I'd like to do the following exercise from my book. The statement is as follows.

Let $\gamma \epsilon { C }^{ 1 }(R\times { R }^{ n })$,$c>0$ and ${ \gamma }_{ 0 }\epsilon { L }^{ 1 }({ R }^{ n })$ such that $\left| { \gamma }(z,x) \right| \le { \gamma }_{ 0 }(x)+c\left| z \right| $.

And let $G(f);=\int _{ { R }^{ n } }^{ }{ { \gamma }(f(x),x)dx } $

Show that $G({ L }^{ 1 }({ R }^{ n }))\rightarrow R$ is Frechet differentiable and calculate the derivative.

It's the first exercise for Frechet differentiability after it was introduced so im not very fermiliar with it.

We need to show that $\lim _{ h\rightarrow 0 }{ \frac { \left| G(u+h)-G(u)-G'(u)h \right| }{ { \left\| h \right\| }_{ { L }^{ 1 } } } } =0$

I started looking at the numerator so we have

$\left| G(u+h)-G(u)-G'(u)h \right| =\int _{ { R }^{ n } }^{ }{ \gamma (u(x)+h(x),x)-\gamma (u(x),x)+\gamma (u(x),x)h(x)dx } $

=$\iint _{ u(x)+h(x) }^{ u(x) }{ \gamma (s,x)ds-\gamma (u(x),x)h(x)dx} $ =$\iint _{ u(x)+h(x) }^{ u(x) }{ \gamma (s,x)-\gamma (u(x),x)dsdx } $

but how can i now make this expression smaller than an arbitrary $\varepsilon >0$? is this calculation even usefull? And how can i use the Assumtion for $\gamma $? Any Help is greatly appreciated Thank you

  • Your calculations are completely irrelevant to this exercise. I suggest you make sure you understand the definition of the Frechet derivative, and after that look for an example for the computation of a Frechet derivative of some operator (should appear in any textbook dealing with that subject). – MOMO Dec 6 at 23:32
  • Moreover, I recommend you try and understand why the equality between integrals you wrote are incorrect, although it is not needed to the exercise. – MOMO Dec 6 at 23:37
  • I tried the exercise for well over an hour, I think I know what the derivative is but I couldn't prove it. There is an issue with regards of the vector spaces. I think you need to assume $|\partial_z \gamma|$ is bounded and then the exercise would follow in a rather straight-forward way using derivative rules (chain rule, mostly). – Will M. Dec 8 at 6:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.