Convex Functions
Contents
9.8. Convex Functions#
Throughout this section, we assume that
We suggest the readers to review the notions of graph, epigraph, sublevel sets of real valued functions in Real Valued Functions. Also pay attention to the notion of extended real valued functions, their effective domains, graphs and level sets.
9.8.1. Convexity of a Function#
(Convex function)
Let
An extended valued function
For a convex function, every chord lies above the graph of the function.
9.8.1.1. Strictly Convex Functions#
(Strictly convex function)
Let
In other words, the inequality is a strict inequality
whenever the point
9.8.1.2. Concave Functions#
(Concave function)
We say that a function
(Linear functional)
A linear functional on
i.e. the inner product of
All linear functionals on a real vector space are convex as well as concave.
Proof. For any
Thus,
9.8.1.3. Arithmetic Mean#
(Arithmetic mean is convex and concave)
Let
Arithmetic mean is a linear function. In fact
for all
Thus, for
Thus, arithmetic mean is both convex and concave.
9.8.1.4. Affine Function#
(Affine functional)
An affine functional is a special type of
affine function
which maps a vector from
On real vector spaces, an affine functional on
i.e. the inner product of
All affine functionals on a real vector space are convex as well as concave.
Proof. A simple way to show this is to recall that for any
affine function
holds true for any
In the particular case of real affine functionals,
for any
This establishes that
Another way to prove this is by following the definition
of
9.8.1.5. Absolute Value#
(Absolute value is convex)
Let
with
Recall that
In particular, for any
Thus
Hence,
9.8.1.6. Norms#
9.8.1.7. Max Function#
(Max function is convex)
Let
with
Let
Let
and .Then, for every
, we have: and .Thus,
.Taking maximum on the left hand side, we obtain:
But
.Thus,
satisfies the convexity defining inequality (9.1):Thus,
is convex.
9.8.1.8. Geometric Mean#
(Geometric mean is concave)
Let
with
Recall from AM-GM inequality:
Let
and .Let
.Note that
too, hence every component of is positive.With
, and applying AM-GM inequality:With
, and applying AM-GM inequality:Multiplying the first inequality by
and the second by and adding them, we get:Recall that
. Thus, the inequality above simplifies to:Thus,
is concave.
9.8.1.9. Powers#
(Powers of absolute value)
Let
with
For
Consider the case where
Let
By triangle inequality:
We can write this as:
By Hölder's inequality, for some
Let
, , and .Applying Hölder’s inequality,
Taking power of
on both sides, we get:Which is the same as
Thus,
is convex.
9.8.1.10. Empty Function#
(Empty function is convex)
Let
Proof. The empty set
This observation may sound uninteresting however
it is important in the algebra of convex functions.
Definition 2.48 provides
definitions for sum of two partial functions
and scalar multiplication with a partial
function. If two convex functions
Jensen's inequality, discussed later in the section, generalizes the notion of convexity of a function to arbitrary convex combinations of points in its domain.
9.8.2. Convexity on Lines in Domain#
Let
In other words,
Proof. For any arbitrary
we have defined
with the domain:
Assume
Let
.It means that there are
and ,and
and .Let
and .Note that:
Since
is convex, hence .Thus,
is defined at for all .We have shown that if
, then for all .Thus,
is convex.Now,
We showed that for any
and ,Thus,
is convex.
For the converse, we need to show that
if for any
Assume for contradiction that
We show that both of these conditions lead to contradictions.
Assume
is not convex.Then, there exists
such that for some :Let
.Note that
.Picking
for this particular choice of and , we note that is defined at and since they corresponding to the points and respectively in .Since
is convex by hypothesis, hence, is defined at every .This contradicts the fact that
is not defined at some .Thus,
must be convex.
We now accept that
Again, let
and note that .Pick
for this choice of and .We have
.In particular,
and .Since
is convex, hence for every ,We have a contradiction.
Thus,
must be convex.
A good application of this result is in showing the concavity of the log determinant function in Example 9.49 below.
9.8.3. Epigraph#
The epigraph
of a function
The definition of epigraph also applies for extended
real valued functions
9.8.3.1. Convex Functions#
(Function convexity = Epigraph convexity)
Let
This statement is also valid for extended real valued functions.
Proof. Let
Assume
Let
.Then,
and .Let
.Consider the point
.We have
.And
.Then,
.Since
is convex, hence .But then,
.Thus,
.Thus,
implies for all .Thus,
is convex.
Assume
Let
.Let
and .Then,
.Let
.Let
.Since
is convex, hence .i.e.,
.Thus,
.Thus,
is convex.And,
.But
.Thus,
.Thus,
is convex as and were arbitrary.
Note that
(Convex function from convex set by minimization)
Let
Then
Proof. We show the convexity of
Let
and be two points in .Then
and .By the definition of
(infimum rule), for every , there exists such that .Similarly, for every
, there exists such that .Consider the sequences
and .By the convexity of
, for every and everyHence for every
Taking the limit
, we haveHence
for every .Hence
is convex.Hence
is convex.
9.8.3.2. Nonnegative Homogeneous Functions#
Recall that a real valued function
(Nonnegative homogeneity = Epigraph is cone)
A function
Proof. Let
Let
.Let
.Then,
.But
since is nonnegative homogeneous.Thus,
is closed under nonnegative scalar multiplication.Thus,
is a cone.
Assume
Let
.Then,
.Since
as is a cone, hence .Now, let
.Since
is a cone then, .Then,
.By definition of epigraph,
.We claim that
must hold true.Assume for contradiction that
where .Then,
.Since
is a cone, hence dividing by , we get, .But then,
which is a contradiction.Hence,
must hold true.
9.8.3.3. Nonnegative Homogeneous Convex Functions#
(Nonnegative homogeneous convex function epigraph characterization)
Let
Proof. By Theorem 9.73,
By Theorem 9.71,
Thus,
(Nonnegative homogeneous convex function is subadditive)
Let
Proof. Assume that
By Theorem 9.74,
is a convex cone.Then, by Theorem 9.47
is closed under addition and nonnegative scalar multiplication.Pick any
.Then
.Then, their sum
.This means that
by definition of epigraph.
Now for the converse, assume that
By Theorem 9.73,
is a cone.Consequently, it is closed under nonnegative scalar multiplication.
Pick any
.Let
.Then,
and .Now,
.Since
is subadditive, hence .Thus,
.We have shown that for any
, .Thus,
is closed under vector addition.Since
is closed under vector addition and nonnegative scalar multiplication, hence is a convex cone.But then, by Theorem 9.74,
is convex.
Let
Proof. Since
Thus,
But,
Thus,
Thus,
(Linearity of nonnegative homogeneous functions)
Let
If
Proof. We are given that
Assume that
Now, for the converse, assume that
Let
.Then,
since is subadditive.Also,
But
Thus,
Combining, we get
. Thus, is additive over .For any
,Thus, for any
, .Thus,
is homogeneous over .Since
is additive and homogeneous over , hence is linear.
Finally, assume that
By previous argument
is homogeneous on the basis vectors; i.e., for any and any basis vector.Let
.Then,
.Since
is subadditive, henceThis can hold only if all the inequalities are equalities in the previous derivation.
Thus,
or for every .Then, following the previous argument,
is linear over .
9.8.4. Extended Value Extensions#
Tracking domains of convex functions is difficult.
It is often convenient to extend a convex function
(Extended value extension)
The extended value extension
of a convex function
We mention that the defining inequality (9.1)
for a convex function
At
and , this is always an equality.Since
is convex, hence if both , then too.The inequality then reduces to (9.1).
If either
or is not in , then the R.H.S. becomes and the inequality stays valid.
(Effective domain)
The effective domain
of an extended valued function
An equivalent way to define the effective domain is:
(Effective domain and sum of functions)
If
Proof. We proceed as follows:
At any
, both and are finite.Thus,
is finite.Hence,
.At any
, either or .Thus,
.Thus,
.
Several commonly used convex functions have vastly different domains. If we work with these functions directly, then we have to constantly worry about identifying the domain and manipulating the domain as per the requirements of the operations we are considering. This quickly becomes tedious.
An alternative approach is to work with the
extended value extensions of all functions.
We don’t have to worry about tracking the function
domain. The domain can be identified whenever needed
by removing the parts from
In this book, unless otherwise specified, we shall assume that the functions are being treated as their extended value extensions.
9.8.5. Proper Functions#
(Proper function)
An extended real-valued function
In other words,
Putting another way, a proper function
is obtained by taking a real valued function
It is easy to see that the codomain for a proper
function can be changed from
(Improper function)
An extended real-valued function
For an improper function
may be empty. might take a value of at some .
Most of our study is focused on proper functions. However, improper functions sometimes do arise naturally in convex analysis.
(An improper function)
Consider a function
Then,
9.8.6. Indicator Functions#
(Indicator function)
Let
The extended value extension of an indicator function is given by:
(Indicator functions and convexity)
An indicator function is convex if and only if its domain is a convex set.
Proof. Let
since
Thus,
If
(Restricting the domain of a function)
Let
Also,
The statement is obvious. And quite powerful.
The problem of minimizing a function
over a set is same as minimizing over .
9.8.7. Sublevel Sets#
Recall from Definition 2.53
that the
The strict sublevel sets for a real valued function
The sublevel sets can be shown to be intersection of a set of strict sublevel sets.
(Sublevel set as intersection)
Let
denote the strict sublevel set of
denote the sublevel set of
Proof. We show that
Let
.Then,
.Thus,
for every .Thus,
for every .Thus,
.
We now show that
Let
.Then,
for every .Taking the infimum on the R.H.S. over the set
, we obtainThus,
.Thus,
.
(Convexity of sublevel sets)
If
are convex.
Proof. Assume
Let
.Then,
and .Let
.Let
.Since
is convex, hence:Thus,
.Thus,
.Thus,
is convex.
The converse is not true. A function need not be convex even if all its sublevel sets are convex.
Consider the function
Its sublevel sets are convex as they are intervals.
(Convexity of strict sublevel sets)
If
are convex.
Proof. Assume
Let
.Then,
and .Let
.Let
.Since
is convex, hence:Thus,
.Thus,
.Thus,
is convex.
An alternate proof for showing the convexity of the sublevel sets is to show it as an intersection of strict sublevel sets.
(Intersection of sublevel sets of convex functions)
Let
is a convex set.
Proof. For each
Then,
Thus,
(Convexity of sublevel sets of the quadratic)
Let
Consider the sets of the form
This is a sublevel set of the quadratic function
Since
Sets of this form include the solid ellipsoids, paraboloids as well as spherical balls. Here is an example of the spherical ball of the norm induced by the inner product.
9.8.8. Hypograph#
The hypograph
of a function
Just like function convexity is connected to epigraph convexity, similarly function concavity is connected to hypograph convexity.
(Function concavity = Hypograph convexity)
A function
Proof.
9.8.9. Super-level Sets#
Recall from Definition 2.56
that the
If
Proof. Assume
Let
.Then,
and .Let
.Let
.Since
is concave, hence:Thus,
.Thus,
.Thus,
is convex.
The converse is not true. A function need not be concave even if all its super-level sets are convex.
Let geometric mean be given by:
with
Let arithmetic mean be given by:
Consider the set:
In Example 9.30 we establish that
is concave.In Example 9.26 we established that
is convex as well as concave.Thus,
is concave.Thus,
is concave (with ).Note that
can be redefined as:Thus,
is a -super-level set of .Since
is concave, hence is convex.
9.8.10. Closed Convex Functions#
Recall from Real Valued Functions that a function is closed if all its sublevel sets are closed. A function is closed if and only if its epigraph is closed. A function is closed if and only if it is lower semicontinuous.
In general, if a function is continuous, then it is lower semicontinuous and hence it is closed.
In this subsection, we provide examples of convex functions which are closed.
9.8.10.1. Affine Functions#
(Affine functions are closed)
Let
where
Proof. We prove closedness by showing that the epigraph of
Let
be a converging sequence of .Let
.We have
for every .In other words
Taking the limit on both sides, we get
Hence
.Hence
.Hence
is closed.
9.8.10.2. Norms#
(All Norms are closed)
Let
Proof. The sublevel sets are given by
9.8.11. Support Functions#
(Support function for a set)
Let
Since
(Finite sets)
Let
This follows directly from the definition.
9.8.11.1. Convexity#
(Convexity of support function)
Let
Proof. Fix a
is a pointwise supremum of convex functions.
By Theorem 9.114,
We note that the convexity of the support function
9.8.11.2. Closedness#
(Closedness of support function)
Let
Proof. Recall that a function is closed if all its sublevel sets are closed.
Let
.Consider the sublevel set
.Then,
Thus,
Define
asThen,
Now,
is a closed set since is a continuous function.Thus,
is an intersection of closed sets.Thus,
is closed.Thus, all sublevel sets of
are closed.Thus,
is closed.
9.8.11.3. Equality of Underlying Sets#
(Equality of underlying sets for support functions)
Let
Proof. If
For contradiction, assume that
.Without loss of generality, assume that there exists
such that .Since
and is a closed convex set, hence, by Theorem 9.169, there exists a hyperplane strongly separating from .Thus, there exists
and such thatTaking supremum over
on the L.H.S., we obtainThus, there exists
such thatThis contradicts our hypothesis that
and are identical.Thus,
must hold.
9.8.11.4. Closure and Convex Hull#
The next result shows that support function
for a set and its closure or its convex hull
are identical. This is why, we required
(Support functions and closure or convex hull of underlying set)
Let
. .
Proof. We first consider the case of closure.
.Thus,
.Let us now show the reverse inequality.
Let
.Then, there exists a sequence
of such thatNow for every
, there exists a point such that .Thus,
.Since
, henceTaking the limit
on the R.H.S., we obtainThus,
must hold true.Since this is true for every
, hence .
Now, consider the case of convex hull.
By definition,
.Thus,
.Let
.Then, there exists a sequence
of such thatSince
, hence, there exists and such thatBy linearity of the inner product
Taking the limit
on the L.H.S., we obtainThus,
must hold true.Since this is true for every
, hence .
9.8.11.5. Arithmetic Properties#
Following properties of support functions are useful in several applications.
(Arithmetic properties of support functions)
(Nonnegative homogeneity) For any nonempty set
and a vector and ,(Subadditivity) For any nonempty set
and a vector ,(Nonnegative scaling of the underlying set) For any nonempty set
and a vector and(Additivity over Minkowski sum of sets) For any two nonempty subsets
and
Proof. (1) Nonnegative homogeneity
Here, we used the fact that
(2) Subadditivity
(3) Nonnegative scaling of the underlying set
(4) Minkowski sum
9.8.11.6. Cones#
Recall from Definition 9.22 that a set
(Support function of a cone)
Let
In words, the support function of a cone
Proof. We proceed as follows
Assume that
.Then,
.In particular,
since is a cone.Accordingly,
.Thus,
Now consider
.Then, there exists
such that .Since
is a cone, hence for all .Accordingly,
Taking the limit
, we see thatThus,
for all and otherwise.Thus,
.
(Support function of nonnegative orthant)
Let
which is the nonpositive orthant
9.8.11.7. Affine Sets#
(Support function for an affine set)
Let
Assume that
Proof. We proceed as follows.
By definition of support function
Introduce a variable
.Then
.Accordingly
where
.We note that the statement
is equivalent to and .In other words,
where .The set
is a convex polyhedral cone.By Theorem 9.93, the support function of a cone is the indicator function of its polar cone.
By Theorem 9.66, the polar cone is given by
It is easy to see that
.Every vector
can be split into two vectors such that .Accordingly
.
This gives us
9.8.11.8. Norm Balls#
(Support functions for unit balls)
Let
Then, the support function is given by
where
Proof. This flows directly from the definitions of support function and dual norm.
9.8.12. Gauge Functions#
(Gauge function for a set)
Let
If
The gauge function is also known as Minkowski functional.
(Nonnegativity)
The Gauge function is always nonnegative.
(Value at origin)
(Subadditive)
If
Proof. We proceed as follows.
If
or , then the inequality is satisfied trivially. So assume that both are finite.Then, the sets
and are not empty.Thus, we can choose some
from and from .If
or , then . Consequently,and the inequality is satisfied.
Now, consider the case where
and .Then,
and .Now,
Thus,
is a convex combination of and .Since
is convex, hence .Thus,
.Thus,
.Thus, for every
and every , .Taking infimum on the R.H.S. over
and , we obtain, .
(Homogeneous)
The Gauge function is homogeneous.
(Seminorm)
The Gauge function is a seminorm.
(Norm as a gauge function)
Let
Let
Then,
The gauge function for the closed unit ball is simply the norm itself.
9.8.13. Jensen’s Inequality#
Jensen’s inequality stated below is another formulation for convex functions.
(Jensen’s inequality)
A proper function
holds true for every
Proof. The Jensen’s inequality reduces to (9.1)
for
Assume
Let
.If any of
for some , then and the Jensen’s inequality holds trivially.Thus, we shall assume that
.Since
is convex, hence their convex combination .Inductively, assume that the Jensen’s inequality holds for
; i.e.,holds true whenever
and .WLOG, assume that
. Thus, .Define
where .Note that
. Also, since .We can now write:
Thus,
satisfies Jensen’s inequality.
For the converse, assume that
Thus,
Jensen’s inequality is essential in proving a number of famous inequalities.
(Logarithm and Jensen’s inequality)
In Example 9.45, we show that
Now, let
Multiplying by
For a particular choice of
which is the AM-GM inequality suggesting that arithmetic mean is greater than or equal to the geometric mean for a group of positive real numbers.
(Jensen’s inequality for nonnegative homogeneous convex functions)
If
holds true for every
Proof. Let
By Theorem 9.75,
The nonnegative homogeneity gives us
We are done.
9.8.14. Quasi-Convex Functions#
(Quasi convex function)
Let
If the sublevel sets