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

kingwinner
Messages
1,266
Reaction score
0

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.[/color]

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:
Physics news on Phys.org
Can someone help me, please?

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.[/color]

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
 
Back
Top