section introduces these concepts and focuses on the interplay of these concepts. For example closedness of functions (the notion that all sublevel sets are closed) is equivalent to closed epigraphs or lower semicontinuity.

When we discuss the closedness of the sublevel sets and epigraphs of a function, the closedness is with respect to the subspace topology of $$(S, d)$$ where $$S = \dom f$$.

Recall from Subspace Topology that for a metric space $$(S, d)$$

1. $$S$$ is open as well as closed in the subspace topology $$(S, d)$$.

2. A set $$A$$ is open in $$(S, d)$$ if and only if $$A = S \cap B$$ for some set $$B$$ which is open in $$(X, d)$$.

3. A set $$A$$ is closed in $$(S, d)$$ if and only if $$A = S \cap B$$ for some set $$B$$ which is closed in $$(X, d)$$.

4. If a sequence $$\{x_n \}$$ of $$S$$ is convergent w.r.t. the subspace topology $$(S,d)$$, then its limit $$x = \lim x_n \in S$$.

## 3.9.1. Extreme Values#

Definition 3.59 (Local extreme value)

We say that $$f(a)$$ is a local extreme value of $$f$$ at $$a \in \dom f$$ if there exists $$\delta > 0$$ such that $$f(x) - f(a)$$ doesn’t change sign on $$B(a, \delta) \cap \dom f$$.

More specifically,

1. $$f(a)$$ is a local maximum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \leq f(a) \Forall x \in B(a, \delta) \cap \dom f.$
2. $$f(a)$$ is a local minimum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \geq f(a) \Forall x \in B(a, \delta) \cap \dom f.$

The point $$x=a$$ is called a local extreme point of $$f$$ or more specifically, a local maximum or a local minimum point of $$f$$.

Definition 3.60 (Global maximum)

We say that $$f : X \to \RR$$ attains a global maximum at some $$a \in \dom f$$, if:

$f(x) \leq f(a) \Forall x \in \dom f.$

Definition 3.61 (Global minimum)

We say that $$f : X \to \RR$$ attains a global minimum at some $$a \in \dom f$$, if:

$f(x) \geq f(a) \Forall x \in \dom f.$

Definition 3.62 (Strict global maximum)

We say that $$f : X \to \RR$$ attains a strict global maximum at some $$a \in \dom f$$, if:

$f(x) < f(a) \Forall x \in \dom f, x \neq a.$

Definition 3.63 (Strict global minimum)

We say that $$f : X \to \RR$$ attains a strict global minimum at some $$a \in \dom f$$, if:

$f(x) > f(a) \Forall x \in \dom f, x \neq a.$

Theorem 3.91 (Extreme value theorem)

Let $$f : X \to \RR$$ be continuous. Let $$A$$ be a nonempty compact subset of $$\dom f$$. Then, the set $$f(A)$$ is closed and bounded. Also, there exists $$a$$ and $$b$$ in $$A$$ such that

$f(a) = \inf f(A) \text{ and } f(b) = \sup f(A);$

i.e., $$f$$ attains its supremum and infimum over the values in $$f(A)$$.

Proof. Recall from Theorem 3.78 that continuous image of a compact set is compact. Hence, $$f(A)$$ is compact.

But the compact subsets of $$\RR$$ are closed and bounded. Hence, $$f(A)$$ is closed and bounded.

Since $$f(A)$$ is closed and bounded, hence it contains a supremum and an infimum.

Let $$y = \inf f(A)$$. Since $$y \in f(A)$$, hence there exists $$a \in \dom f$$ such that $$f(a) = y$$.

Let $$z = \sup f(A)$$. Since $$z \in f(A)$$, hence there exists $$b \in \dom f$$ such that $$f(b) = z$$.

Extreme value theorem is useful in optimization problems. If it is possible to identify the feasible set of input values as a closed and bounded set, then it is possible to indicate if the optimization problem has a solution or not. Although, the theorem doesn’t help in identifying the solution as such.

Example 3.26

Consider the optimization problem of maximizing the volume of a box with the constraints:

$w + h + d \leq 6$

where $$w$$ indicates the width, $$h$$ indicates the height and $$d$$ indicates the depth of the box.

We can define volume as a function $$v : \RR^3 \to \RR$$ as

$v = w h d$

where each input vector $$\bx \in \RR^3$$ is a triplet $$(w, h, d)$$.

Now, note the implicit assumption that width, height and depth cannot be negative.

Thus, we have the following constraints:

$w \geq 0, h \geq 0, d \geq 0, w + h + d \leq 6.$

These constraints define a simplex in $$\RR^3$$ which is a closed and bounded set (thus compact). Hence, the range of function values is also closed and bounded. Hence, it is possible to choose a configuration which maximizes the volume.

## 3.9.2. Closed Functions#

Definition 3.64 (Closed function)

A real valued function $$f : X \to \RR$$ with $$S = \dom f$$ is closed if for each $$\alpha \in \RR$$, the corresponding sublevel set is closed with respect to the subspace topology $$(S,d)$$.

In other words, the sublevel set $$\{ x \in S \ST f(x) \leq \alpha \}$$ is closed for every $$\alpha \in \RR$$ in the subspace topology $$(S,d)$$.

### 3.9.2.1. Closed Functions on Non-Closed Domains#

Although every sublevel set of a closed function is closed, it doesn’t imply that the domain of the function itself is closed. We can very well have functions which are closed but their domain is open or semi-open or neither open nor closed.

Example 3.27 (A closed function need not have closed domain)

Let $$f: \RR \to \RR$$ be defined as $$f(x) = \frac{1}{x}$$ with $$S = \dom f = (0, \infty)$$.

1. The domain of $$f$$ is an open set.

2. Let $$\sublevel(f, \alpha)= \{ x \in S \ST f(x) \leq \alpha \}$$ denote the sublevel set for $$\alpha$$.

3. Then, $$\sublevel(f, \alpha) = [\frac{1}{\alpha}, \infty)$$ for every $$\alpha > 0$$. Thus, it is closed.

4. $$\sublevel(f, \alpha) = \EmptySet$$ for every $$\alpha \leq 0$$ since $$f(x)$$ is always positive. Thus, it is closed.

5. Thus, $$\sublevel(f, \alpha)$$ is closed for every $$\alpha \in \RR$$.

6. Thus, $$f$$ is a closed function.

We have shown a counter example where the function is closed but its domain is not closed.

While the domain of a closed function may not be closed, its epigraph indeed is closed.

### 3.9.2.2. Epigraphs#

Theorem 3.92 (Closed function = closed epigraph)

The epigraph of a function $$f : X \to \RR$$ with $$S = \dom f$$ is closed if and only if $$f$$ is closed.

Proof. Let $$f : X \to \RR$$ with $$S = \dom f$$. The epigraph of $$f$$ is given by

$\epi f = \{ (x, r) \in X \times \RR \ST x \in S, f(x) \leq r \}.$

By $$T_{\alpha}$$, we shall denote the sublevel set given as

$T_{\alpha} = \{ x \in S \ST f(x) \leq \alpha \}.$

Assume that $$\epi f$$ is closed.

1. Pick any $$\alpha \in \RR$$.

2. Let $$T_{\alpha} = \{ x \in S \ST f(x) \leq \alpha \}$$ be the corresponding sublevel set.

3. Let $$\{ x_n \}$$ be a convergent sequence of $$T_{\alpha}$$.

4. Let $$x = \lim_{n \to \infty} x_n$$. We need to show that $$x \in T_{\alpha}$$.

5. By definition of $$T_{\alpha}$$, for every $$x_n$$, we have $$f(x_n) \leq \alpha$$.

6. Thus, $$p_n = (x_n, \alpha) \in \epi f$$.

7. Now, we see that the sequence $$\{ p_n \}$$ of $$\epi f$$ is convergent and

$p = \lim p_n = \lim (x_n, \alpha) = (\lim x_n, \alpha) = (x, \alpha).$
8. Since $$\epi f$$ is closed, hence $$(x, \alpha) \in \epi f$$.

9. Thus, $$x \in S$$.

10. Also, by definition, $$(x, f(x)) \in \epi f$$ and $$f(x) \leq \alpha$$.

11. Thus, $$x \in S$$ and $$f(x) \leq \alpha$$.

12. Thus, $$x \in T_{\alpha}$$.

13. This, every convergent sequence of $$T_{\alpha}$$ converges in $$T_{\alpha}$$.

14. Thus, $$T_{\alpha}$$ is closed.

15. Since $$\alpha$$ was arbitrary, hence every sublevel set of $$f$$ is closed.

16. Thus, $$f$$ is a closed function.

Assume that $$f$$ is closed.

1. Thus, every sublevel set of $$f$$ is closed.

2. Let $$\{p_n\}$$ be a convergent sequence of $$\epi f$$.

3. Let $$p_n = (x_n, r_n)$$.

4. Then, $$f(x_n) \leq r_n$$ for all $$n \in \Nat$$.

5. Let $$p = \lim p_n$$. Let $$p = (x, r)$$.

6. Then, $$\lim x_n = x$$ and $$\lim r_n = r$$.

7. We need to show that $$p \in \epi f$$.

8. Recall from Theorem 2.5 that every convergent sequence of real numbers is bounded.

9. Since $$\{ r_n \}$$ is convergent, hence it is bounded.

10. Let $$M \in \RR$$ such that $$r_n \leq M$$ for all $$n \in \Nat$$.

11. Then, $$f(x_n) \leq r_n \leq M$$ for all $$n \in \Nat$$.

12. Consider the sublevel set $$T_M = \{ x \in S \ST f(x) \leq M \}$$.

13. Then, $$x_n \in T_M$$ for all $$n \in \Nat$$.

14. Then $$\{x_n \}$$ is a convergent sequence of $$T_M$$.

15. But every sublevel set of $$f$$ is closed. Hence $$T_M$$ is closed.

16. Every convergent sequence of a closed set converges in the set. Hence, $$x = \lim x_n \in T_M$$.

17. Thus, $$(x, f(x)) \in \epi f$$.

18. To show that $$p = (x, r) \in \epi f$$, we need to show that $$f(x) \leq r$$.

19. Let $$\limsup_{n \to \infty} f(x_n) = u$$.

20. Then, by Theorem 2.24, for any $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that

$f(x_n) < u + \epsilon \Forall n \geq n_0.$
21. Thus, $$x_n \in T_{u + \epsilon}$$ for every $$n \geq n_0$$ where $$T_{u + \epsilon}$$ is the sublevel set for $$u + \epsilon$$.

22. Since $$\{x_n \}$$ (after dropping the finite $$n_0$$ terms) is a convergent sequence of $$T_{u + \epsilon}$$ and $$T_{u + \epsilon}$$ is closed, hence $$x \in T_{u + \epsilon}$$.

23. Thus, $$f(x) \leq u + \epsilon$$.

24. Since, this is true for every $$\epsilon > 0$$, hence $$f(x) \leq u = \limsup_{n \to \infty} f(x_n)$$.

25. Recall that $$f(x_n) \leq r_n$$ for all $$n \in \Nat$$.

26. Then, by Theorem 2.29,

$f(x) \leq \limsup f(x_n) \leq \limsup r_n = r.$
27. Thus, $$f(x) \leq r$$.

28. Thus, $$p = (x, r) \in \epi f$$.

29. Thus, $$\{p_n\}$$ converges in $$\epi f$$.

30. Since $$\{p_n\}$$ was an arbitrary convergent sequence, hence every convergent sequence of $$\epi f$$ converges in $$\epi f$$.

31. Thus, $$\epi f$$ is closed.

A nice application of this result is the fact that pointwise supremum of closed functions is closed.

Theorem 3.93 (Pointwise supremum of closed functions)

Let $$f_i : X \to \RR$$ for $$i \in I$$ with $$S_i = \dom f_i$$ be a family of closed functions where $$I$$ is an index set.

The function

$f(x) = \sup_{i \in I} f_i(x)$

with $$\dom f = \bigcap_{i \in I} S_i$$ is closed.

Proof. Recall from Theorem 2.34 that the epigraph of maximum of two functions is the intersection of epigraphs.

1. Since $$f_i$$ are closed, hence $$\epi f_i$$ are closed for every $$i \in I$$.

2. The epigraph of $$f$$ is given by

$\epi f = \bigcap_{i \in I} \epi f_i.$
3. Since $$\epi f_i$$ are closed, hence $$\epi f$$ is closed due to Theorem 3.9.

4. Since $$\epi f$$ is closed, hence $$f$$ is closed due to Theorem 3.92.

### 3.9.2.3. Nonnegative Scaling#

Theorem 3.94 (Nonnegative scaling of closed function)

Let $$f: X \to \RR$$ be a closed function. Let $$t \geq 0$$. Then a function $$g : X \to \RR$$ given by

$g(x) = t f(x)$

is closed.

Proof. Note that $$\dom g = \dom f$$. Consider first the case of $$t=0$$.

1. Then $$g(x) = 0$$ for every $$x \in \dom f$$.

2. Thus for any $$s \geq 0$$, $$\sublevel(g, s) = \dom f$$.

3. And for $$s < 0$$, $$\sublevel(g, s) = \EmptySet$$.

4. Both $$\dom f$$ and $$\EmptySet$$ are closed set w.r.t. the subspace topology of $$\dom f$$.

5. Hence $$g$$ is closed.

Now consider the case where $$t > 0$$.

1. Pick any $$s \in \RR$$.

2. Then

$\begin{split} \sublevel(g, s) &= \{ x \in \dom g \ST g(x) \leq s \}\\ &= \{ x \in \dom f \ST t f(x) \leq s \} \\ &= \{ x \in \dom f \ST f(x) \leq \frac{s}{t} \} \\ &= \sublevel(f, \frac{s}{t}). \end{split}$
3. Since $$f$$ is closed, hence $$\sublevel(f, \frac{s}{t})$$ is closed, hence $$\sublevel(g, s)$$ is closed.

4. Since this is true for every $$s \in \RR$$, hence $$g$$ is closed.

### 3.9.2.4. Sum Rule#

Theorem 3.95 (Sum of closed functions)

Let $$f, g: X \to \RR$$ be closed functions. Then $$h = f + g$$ with $$\dom h = \dom f \cap \dom g$$ is also a closed function.

Proof. We make use of the fact that closed functions are lower semicontinuous. See later in Theorem 3.119.

1. Since $$f$$ and $$g$$ are closed, hence due to Theorem 3.119 they are l.s.c..

2. By Theorem 3.113, $$h = f + g$$ is l.s.c..

3. Again due to Theorem 3.119, $$h$$ is closed.

### 3.9.2.5. Continuous Functions#

Theorem 3.96 (Continuity + closed domain implies closedness)

If $$f: X \to \RR$$ is continuous and $$\dom f$$ is closed, then $$f$$ is closed.

Proof. We shall prove this by showing that the sublevel sets are closed.

1. Let $$S = \dom f$$.

2. Pick $$t \in \RR$$.

3. Let $$T = \sublevel(f, t)$$.

4. By definition, $$T \subseteq S$$.

5. Let $$\{ x_n \}$$ be a convergent sequence of $$T$$.

6. Let $$x = \lim x_n$$.

7. Since $$S$$ is closed, hence $$\{ x_n \}$$ converges in $$S$$.

8. Hence $$x \in S$$ and $$f(x)$$ is well defined.

9. Since $$\lim x_n = x$$ and $$f$$ is continuous, hence due to Theorem 3.42 (3), $$\lim f(x_n) = f(x)$$.

10. By sublevel property of $$T$$, $$f(x_n) \leq t$$ for every every $$n$$.

11. Consequently,

$f(x) = \lim_{n \to \infty} f(x_n) \leq t.$
12. Since $$f(x) \leq t$$, hence $$x \in T$$.

13. Thus every convergent sequence of $$T$$ converges in $$T$$.

14. Hence $$T$$ is closed.

15. Since $$t$$ was arbitrarily chosen, hence every sublevel set of $$f$$ is closed.

16. Hence $$f$$ is a closed function.

Theorem 3.97 (Closedness conditions for continuity + open domain)

If $$f: X \to \RR$$ is continuous and $$\dom f$$ is open, then $$f$$ is closed if and only if $$f$$ converges to $$\infty$$ along every sequence converging to a boundary point of $$\dom f$$.

Proof. To show that a function is closed, we need to show that all its sublevel sets are closed. To show that a sublevel set is closed we need to show that every convergent sequence of a sublevel set converges in the set itself.

1. Let $$S = \dom f$$.

2. Let $$C = \closure S$$.

3. It is given that $$S$$ is open. Hence $$S = \interior C$$.

4. Let $$B = \boundary S = C \setminus \interior C$$.

5. Then $$B = C \setminus S$$. In other words, $$B \cap S = \EmptySet$$.

6. Let $$\{ x_n \}$$ be a convergent sequence of $$S$$.

7. Then $$x = \lim x_n \in C$$.

8. So either $$x \in S$$ or $$x \in B$$.

9. If $$x \in S$$ then $$f(x)$$ is well defined. If $$x \in B$$ then $$f(x)$$ is not defined.

Assume that $$f$$ converges to infinity along any sequence converging to $$B$$.

1. Pick $$t \in \RR$$.

2. Let $$T = \sublevel (f, t)$$.

3. Suppose $$\{ x_n \}$$ is a convergent sequence of $$T$$ with $$x = \lim x_n$$.

4. Then $$f(x_n) \leq t$$ for every $$n$$.

5. For contradiction, assume that $$x \in B$$.

6. Then $$\lim f(x_n) = \infty$$.

7. But then there exists $$n_0$$ such that for every $$n > n_0$$, $$f(x_n) > t$$.

8. This contradicts the assumption that $$f(x_n) \leq t$$ for every $$n$$.

9. Hence $$x \in S$$.

10. But then $$f(x)$$ is well defined.

11. By continuity of $$f$$, $$f(x) = \lim f(x_n) \leq t$$.

12. Hence $$x \in T$$.

13. Hence $$T$$ is closed.

14. Since every sublevel set is closed, hence $$f$$ is closed.

Now for the converse, assume that $$f$$ does not converge to infinity along some sequence converging to $$B$$.

1. Let $$\{ x_n \}$$ be such a convergent sequence such that $$x = \lim x_n \in B$$ and $$\lim f(x_n) = r \in \RR$$.

2. Pick some $$\epsilon > 0$$.

3. Then there exists $$n_0$$ such that for all $$n > n_0$$, \$| r - f(x_n)| < \epsilon.

4. Thus for all $$n > n_0$$, $$f(x_n) < r + \epsilon$$.

5. Consider the sublevel set $$R = \sublevel(f, r+\epsilon)$$.

6. By dropping the first $$n_0$$ points of $$\{ x_n \}$$, the remaining sequence $$\{ y_n \}$$ where $$y_n = x_{n + n_0}$$ belongs to $$R$$.

7. Thus we have a convergent sequence of $$R$$ which doesn’t converge in $$R$$ since $$R \subseteq S$$, $$x = \lim y_n \in B$$ and $$B \setminus R = \EmptySet$$.

8. Thus $$R$$ is not closed.

9. Since there are sublevel sets of $$f$$ which are not closed, hence $$f$$ is not closed.

TODO, is it possible that for a convergent sequence $$\{ x_n \}$$, the corresponding sequence $$f(x_n)$$ doesn’t converge to anything?

## 3.9.3. Limit Superiors and Inferiors#

Definition 3.65 (Limit superior and limit inferior for functions)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$.

For some $$\delta > 0$$, let

$u_{\delta} = \sup_{x \in B_d(a,r) \cap S} f(x).$

Then, the limit superior of the function $$f$$ at $$a$$ is defined by

$\limsup_{x \to a } f(x) \triangleq \inf_{\delta > 0} u_{\delta} = \inf_{\delta > 0} \sup_{x \in B_d(a,r) \cap S} f(x).$

Similarly, let

$l_{\delta} = \inf_{x \in B_d(a,r) \cap S} f(x).$

Then, the limit inferior of the function $$f$$ at $$a$$ is defined by

$\liminf_{x \to a } f(x) \triangleq \sup_{\delta > 0} l_{\delta} = \sup_{\delta > 0} \inf_{x \in B_d(a,r) \cap S} f(x).$

We note that limit superior and inferior is also defined for points which are not necessarily in $$S$$ but are on the boundary of $$S$$ as some accumulation points of $$S$$ may be on its boundary outside $$S$$. E.g., $$\tan(x)$$ is not defined at $$x = \frac{\pi}{2}$$ but $$\frac{\pi}{2}$$ is an accumulation point for $$\dom \tan$$. Hence, $$\limsup$$ and $$\liminf$$ can be computed there.

$$B_d(a,r) \cap S$$ is simply the part of deleted neighborhood at $$a$$ of radius $$r$$ which intersects with the domain of $$f$$. Since $$a$$ is an accumulation point of $$S$$, hence $$B_d(a,r) \cap S$$ is not empty. Thus, we are evaluating $$f$$ only at points at which it is defined.

$$u_{\delta}$$ is the supremum value of $$f$$ in the deleted neighborhood $$B_d(a, \delta) \cap S$$.

It is clear that as $$\delta$$ increases, $$u_{\delta}$$ also increases.

If we define a function $$g : (0, \infty) \to \RERL$$ as

$g(\delta) = u_{\delta} = \sup_{x \in B_d(a,\delta) \cap S} f(x),$

then $$g$$ is a nondecreasing function. Then,

$\limsup_{x \to a } f(x) = \inf_{\delta > 0} g(\delta).$

Definition 3.66 (Locally bounded function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$.

We say that $$f$$ is locally bounded above around $$a$$ if there exists $$r > 0$$ and $$M \in \RR$$ such that

$f(x) \leq M \Forall x \in B(x, r) \cap S.$

We say that $$f$$ is locally bounded below around $$a$$ if there exists $$r > 0$$ and $$m \in \RR$$ such that

$f(x) \geq m \Forall x \in B(x, r) \cap S.$

Remark 3.11 (Locally bounded functions and limit superior and inferior)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a \in S$$.

If $$f$$ is locally bounded above at $$a$$, then $$\limsup_{x \to a} f(x)$$ is finite. Otherwise, $$\limsup_{x \to a} f(x) = \infty$$.

Similarly, if $$f$$ is locally bounded below at $$a$$, then $$\liminf_{x \to a} f(x)$$ is finite. Otherwise, $$\liminf_{x \to a} f(x) = -\infty$$.

### 3.9.3.1. Limit Superior#

Theorem 3.98 (Characterization of function limit superior)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. Then, $$u = \limsup_{x \to a} f(x)$$ if and only if the following two conditions hold:

1. For every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x) < u + \epsilon \Forall x \in B_d(a, \delta) \cap S.$
2. For every $$\epsilon > 0$$ and for every $$\delta > 0$$, there exists $$x_{\delta} \in B_d(a, \delta) \cap S$$ such that

$u - \epsilon < f(x_{\delta}).$

Proof. Let $$g : (0, \infty) \to \RERL$$ be defined as

$g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x).$

Suppose that $$u = \limsup_{x \to a} f(x)$$.

1. Then, $$u = \inf_{\delta > 0} g(\delta)$$.

2. Then, for ever $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$u \leq g(\delta) < u + \epsilon.$

Otherwise, $$u$$ won’t be the infimum of $$g(\delta)$$.

3. Thus, for ever $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x) < u + \epsilon \Forall x \in B_d(a,\delta) \cap S$

which proves condition (1).

4. Now, for every $$\epsilon > 0$$ and every $$\delta > 0$$, we have

$u - \epsilon < u \leq g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x).$
5. By definition of the supremum, there exists $$x \in B_d(a,\delta) \cap S$$ such that

$u - \epsilon < f(x_{\delta}).$

Otherwise, $$g(\delta)$$ would be smaller than $$u$$. This proves condition (2).

For the converse, we assume that conditions (1) and (2) are satisfied.

1. Let $$\epsilon > 0$$. Choose $$\delta > 0$$ that satisfies condition (1).

2. Then, we get

$g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x) \leq u + \epsilon.$
3. Consequently,

$\limsup_{x \to a} f(x) = \inf_{\delta > 0} g(\delta) \leq u + \epsilon.$
4. Since $$\epsilon > 0$$ can be arbitrarily small, hence

$\limsup_{x \to a} f(x) \leq u.$
5. Again, fix any $$\epsilon > 0$$ and pick any $$\delta > 0$$. From condition (2), there exists $$x_{\delta} \in B_d(a,\delta) \cap S$$ such that

$u - \epsilon < f(x_{\delta}).$
6. But,

$f(x_{\delta}) \leq \sup_{x \in B_d(a,\delta) \cap S} f(x) = g(\delta).$
7. Thus, $$u - \epsilon < g(\delta)$$.

8. Taking infimum on the R.H.S., over all $$\delta > 0$$,

$u - \epsilon \leq \inf_{\delta > 0} g(\delta) = \limsup_{x \to a} f(x).$
9. Since $$\epsilon > 0$$ can be arbitrarily small, hence

$u \leq \limsup_{x \to a} f(x).$

Combining, these inequalities, we get $$u = \limsup_{x \to a} f(x)$$.

Theorem 3.99 (Existence of convergent sequence to the limit superior of a function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. If $$u = \limsup_{x \to a} f(x)$$, then there exists a sequence $$\{ x_n \}$$ of $$S$$ converging to $$a$$ with $$x_n \neq a$$ for every $$n$$ such that

$\lim_{n \to \infty} f(x_n) = u.$

Moreover, if $$\{ y_n \}$$ is any sequence of $$S$$ converging to $$a$$ with $$y_n \neq a$$ for every $$n$$, and the sequence $$\{ f(y_n ) \}$$ converges, then

$\lim_{n \to \infty} f(y_n) = u' \leq u.$

In other words, for any sequence $$\{ x_n \}$$ of $$S \setminus \{ a \}$$ converging to $$a$$, $$u$$ is the least upper bound on $$\lim_{n \to \infty} f(y_n)$$ if $$\{ f(y_n) \}$$ converges.

Proof. $$u$$ as the least upper bound.

1. Let $$\{ y_n \}$$ be a sequence of $$S \setminus \{a\}$$ such that $$\lim_{n \to \infty} y_n = a$$ and $$\{ f(y_n) \}$$ converges.

2. Let $$\epsilon > 0$$.

3. Then, due to Theorem 3.98, there exists $$\delta > 0$$ such that

$f(x) < u + \epsilon \Forall x \in B_d(a, \delta) \cap S.$
4. Since $$\{ y_n \}$$ is convergent, hence there exists $$n_0 \in \Nat$$ such that for every $$n > n_0$$, $$d(a, y_n) < \delta$$.

5. Thus, for every $$n > n_0$$, $$y_n \in B_d(a, \delta) \cap S$$.

6. Thus, for every $$n > n_0$$, $$f(y_n) < u + \epsilon$$.

7. Thus,

$\lim_{n \to \infty} f(y_n) \leq u + \epsilon.$
8. Since this is true for any $$\epsilon > 0$$ and $$\epsilon$$ can be made arbitrarily small, hence

$\lim_{n \to \infty} f(y_n) \leq u = \limsup_{ x \to a} f(x).$

Construction of $$\{ x_n \}$$ such that $$\lim_{n \to \infty} f(x_n) = u$$.

1. Let $$\epsilon_n = \frac{1}{n}$$.

2. Then, due to Theorem 3.98 (1), there exists $$\delta_n > 0$$ such that

$f(x) < u + \epsilon_n \Forall x \in B_d(a, \delta_n) \cap S.$
3. Now, let $$\delta'_n = \min\{\delta_n, \frac{1}{n} \}$$. Clearly, $$\delta'_n > 0$$.

4. Due to Theorem 3.98 (2), there exists $$x_n \in B_d(a, \delta'_n) \cap S$$ such that

$u - \epsilon_n < f(x_n).$
5. Consider the sequence so constructed $$\{ x_n \}$$.

6. Since $$\delta'_n \leq \frac{1}{n}$$, hence

$\lim_{n \to \infty} x_n = a.$
7. For every $$n$$, $$u - \epsilon_n < f(x_n) < u + \epsilon_n$$.

8. $$\lim_{n \to \infty} \epsilon_n = 0$$.

9. Thus, by squeeze theorem,

$\lim_{n \to \infty} f(x_n) = u.$

Theorem 3.100 (Divergence of limit superior of a function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. Then,

$\limsup_{x \to a} f(x) = \infty$

if and only if there exists a sequence $$\{x_n \}$$ of $$S \setminus \{ a \}$$ such that $$\lim x_n = a$$ and $$\lim f(x_n) = \infty$$.

Proof. Let $$g : (0, \infty) \to \RERL$$ be defined as

$g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x).$

Assume that $$\limsup_{x \to a} f(x) = \infty$$.

1. Then, $$\inf_{\delta > 0} g(\delta) = \infty$$.

2. Thus, $$g(\delta) = \infty$$ for every $$\delta > 0$$.

3. For each $$n \in \Nat$$, let $$\delta_n = \frac{1}{n}$$.

4. We have

$g(\delta_n) = \sup_{x \in B_d(a,\delta_n) \cap S} f(x) = \infty.$
5. Thus, there exists $$x_n \in B_d(a,\delta_n) \cap S$$ such that $$f(x_n) > n$$.

6. We note that $$\{ x_n \}$$ converges to $$a$$ since $$d(x_n, a) < \frac{1}{n}$$.

7. At the same time $$\lim f(x_n) = \infty$$ as $$\{f(x_n) \}$$ is unbounded.

For the converse, assume that there exists a sequence $$\{x_n \}$$ of $$S \setminus \{ a \}$$ such that $$\lim x_n = a$$ and $$\lim f(x_n) = \infty$$.

1. Let $$\delta > 0$$.

2. Since $$\{ x_n \}$$ is convergent, hence there exists $$m \in \Nat$$ such that for all $$n \geq m$$, we have $$x_n \in B_d(a,\delta) \cap S$$.

3. Since $$\lim f(x_n) = \infty$$, hence for every $$M > 0$$, there exists $$k \in \Nat$$ such that $$f(x_n) \geq M$$ for all $$n \geq k$$.

4. Let $$p = \max(k, m)$$.

5. Then, for all $$n \geq p$$, $$x_n \in B_d(a,\delta) \cap S$$ and $$f(x_n) \geq M$$.

6. Thus,

$g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x) \geq M.$
7. Since $$M$$ can be made arbitrarily large, hence $$g(\delta) = \infty$$.

8. Since $$g(\delta) = \infty$$ for every $$\delta > 0$$, hence

$\limsup_{x \to a} f(x) = \inf_{\delta > 0} g(\delta) = \infty.$

### 3.9.3.2. Limit Inferior#

Theorem 3.101 (Characterization of function limit inferior)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. Then, $$l = \liminf_{x \to a} f(x)$$ if and only if the following two conditions hold:

1. For every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$l - \epsilon < f(x) \Forall x \in B_d(a, \delta) \cap S.$
2. For every $$\epsilon > 0$$ and for every $$\delta > 0$$, there exists $$x_{\delta} \in B_d(a, \delta) \cap S$$ such that

$f(x_{\delta}) < l + \epsilon.$

Theorem 3.102 (Existence of convergent sequence to the limit inferior of a function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. If $$l = \liminf_{x \to a} f(x)$$, then there exists a sequence $$\{ x_n \}$$ of $$S$$ converging to $$a$$ with $$x_n \neq a$$ for every $$n$$ such that

$\lim_{n \to \infty} f(x_n) = l.$

Moreover, if $$\{ y_n \}$$ is any sequence of $$S$$ converging to $$a$$ with $$y_n \neq a$$ for every $$n$$, and the sequence $$\{ f(y_n ) \}$$ converges, then

$\lim_{n \to \infty} f(y_n) = l' \geq l.$

In other words, for any sequence $$\{ x_n \}$$ of $$S \setminus \{ a \}$$ converging to $$a$$, $$l$$ is the greatest lower bound on $$\lim_{n \to \infty} f(y_n)$$ if $$\{ f(y_n) \}$$ converges.

Theorem 3.103 (Divergence of limit inferior of a function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$. Then,

$\liminf_{x \to a} f(x) = -\infty$

if and only if there exists a sequence $$\{x_n \}$$ of $$S \setminus \{ a \}$$ such that $$\lim x_n = a$$ and $$\lim f(x_n) = -\infty$$.

### 3.9.3.3. Existence of Function Limit#

Theorem 3.104 (Function limit = limit superior = limit inferior)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a$$ be an accumulation point of $$S$$.

Then,

$\lim_{x \to a} f(x) = l$

if and only if

$\limsup_{x \to a} f(x) = \liminf_{x \to a} f(x) = l.$

Proof. Suppose $$\lim_{x \to a} f(x) = l$$.

1. For every $$\epsilon > 0$$, there exists $$r > 0$$ such that

$l - \epsilon < f(x) < l + \epsilon \Forall x \in B_d(a, r) \cap S.$
2. Note that this holds true for every $$\delta \in (0, r]$$.

3. Thus, for every $$\delta \in (0, r]$$

$l - \epsilon < g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x) \leq l + \epsilon.$
4. Recall that $$g(\delta)$$ is a nondecreasing function.

5. Thus, taking infimum over $$\delta > 0$$

$l - \epsilon \leq \inf_{\delta > 0}g(\delta) \leq l + \epsilon.$
6. Since, $$\epsilon$$ can be made arbitrarily small, hence

$\limsup_{x \to a} f(x) = \inf_{\delta > 0}g(\delta) = l.$
7. An identical reasoning shows that $$\liminf_{x \to a} f(x) = l$$.

For the converse, assume that $$\limsup_{x \to a} f(x) = \liminf_{x \to a} f(x) = l$$.

1. Let $$\epsilon > 0$$.

2. From Theorem 3.98, there exists $$\delta_1 > 0$$ such that

$f(x) < l + \epsilon \Forall x \in B_d(a, \delta_1) \cap S.$
3. From Theorem 3.101, there exists $$\delta_2 > 0$$ such that

$l - \epsilon < f(x) \Forall x \in B_d(a, \delta_2) \cap S.$
4. Let $$\delta = \min\{\delta_1, \delta_2 \}$$. Then,

$l - \epsilon < f(x) < l + \epsilon \Forall x \in B_d(a, \delta) \cap S.$
5. Thus, for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$|f(x) - l | < \epsilon \Forall x \in B_d(a, \delta) \cap S.$
6. Thus, $$\lim_{x \to a} f(x) = l$$.

## 3.9.4. Semicontinuity#

The concept of semicontinuity is useful for the study of extreme values of some discontinuous functions.

We start with the notion of limit superior and limit inferior at a point for functions. We then proceed to define the notion of semicontinuity.

It is conventional to abbreviate “lower semicontinuous” as “l.s.c.” and “upper semicontinuous” as “u.s.c.”. We will use these abbreviations liberally.

Definition 3.67 (Lower and upper semicontinuity)

A (partial) function $$f : X \to \RR$$ with $$S = \dom f \subseteq X$$ is said to be lower-semicontinuous at $$a \in S$$ if for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(a) - \epsilon < f(x) \text{ for every } x \in B(a, \delta) \cap S.$

Similarly, $$f$$ is said to be upper-semicontinuous at $$a \in S$$ if for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x) < f(a) + \epsilon \text{ for every } x \in B(a, \delta) \cap S.$

We say that $$f$$ is lower semicontinuous (l.s.c.) if $$f$$ is l.s.c. at every point of $$S$$.

Similarly, we say that $$f$$ is upper semicontinuous (u.s.c.) if $$f$$ is u.s.c. at every point of $$S$$.

Example 3.28 (Semicontinuous functions)

Consider the function $$f : \RR \to \RR$$ defined as

$\begin{split} f(x) = \begin{cases} 0, & x < 0\\ 1, & x \geq 0. \end{cases} \end{split}$

$$f$$ is upper semicontinuous at $$x=0$$.

1. We have $$f(0)= 1$$.

2. Let $$\epsilon > 0$$.

3. For any $$\delta > 0$$

1. $$f(\delta) = 1 < 1 + \epsilon$$.

2. $$f(-\delta) = 0 < 1 + \epsilon$$.

4. Thus, we can pick any $$\delta > 0$$ and for any $$x \in (-\delta, \delta)$$, $$f(x) < f(0) + \epsilon$$.

5. Thus, $$f$$ is upper semicontinuous at $$x=0$$.

We can easily show that $$f$$ is not lower semicontinuous at $$x=0$$.

1. Let $$\epsilon = \frac{1}{2}$$.

2. $$f(0)-\epsilon = \frac{1}{2}$$.

3. For any $$\delta > 0$$, $$f(0 - \delta) = 0 \not> \frac{1}{2} = f(0) - \epsilon$$.

4. thus, for this choice of $$\epsilon$$, there is no $$\delta > 0$$ satisfying the lower semicontinuity inequality.

Consider the function $$g : \RR \to \RR$$ defined as

$\begin{split} g(x) = \begin{cases} 0, & x \leq 0\\ 1, & x > 0. \end{cases} \end{split}$

$$g$$ is lower semicontinuous at $$x=0$$.

The ceiling function $$f(x) = \lceil x \rceil$$ is lower semicontinuous.

The floor function $$f(x) = \lfloor x \rfloor$$ is upper semicontinuous.

Theorem 3.105 (Semicontinuity at isolated points)

Let $$f : X \to \RR$$ with $$S = \dom f \subseteq X$$.

Let $$a \in S$$ be an isolated point of $$S$$. Then, $$f$$ is lower semicontinuous as well as upper semicontinuous at $$a$$.

Proof. Recall from Definition 3.23 that $$a$$ is isolated if there exists $$\delta > 0$$ such that $$B(a, r) \cap S = \{ a \}$$.

1. We are given that $$a$$ is isolated.

2. Let $$\epsilon > 0$$.

3. Choose $$\delta > 0$$ such that $$B(a, r) \cap S = \{ a \}$$

4. Since $$\epsilon > 0$$, hence $$f(a) - \epsilon < f(a)$$ and $$f(a) < f(a) + \epsilon$$.

5. Thus, $$f$$ is l.s.c. as well as u.s.c. at $$a$$.

### 3.9.4.1. Continuity#

Theorem 3.106 (Lower + upper semicontinuity = Continuity)

A (partial) function $$f : X \to \RR$$ with $$S = \dom f \subseteq X$$ is continuous at $$a \in S$$ if and only if $$f$$ is both lower semicontinuous and upper semicontinuous at $$a$$.

Proof. Assume that $$f$$ is continuous at $$a \in S$$.

1. Let $$\epsilon >0$$.

2. There exists $$\delta > 0$$ such that for every $$x \in B(a, \delta) \cap S$$, $$|f(x) - f(a)| < \epsilon$$.

3. Consequently, $$f(x) - f(a) < \epsilon$$ means that $$f(x) < f(a) + \epsilon$$ for every $$x \in B(a, \delta) \cap S$$.

4. Thus, $$f$$ is upper semicontinuous at $$a$$.

5. Similarly, $$f(a) - f(x) < \epsilon$$ means that $$f(a) - \epsilon < f(x)$$ for every $$x \in B(a, \delta) \cap S$$.

6. Thus, $$f$$ is lower semicontinuous at $$a$$.

Assume $$f$$ is lower and upper semicontinuous at $$a \in S$$.

1. Let $$\epsilon > 0$$.

2. There exists $$\delta_1 > 0$$ such that $$f(a) - \epsilon < f(x)$$ for every $$x \in B(a, \delta_1) \cap S$$.

3. There exists $$\delta_2 > 0$$ such that $$f(x) < f(a) + \epsilon$$ for every $$x \in B(a, \delta_2) \cap S$$.

4. Let $$\delta = \min(\delta_1, \delta_2)$$.

5. Then, for every $$x \in B(a, \delta) \cap S$$, $$f(a) - f(x) < \epsilon$$ as well as $$f(x) - f(a) < \epsilon$$.

6. Thus, for every $$x \in B(a, \delta) \cap S$$, $$|f(x) - f(a)| < \epsilon$$.

7. Thus, $$f$$ is continuous at $$a$$.

### 3.9.4.2. Limit Superior and Inferior#

Theorem 3.107 (Semicontinuity and function limits)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a \in S$$ be an accumulation point of $$S$$. Then, $$f$$ is lower semicontinuous at $$a$$ if and only if

$\liminf_{x \to a} f(x) \geq f(a).$

Similarly, $$f$$ is upper semicontinuous at $$a$$ if and only if

$\limsup_{x \to a}f(x) \leq f(a).$

For a function $$f$$ and at an accumulation point $$a \in \dom f$$, we define a function $$g : (0, \infty) \to \RERL$$ as

$g(\delta) = \sup_{x \in B_d(a,\delta) \cap S} f(x).$

$$g$$ is a nondecreasing function of $$\delta$$.

Similarly, we define a function $$h : (0, \infty) \to \LERL$$ as

$h(\delta) = \inf_{x \in B_d(a,\delta) \cap S} f(x).$

$$h$$ is a nonincreasing function of $$\delta$$.

We note that

$\limsup_{x \to a} f(x) = \inf_{\delta > 0} g(\delta) \text{ and } \liminf_{x \to a} f(x) = \sup_{\delta > 0} h(\delta).$

Proof. Let $$f$$ be lower semicontinuous at $$a$$.

1. Let $$\epsilon > 0$$.

2. Since $$f$$ is l.s.c. at $$a$$, hence there exists $$\delta_0 > 0$$ such that

$f(a) - \epsilon < f(x) \Forall x \in B(a, \delta_0) \cap S.$
3. Taking infimum in the R.H.S. over the set $$B_d(a, \delta_0) \cap S$$,

$f(a) - \epsilon \leq h(\delta_0).$
4. Thus,

$\liminf_{x \to a} f(x) = \sup_{\delta > 0} h(\delta) \geq h(\delta_0) \geq f(a) - \epsilon.$
5. Since $$\epsilon$$ is arbitrary, hence

$\liminf_{x \to a} f(x) \geq f(a).$

For the converse, assume that $$\liminf_{x \to a} f(x) \geq f(a)$$.

1. We can write this as

$\liminf_{x \to a} f(x) = \sup_{\delta > 0} h(\delta) \geq f(a).$
2. Let $$\epsilon > 0$$.

3. Then,

$\sup_{\delta > 0} h(\delta) > f(a) - \epsilon.$
4. Thus, there exists $$\delta > 0$$ such that $$h(\delta) > f(a) - \epsilon$$.

5. Thus,

$f(x) > f(a) - \epsilon \Forall x \in B_d(a, \delta) \cap S.$
6. Also, $$f(a) > f(a) - \epsilon$$ trivially.

7. Thus, for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x) > f(a) - \epsilon \Forall x \in B(a, \delta) \cap S.$
8. Thus, $$f$$ is l.s.c. at $$a$$.

The proof for upper semicontinuity is analogous.

### 3.9.4.3. Converging Sequences#

Recall from Definition 2.39 that the limit superior and limit inferior of a sequence of real numbers is defined as

$\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup \{ x_k \ST k \geq n\}$

and

$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} \inf \{ x_k \ST k \geq n\}.$

Theorem 3.108 (Semicontinuity and converging sequences)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a \in S$$. Then, $$f$$ is l.s.c. at $$a$$ if and only if for every sequence $$\{x_n \}$$ of $$S$$ that converges to $$a$$,

$\liminf_{n \to \infty} f(x_n) \geq f(a).$

Similarly, $$f$$ is upper semicontinuous at $$a$$ if and only if every sequence $$\{x_k \}$$ of $$S$$ that converges to $$a$$,

$\limsup_{n \to \infty}f(x_n) \leq f(a).$

Proof. If $$a \in S$$ is an isolated point of $$S$$, then the only sequence that converges to $$a$$ is $$\{ x_n \}$$ where $$x_n = a$$ for all terms after a finitely many terms. For such sequences,

$\liminf_{n \to \infty} f(x_n) = \limsup_{n \to \infty} f(x_n) = f(a).$

Also, due to Theorem 3.105, $$f$$ is l.s.c. as well as u.s.c. at isolated points. Thus, this result holds trivially at isolated points.

We are now left with the case where $$a \in S$$ is an accumulation point of $$S$$.

Let $$f$$ be lower semicontinuous at $$a$$.

1. Let $$\epsilon > 0$$.

2. Since $$f$$ is l.s.c. at $$a$$, hence there exists $$\delta > 0$$ such that

$f(a) - \epsilon < f(x) \Forall x \in B(a, \delta) \cap S.$
3. Let $$\{x_n \}$$ be a sequence of $$S$$ that converges to $$a$$.

4. Then, there exists $$n_0 \in \Nat$$ such that for all $$n > n_0$$, $$x_n \in B(a, \delta) \cap S$$.

5. Thus, for all $$n > n_0$$, $$f(a) - \epsilon < f(x_n)$$.

6. It follows that $$f(a) - \epsilon \leq \liminf_{n \to \infty} f(x_n)$$.

7. Since $$\epsilon$$ can be arbitrarily small, hence $$f(a) \leq \liminf_{n \to \infty} f(x_n)$$.

For the converse, we assume that if $$\lim x_n = a$$, then $$\liminf_{n \to \infty} f(x_n) \geq f(a)$$.

1. By way of contradiction, assume that $$f$$ is not l.s.c. at $$a$$.

2. Then, there exists $$\epsilon > 0$$ such that for every $$\delta > 0$$, there exists $$x_{\delta} \in B(a, \delta) \cap S$$ such that

$f(a) - \epsilon \geq f(x_{\delta}).$
3. Let $$\delta_n = \frac{1}{n}$$.

4. We can construct a sequence $$\{ x_n \}$$ such that for every $$n$$, $$x_n \in B(a, \frac{1}{n}) \cap S$$ such that

$f(a) - \epsilon \geq f(x_n).$
5. This implies that

$f(a) - \epsilon \geq \liminf_{n \to \infty} f(x_n).$

A similar argument can be used for limit superior.

## 3.9.5. Extended Real Valued Functions#

Often, it is easier to work with extended real valued functions $$f : X \to \ERL$$. In this case, $$f$$ is defined at every $$x \in X$$ with $$f$$ taking the value of $$\infty$$ or $$-\infty$$ outside its effective domain.

$\dom f = \{x \in X \ST f(x) \in \RR \}.$

We don’t have to think in terms of subspace topology w.r.t. $$(S, d)$$ where $$S$$ is the effective domain. All the definitions and results can be presented w.r.t. the topology of $$(X, d)$$ itself.

### 3.9.5.1. Proper Functions#

Definition 3.68 (Proper function)

Let $$(X, d)$$ be a metric space. An extended real-valued function $$f : X \to \ERL$$ is called proper if its domain is nonempty, it never takes the value $$-\infty$$ and is not identically equal to $$\infty$$.

$\exists x \in X \text{ such that } f(x) < \infty \text{ and } f(x) > -\infty \Forall x \in X.$

Putting another way, a proper function is obtained by taking a real valued function $$f$$ defined on a nonempty set $$C \subseteq X$$ and then extending it to all of $$X$$ by setting $$f(x) = +\infty$$ for all $$x \notin C$$.

It is easy to see that the codomain for a proper function can be changed from $$\ERL$$ to $$\RERL$$ to clarify that it never takes the value $$-\infty$$.

Definition 3.69 (Improper function)

Let $$(X, d)$$ be a metric space. An extended real-valued function $$f : X \to \ERL$$ is called improper if it is not proper.

For an improper function $$f$$:

• $$\dom f$$ may be empty.

• $$f$$ might take a value of $$-\infty$$ at some $$x \in X$$.

Definition 3.70 (Indicator function)

Let $$(X, d)$$ be a metric space. Let $$C \subseteq X$$. Then, its indicator function is given by $$I_C(x) = 0 \Forall x \in C$$. Here, $$\dom I_C = C$$.

The extended value extension of an indicator function is given by:

$\begin{split} \tilde{I_C}(x) \triangleq \begin{cases} 0 & \text{for} & x \in C \\ \infty & \text{for} & x \notin C. \end{cases} \end{split}$

### 3.9.5.2. Extreme Values#

Let $$f : X \to \ERL$$ be an extended real valued function with $$S = \dom f$$.

1. For some $$a \in \dom f$$, $$f(a)$$ is a local maximum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \leq f(a) \Forall x \in B(a, \delta).$
2. For some $$a \in \dom f$$, $$f(a)$$ is a local minimum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \geq f(a) \Forall x \in B(a, \delta).$
3. We say that $$f$$ attains a global maximum at some $$a \in \dom f$$, if:

$f(x) \leq f(a) \Forall x \in X.$
4. We say that $$f$$ attains a global minimum at some $$a \in \dom f$$, if:

$f(x) \geq f(a) \Forall x \in X.$
5. We say that $$f$$ attains a strict global maximum at some $$a \in \dom f$$, if:

$f(x) < f(a) \Forall x \in X, x \neq a.$
6. We say that $$f$$ attains a strict global minimum at some $$a \in \dom f$$, if:

$f(x) > f(a) \Forall x \in X, x \neq a.$

Let $$f : X \to \ERL$$ be continuous. Let $$K$$ be a nonempty compact subset of $$X$$. Then, the set $$f(K)$$ is closed and bounded. Also, there exists $$a$$ and $$b$$ in $$K$$ such that

$f(a) = \inf f(K) \text{ and } f(b) = \sup f(K);$

i.e., $$f$$ attains its supremum and infimum over the values in $$f(K)$$.

### 3.9.5.3. Closed Functions#

Definition 3.71 (Closed Extended Real Valued Functions)

$$f : X \to \ERL$$ is closed if for each $$\alpha \in \RR$$, the sublevel set $$T_{\alpha} = \{ x \in X \ST f(x) \leq \alpha \}$$ is closed.

We note that $$T_{\infty} = X$$ is closed.

### 3.9.5.4. Limits#

Definition 3.72 (Limit superior and limit inferior)

Let $$f : X \to \ERL$$. Let $$a$$ be an accumulation point of $$X$$.

For some $$\delta > 0$$, let

$u_{\delta} = \sup_{x \in B_d(a,r)} f(x).$

Then, the limit superior of the function $$f$$ at $$a$$ is defined by

$\limsup_{x \to a } f(x) = \inf_{\delta > 0} u_{\delta} = \inf_{\delta > 0} \sup_{x \in B_d(a,r)} f(x).$

Similarly, let

$l_{\delta} = \inf_{x \in B_d(a,r)} f(x).$

Then, the limit inferior of the function $$f$$ at $$a$$ is defined by

$\liminf_{x \to a } f(x) \triangleq \sup_{\delta > 0} l_{\delta} = \sup_{\delta > 0} \inf_{x \in B_d(a,r)} f(x).$
• Even if $$a \notin \dom f$$, the $$\limsup_{x \to a} f(x)$$ and $$\liminf_{x \to a} f(x)$$ may still be finite as long as there is a deleted neighborhood of $$a$$ which is entirely contained in $$\dom f$$.

• If there is a deleted neighborhood at $$a$$ such that $$B_d(a, \delta) \cap \dom f = \EmptySet$$, then both limits will diverge.

Theorem 3.109 (Characterization of function limit superior)

Let $$f : X \to \ERL$$. Let $$a$$ be an accumulation point of $$X$$. Then, $$u = \limsup_{x \to a} f(x)$$ if and only if the following two conditions hold:

1. For every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x) < u + \epsilon \Forall x \in B_d(a, \delta).$
2. For every $$\epsilon > 0$$ and for every $$\delta > 0$$, there exists $$x_{\delta} \in B_d(a, \delta)$$ such that

$u - \epsilon < f(x_{\delta}).$

Theorem 3.110 (Characterization of function limit inferior)

Let $$f : X \to \ERL$$. Let $$a$$ be an accumulation point of $$X$$. Then, $$l = \liminf_{x \to a} f(x)$$ if and only if the following two conditions hold:

1. For every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that

$l - \epsilon < f(x) \Forall x \in B_d(a, \delta).$
2. For every $$\epsilon > 0$$ and for every $$\delta > 0$$, there exists $$x_{\delta} \in B_d(a, \delta)$$ such that

$f(x_{\delta}) < l + \epsilon.$

### 3.9.5.5. Semicontinuity#

Definition 3.73 (Lower and upper semicontinuity)

A function $$f : X \to \ERL$$ is said to be lower-semicontinuous at $$a \in X$$ if for every real $$t < f(a)$$, there exists $$\delta > 0$$ such that

$t < f(x) \text{ for every } x \in B(a, \delta).$

Similarly, $$f$$ is said to be upper-semicontinuous at $$a \in X$$ if for every real $$t > f(a)$$, there exists $$\delta > 0$$ such that

$f(x) < t \text{ for every } x \in B(a, \delta).$

We say that $$f$$ is lower semicontinuous (l.s.c.) if $$f$$ is l.s.c. at every point of $$X$$.

Similarly, we say that $$f$$ is upper semicontinuous (u.s.c.) if $$f$$ is u.s.c. at every point of $$X$$.

Theorem 3.111 (Semicontinuity and function limits)

Let $$f : X \to \ERL$$. Let $$a \in X$$ be an accumulation point of $$X$$. Then, $$f$$ is lower semicontinuous at $$a$$ if and only if

$\liminf_{x \to a} f(x) \geq f(a).$

Similarly, $$f$$ is upper semicontinuous at $$a$$ if and only if

$\limsup_{x \to a}f(x) \leq f(a).$

Theorem 3.112 (Semicontinuity and converging sequences)

Let $$f : X \to \ERL$$. Let $$a \in X$$. Then, $$f$$ is l.s.c. at $$a$$ if and only if for every sequence $$\{x_n \}$$ that converges to $$a$$,

$\liminf_{n \to \infty} f(x_n) \geq f(a).$

Similarly, $$f$$ is upper semicontinuous at $$a$$ if and only if every sequence $$\{x_k \}$$ that converges to $$a$$,

$\limsup_{n \to \infty}f(x_n) \leq f(a).$

## 3.9.6. Lower Semicontinuity#

The topological properties of convex sets can be studied in terms of lower semicontinuity. In this subsection, we study the implications of lower semicontinuity under the subspace topology.

### 3.9.6.1. Sum Rules#

Theorem 3.113 (Sum of lower semicontinuous functions)

Let $$f, g : X \to \RR$$ be lower semicontinuous functions. Then their sum $$h = f + g$$ with $$\dom h = \dom f \cap \dom g$$ is lower semicontinuous.

Similarly, if $$f, g: X \to \ERL$$ are l.s.c., then their sum $$h = f + g$$ is l.s.c. if $$h$$ is well defined at every $$x \in X$$ (i.e., the sum doesn’t take any indeterminate form).

Proof. We proceed as follows.

1. Let $$F = \dom f$$, $$G = \dom g$$ and $$H = \dom h$$.

2. Then $$H = F \cap G$$.

3. Let $$a \in H$$.

4. Since $$a \in F$$ and $$x \in G$$ hence both $$f$$ and $$g$$ are l.s.c. at $$a$$.

5. Choose $$\epsilon > 0$$.

6. Since $$f$$ is l.s.c. at $$a$$, there exists $$r_1 > 0$$ such that for every $$x \in B(a, r_1)$$, $$f(a) - \epsilon < f(x)$$.

7. Since $$g$$ is l.s.c. at $$a$$, there exists $$r_2 > 0$$ such that for every $$x \in B(a, r_2)$$, $$g(a) - \epsilon < g(x)$$.

8. Let $$r = \min(r_1, r_2)$$.

9. Then for every $$x \in B(a, r)$$

$f(a) + g(a) - 2\epsilon < f(x) + g(x).$
10. In other words, $$h(a) - 2 \epsilon < h(x)$$ for every $$x \in B(a, r)$$.

11. Hence $$h$$ is l.s.c. at $$a$$.

12. Since $$a \in H$$ is arbitrary, hence $$h$$ is l.s.c..

The argument for extended valued functions is similar.

1. Let $$a \in X$$.

2. We are given that both $$f$$ and $$g$$ are l.s.c. at $$a$$.

3. Let $$t \in \RR$$ such that $$t < h(a) = f(a) + g(a)$$.

4. Then there exist $$t_1, t_2 \in \RR$$ such that $$t = t_1 + t_2$$ and $$t_1 < f(a), t_2 < g(a)$$.

5. Since $$f$$ is l.s.c. at $$a$$, there exists $$r_1 > 0$$ such that for every $$x \in B(a, r_1)$$, $$t_1 < f(x)$$.

6. Since $$g$$ is l.s.c. at $$a$$, there exists $$r_2 > 0$$ such that for every $$x \in B(a, r_2)$$, $$t_2 < g(x)$$.

7. Let $$r = \min(r_1, r_2)$$.

8. Then for every $$x \in B(a, r)$$

$t = t_1 + t_2 < f(x) + g(x) = h(x).$
9. In other words, $$t < h(x)$$ for every $$x \in B(a, r)$$.

10. Hence $$h$$ is l.s.c. at $$a$$.

11. Since $$a \in X$$ is arbitrary, hence $$h$$ is l.s.c..

Theorem 3.114 (Positive combinations)

Let $$f_i : X \to \RERL$$, $$i=1,\dots,m$$ be given functions. Let $$t_1, \dots, t_m$$ be positive scalars. Consider the function $$g: X \to \RERL$$ given by

$g(x) = t_1 f_1(x) + \dots + t_m f_m(x) \Forall x \in X.$

If $$f_1, \dots, f_m$$ are l.s.c., then $$g$$ is l.s.c.

Proof. We proceed as follows.

1. Pick some $$x \in X$$.

2. For every sequence $$\{ x_k \}$$ converging to $$x$$, we have

$f_i(x) \leq \liminf_{k \to \infty} f_i (x_k)$

for every $$i$$.

3. Hence

$\begin{split} g(x) &= t_1 f_1(x) + \dots + t_m f_m(x) \\ &\leq \sum_{i=1}^m t_i \liminf_{k \to \infty} f_i (x_k) \\ &\leq \liminf_{k \to \infty} \sum_{i=1}^m t_i f_i (x_k) \\ &= \liminf_{k \to \infty} g(x_k). \end{split}$

This is valid since $$t_i > 0$$ for every $$i$$.

4. Hence $$g$$ is l.s.c. at $$x$$.

5. Since $$x$$ is arbitrarily chosen, hence $$g$$ is l.s.c..

### 3.9.6.2. Composition Rules#

Theorem 3.115 (Composition with a continuous function)

Let $$(X, d_1)$$ and $$(Y, d_2)$$ be metric spaces. Let $$f : X \to Y$$ be a continuous function and let $$g : Y \to \RR$$ be a lower semicontinuous function. Then their composition $$h = g \circ f$$ is lower semicontinuous.

Proof. .

1. Let $$\{ x_k \}$$ be a sequence of points of $$\dom h$$ converging to some $$x \in \dom h$$.

2. Since $$h = g \circ f$$, hence $$f(x_k) \in \dom g$$ for every $$k$$ and $$f(x) \in \dom g$$.

3. By continuity of $$f$$, the sequence $$\{ f(x_k) \}$$ converges to $$f(x)$$ (Theorem 3.43).

4. Note that $$\{ f(x_k) \}$$ is a sequence of $$Y$$ converging to $$f(x)$$.

5. Since $$g$$ is l.s.c., hence due to Theorem 3.111

$\liminf_{k \to \infty} g(f(x_k)) \geq g(f(x)).$
6. Hence $$h$$ is l.s.c..

Theorem 3.116 (Composition with a real function)

Let $$(X, d)$$ be a metric space. Let $$f : X \to \RR$$ be a lower semicontinuous function. Let $$g: \RR \to \RR$$ be a lower semicontinuous and monotonically nondecreasing function. Then their composition $$h = g \circ f$$ is lower semicontinuous.

Proof. Assume for contradiction that $$h$$ is not l.s.c..

1. Then there exists a sequence $$\{ x_n \}$$ of $$\dom h$$ converging to $$x \in \dom h$$ such that

$\liminf_{k \to \infty} g(f(x_k)) < g(f(x)).$
2. Let $$\{ x_l \}$$ be a subsequence of $$\{ x_k \}$$ achieving this limit inferior; i.e.

$\lim_{l \to \infty} g(f(x_l)) = \liminf_{k \to \infty} g(f(x_k)) < g(f(x)).$
3. Without loss of generality, we can assume that

$g(f(x_l)) < g(f(x)) \Forall l.$

We can achieve this by simply dropping the finitely many terms from the sequence for which this condition doesn’t hold.

4. Since $$g$$ is monotonically nondecreasing, hence $$g(f(x_l)) < g(f(x))$$ implies that $$f(x_l) < f(x)$$ for every $$l$$.

5. Taking limit superior, we have

$\limsup_{l \to \infty} f(x_l) \leq f(x).$
6. Since $$x_l \to x$$, hence due to lower semicontinuity of $$f$$

$f(x) \leq \liminf_{l \to \infty} f(x_l) \leq \limsup_{l \to \infty} f(x_l) \leq f(x).$
7. Thus $$\lim_{l \to \infty}f(x_l) = f(x)$$ since limit superior and limit inferior must be identical and equal to $$f(x)$$.

8. Since $$g$$ is lower semicontinuous and $$f(x_l) \to f(x)$$ hence

$\lim_{l \to \infty} g(f(x_l)) = \liminf_{l \to \infty} g(f(x_l)) \geq g(f(x)).$
9. This contradicts our earlier claim that $$\lim_{l \to \infty} g(f(x_l)) < g(f(x))$$.

10. Hence $$h$$ must be lower semicontinuous.

### 3.9.6.3. Convergent Dominating Sequences#

Theorem 3.117 (Lower semicontinuity and convergent dominating sequence)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Let $$a \in S$$. Let $$\{ a_n \}$$ be a sequence of $$S$$. Let $$\{ \mu_n \}$$ be a sequence of real numbers such that $$\mu_n \geq f(a_n)$$.

Then, $$f$$ is l.s.c. at $$a$$ if and only if $$\mu \geq f(a)$$ whenever $$\mu = \lim \mu_n$$ and $$a = \lim a_n$$.

Proof. Assume that $$f$$ is l.s.c. at $$a$$ and $$\mu = \lim \mu_n$$ and $$a = \lim a_n$$.

1. Then, due to Theorem 3.108,

$\liminf_{n \to \infty} f(a_n) \geq f(a).$
2. We are given that $$\mu_n \geq f(a_n)$$ for every $$n$$.

3. By Theorem 2.29,

$\liminf_{n \to \infty} \mu_n \geq \liminf_{n \to \infty} f(a_n).$
4. Since $$\mu_n$$ is convergent, hence

$\lim_{n \to \infty} \mu_n = \liminf_{n \to \infty} \mu_n \geq \liminf_{n \to \infty} f(a_n) \geq f(a).$

For the converse, we are given that for any sequence $$\{ x_n \}$$ converging to $$a$$, and any sequence $$\{ \mu_n \}$$ with $$\mu_n \geq f(x_n)$$ converging to $$\mu$$, we have $$\mu \geq f(a)$$.

1. Pick a sequence $$\{ a_n \}$$ such that $$\lim a_n = a$$.

2. Pick a convergent sequence $$\{ s_n \}$$ such that $$s_n \geq f(a_n)$$.

3. By hypothesis $$s = \lim s_n \geq f(a)$$.

4. By way of contradiction, assume that $$f$$ is not l.s.c. at $$a$$.

5. Then $$\liminf_{n \to \infty} f(a_n) < f(a)$$ due to Theorem 3.107.

6. Let $$\liminf_{n \to \infty} f(a_n) = f(a) - r$$ for some $$r > 0$$.

7. Since $$\{ s_n \}$$ is convergent, hence it is bounded (see Theorem 2.5).

8. Since $$s_n \geq f(a_n)$$, hence $$\{ f(a_n) \}$$ is bounded from above.

9. $$\{ f(a_n) \}$$ cannot be unbounded from below.

1. For contradiction, assume $$\{ f(a_n) \}$$ is unbounded below.

2. We can choose a subsequence $$\{ f(a_{k_n}) \}$$ such that $$\lim_{n \to \infty} f(a_{k_n}) = -\infty$$.

3. Let $$b_n = a_{k_n}$$ and $$t_n = f(b_n)$$.

4. Then, $$\lim b_n = a$$ and $$\lim t_n = -\infty$$ even though $$t_n \geq f(b_n)$$.

5. This contradicts the assumption that $$\lim t_n \geq f(a)$$.

6. Thus, $$\{ f(a_n)\}$$ must be bounded from below.

10. Since $$\{ f(a_n)\}$$ is bounded, hence there is a subsequence that converges to the limit inferior $$f(a) -r$$. (see Remark 2.12).

11. Let $$\{f (a_{k_n}) \}$$ be a subsequence of $$\{ f(a_n) \}$$ such that $$\lim f (a_{k_n}) = f(a) - r$$.

12. Let $$b_n = a_{k_n}$$.

13. Then, $$\{b_n \}$$ is a subsequence of $$\{ a_n \}$$.

14. Since $$\{a_n \}$$ converges to $$a$$, hence $$\{b_n \}$$ converges to $$a$$.

15. Now, choose $$t_n = f(b_n)$$. This is by definition a convergent sequence satisfying $$t_n \geq f(b_n)$$.

16. But then, $$t = \lim t_n = f(a) - r < f(a)$$.

17. This contradicts the hypothesis that $$t \geq f(a)$$.

18. Thus, $$f$$ must be l.s.c. at $$a$$.

### 3.9.6.4. Epigraphs#

Theorem 3.118 (Lower semicontinuity = closed epigraph)

Let $$f : X \to \RR$$ with $$S = \dom f$$. $$f$$ is lower semicontinuous if and only if $$\epi f$$ is closed.

Proof. Recall that $$\epi f$$ is a subset of $$X \times \RR$$ given by

$\epi f = \{(x, y) \ST f(x) \leq y \}.$

Suppose that $$\epi f$$ is closed.

1. Let $$a \in S$$ and let $$\epsilon > 0$$.

2. Let $$b = f(a) - \epsilon$$.

3. Then, $$(a, b) \notin \epi f$$.

4. Since $$\epi f$$ is closed, hence, there is an open ball $$B(a, \delta)$$ around $$a$$ and an $$r > 0$$ such that

$B(a, \delta) \times (b - r, b + r) \cap \epi f = \EmptySet.$
5. By structure of epigraph, $$(a, c) \notin \epi f$$ for any $$c \leq b$$. Thus,

$B(a, \delta) \times (-\infty, b + r) \cap \epi f = \EmptySet.$
6. Thus, $$f(x) \geq b + r$$ for all $$x \in B(a, \delta) \cap S$$.

7. Thus, $$f(x) > b = f(a) - \epsilon$$ for all $$x \in B(a, \delta) \cap S$$.

8. Thus, for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that for every $$x \in B(a, \delta) \cap S$$, $$f(x) > f(a) - \epsilon$$.

9. Thus, $$f$$ is l.s.c. at $$a$$.

10. Since $$a$$ was arbitrary, hence $$f$$ is l.s.c.

For the converse, assume that $$f$$ is l.s.c.

1. Let $$\{p_n \}$$ be a convergent sequence of $$\epi f$$.

2. Let $$p_n = (a_n, b_n)$$.

3. Let $$\lim p_n = p = (a, b)$$.

4. Then, $$\lim a_n = a$$ and $$\lim b_n = b$$.

5. Also, $$f(a_n) \leq b_n$$.

6. Since $$f$$ is l.s.c. at $$a$$, hence by Theorem 3.108

$\liminf_{n \to \infty} f(a_n) \geq f(a).$
7. But then, $$b_n \geq f(a_n)$$ implies that

$b = \lim b_n \geq \liminf_{n \to \infty} f(a_n) \geq f(a).$
8. But then, $$f(a) \leq b$$ implies that $$(a, b) \in \epi f$$.

9. Thus, every convergent sequence of $$\epi f$$ converges in $$\epi f$$.

10. Thus, by Theorem 3.33, $$\epi f$$ is closed.

### 3.9.6.5. Closed Functions#

Theorem 3.119 (Lower semicontinuity = closed function)

Let $$f : X \to \RR$$ with $$S = \dom f$$. $$f$$ is lower semicontinuous if and only if $$f$$ is closed.

Proof. We shall denote the sublevel sets for $$\alpha \in \RR$$ by

$T_{\alpha} = \{ x \in S \ST f(x) \leq \alpha \}.$

Also, define

$U_{\alpha} = \{ x \in S \ST f(x) > \alpha \}.$

Note that $$U_{\alpha} = S \setminus T_{\alpha}$$. Thus, $$T_{\alpha}$$ is closed if and only if $$U_{\alpha}$$ is open with respect to the subspace topology $$(S, d)$$.

Assume that $$f$$ is closed.

1. Let $$a \in S$$ and $$\epsilon > 0$$.

2. Let $$r = f(a) - \epsilon$$.

3. Consider the set $$U_r = \{ x \in S \ST f(x) > r \}$$.

4. Since sublevel sets of $$f$$ are closed, hence $$U_r$$ is open.

5. We note that $$f(a) = r + \epsilon > r$$.

6. Thus, $$a \in U_r$$.

7. Since $$U_r$$ is open, hence $$a$$ is an interior point of $$U_r$$.

8. Thus, there exists an open ball $$B(a, \delta) \cap S \subseteq U_r$$ around $$a$$.

9. Thus, for every $$x \in B(a, \delta) \cap S$$, $$f(x) > r = f(a) - \epsilon$$.

10. Thus, for every $$\epsilon > 0$$, there exists $$\delta > 0$$ such that $$f(x) > f(a) - \epsilon$$ for every $$x \in B(a, \delta) \cap S$$.

11. Thus, $$f$$ is l.s.c. at $$a$$.

12. Since $$a \in S$$ was arbitrary, hence $$f$$ is l.s.c.

For the converse, assume that $$f$$ is l.s.c.

1. Let $$r \in \RR$$.

2. Let $$T_r = \{ x \in S \ST f(x) \leq r \}$$ be the corresponding sublevel set.

3. Let $$\{x_n \}$$ be a convergent sequence of $$T_r$$.

4. Let $$x = \lim x_n$$ with $$x \in S$$ (subspace topology).

5. Then, $$f(x_n) \leq r$$ for every $$n$$.

6. We need to show that $$f(x) \leq r$$.

7. Since $$f$$ is l.s.c. at $$x$$, hence

$f(x) \leq \liminf_{n \to \infty} f(x_n).$
8. But $$f(x_n) \leq r$$.

9. Hence, $$\liminf_{n \to \infty} f(x_n) \leq r$$.

10. Thus, $$f(x) \leq r$$.

11. Thus, $$x \in T_r$$.

12. Since the sequence $$\{x_n \}$$ was arbitrary, hence $$T_r$$ is closed.

### 3.9.6.6. Lower Semicontinuous Hull#

Definition 3.74 (Lower semicontinuous hull of a function)

Let $$f : X \to \RR$$ with $$S = \dom f$$ be a function. There exists a greatest l.s.c. function $$g$$, majorized by $$f$$, namely the function whose epigraph is the closure of the epigraph of $$f$$. This function is known as the lower semicontinuous hull of $$f$$.

$\epi g = \closure \epi f.$

## 3.9.7. Compact Subsets#

Theorem 3.120 (Upper semicontinuity and absolute maximum on a compact set)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Assume that $$f$$ is upper semicontinuous on $$S$$. Let $$A$$ be a compact subset of $$S$$. Then $$f$$ attains a maximum on $$A$$; i.e., there exists $$a \in A$$ such that

$f(x) \leq f(a) \Forall x \in A.$

Proof. We are given that $$A$$ is a compact subset of $$S = \dom f$$. Recall from Theorem 3.76 that compact sets are closed and bounded.

We first establish that $$f(A)$$ is bounded above.

1. For contradiction, assume that $$f(A)$$ is not bounded above.

2. Thus, $$\sup_{x \in A} f(x) = \infty$$.

3. Then, for every $$n \in \Nat$$, there exists $$x_n \in A$$ such that $$f(x_n) \geq n$$.

4. Consider the sequence $$\{x_n \}$$. We have $$\lim_{n \to \infty} f(x_n) = \infty$$.

5. Since $$A$$ is compact, hence due to Theorem 3.75, $$\{x_n \}$$ has a convergent subsequence $$\{ y_n \}$$.

6. Since $$A$$ is closed, hence $$y = \lim y_n \in A$$.

7. Since $$f$$ is u.s.c. at $$y$$, hence

$\limsup_{n \to \infty} f(y_n) \leq f(y).$
8. Since $$f$$ is real valued, hence $$f(y)$$ is finite.

9. But then,

$\infty = \lim_{n \to \infty} f(x_n) \leq \limsup_{n \to \infty} f(y_n) \leq f(y).$

10. Thus, $$f(A)$$ must be bounded from above.

We now show that $$f$$ attains a maximum at some point in $$A$$.

1. Suppose that $$\sup_{x \in A} f(x) = M$$.

2. Then, for each $$n$$, there exists $$x_n \in A$$ such that

$M - \frac{1}{n} \leq f(x_n) \leq M.$
3. Thus, we obtain a sequence $$\{ x_n \}$$ such that $$\lim_{n \to \infty} f(x_n) = M$$.

4. Since $$A$$ is bounded, hence $$\{ x_n \}$$ contains a convergent subsequence $$\{ y_n \}$$.

5. Since $$A$$ is closed, hence $$y = \lim y_n \in A$$.

6. Also $$\lim f(y_n) = M$$.

7. Thus, $$f(y) = M$$.

8. Thus, $$f$$ attains a maximum value of $$M$$ at $$y \in A$$.

Theorem 3.121 (Lower semicontinuity and absolute minimum on a compact set)

Let $$f : X \to \RR$$ with $$S = \dom f$$. Assume that $$f$$ is lower semicontinuous on $$S$$. Let $$A$$ be a compact subset of $$S$$. Then $$f$$ attains a minimum on $$A$$; i.e., there exists $$a \in A$$ such that

$f(x) \geq f(a) \Forall x \in A.$