# Convexity & Strict Convexity of Functionals (function of a function)

• kingwinner
In summary: Examine u(x) = r*u2(x) + (1-r)*u2(x), with 0 <= r <= 1.Since r is between 0 and 1, u(x) is a convex function.
kingwinner

## Homework Statement

Let C be the class of C1 functions on interval [0,1] satisfying u(0)=0=u(1).
Consider the functional F(u)=
1
∫[(u')2 + 3u4 + cosh(u) + (x3-x)u] dx
0
(note: u is a function of x.)
Analyse the functional F term by term. Decide for each term whether it is convex or strictly convex on C.

## Homework Equations

(Strict) Convexity of functionals.

## The Attempt at a Solution

Definition: A functional F is "convex" if for all u,v in C, 0<a<1, we have
F((1-a)u+av) ≤ (1-a)F(u) + aF(v).
F is "strictly convex" if for all u,v in C such that u≠v, and for all 0<a<1, we have
F((1-a)u+av) < (1-a)F(u) + aF(v).

1) I think by linearity of integrals, we can show that the last term is convex, but NOT strictly convex. Am I correct?

2) Each of the first two terms is strictly convex. Am I correct?
[I believe I have a proof for 1) and 2) using strict convexities of f(x)= x2 and g(x)= x4, but there is no answer at the back of the textbook for this problem, and so I'm not sure if I'm correct so far. It would be much appreciated if someone can confirm my answer, or point out if I'm wrong.]

3) I'm really stuck for the third term, G(u)=
1
∫cosh(u)dx
0
How can I prove that this is convex/strictly convex? I really don't have much idea on this part of the problem...
Hopefully someone can explain how to prove this. Thanks a million!

Last edited:

The definitions of convex/strictly convex of functionals (function of a function) are as follows:

Let C be the class of C1 functions on interval [0,1] satisfying u(0)=0=u(1).
A functional F is convex if for all u,v in C, 0<a<1, we have
F((1-a)u+av) ≤ (1-a)F(u) + aF(v).
F is strictly convex if for all u,v in C such that u≠v, and for all 0<a<1, we have
F((1-a)u+av) < (1-a)F(u) + aF(v).

Any help would be much appreciated!

kingwinner said:

## Homework Statement

Let C be the class of C1 functions on interval [0,1] satisfying u(0)=0=u(1).
Consider the functional F(u)=
1
∫[(u')2 + 3u4 + cosh(u) + (x3-x)u] dx
0
(note: u is a function of x.)
Analyse the functional F term by term. Decide for each term whether it is convex or strictly convex on C.

## Homework Equations

(Strict) Convexity of functionals.

## The Attempt at a Solution

Definition: A functional F is "convex" if for all u,v in C, 0<a<1, we have
F((1-a)u+av) ≤ (1-a)F(u) + aF(v).
F is "strictly convex" if for all u,v in C such that u≠v, and for all 0<a<1, we have
F((1-a)u+av) < (1-a)F(u) + aF(v).

1) I think by linearity of integrals, we can show that the last term is convex, but NOT strictly convex. Am I correct?

2) Each of the first two terms is strictly convex. Am I correct?
[I believe I have a proof for 1) and 2) using strict convexities of f(x)= x2 and g(x)= x4, but there is no answer at the back of the textbook for this problem, and so I'm not sure if I'm correct so far. It would be much appreciated if someone can confirm my answer, or point out if I'm wrong.]

3) I'm really stuck for the third term, G(u)=
1
∫cosh(u)dx
0
How can I prove that this is convex/strictly convex? I really don't have much idea on this part of the problem...
Hopefully someone can explain how to prove this. Thanks a million!

Is cosh(u) a convex function of u?

RGV

Ray Vickson said:
Is cosh(u) a convex function of u?

RGV
I believe cosh(u) is actually a strictly convex function of u, and so I claim that
1
∫cosh(u(x))dx := G(u) is also strictly convex, is this a correct implication?
0

Thanks!

Examine u(x) = r*u2(x) + (1-r)*u2(x), with 0 <= r <= 1.

RGV

## 1. What is convexity of functionals?

Convexity of functionals refers to the property of a functional (a function that takes in a function as input) where the graph of the functional lies above the chord connecting any two points on the graph. In other words, the functional preserves the idea of "straightness" and does not curve downwards in any section.

## 2. How is convexity different from strict convexity?

Strict convexity is a stronger version of convexity, where the graph of the functional lies strictly above the chord connecting any two points on the graph. This means that the functional cannot be flat or level at any point, and must always be increasing.

## 3. Why is convexity important in functional analysis?

Convexity is important in functional analysis because it allows for the use of powerful mathematical tools and techniques to analyze and optimize functions. It also has applications in fields such as economics, optimization, and machine learning.

## 4. Can a functional be both convex and concave?

No, a functional cannot be both convex and concave. This is because convexity and concavity are opposite properties - a function is either convex or concave, but not both. A function can only be strictly convex or strictly concave, but never both.

## 5. How can we determine if a functional is convex or not?

To determine if a functional is convex, we can use the property of convexity where the graph of the functional lies above the chord connecting any two points. This can be checked by taking the second derivative of the functional and ensuring that it is always positive. If the second derivative is positive, then the functional is convex. Otherwise, it is not convex.

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