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

  • $\begingroup$ 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). $\endgroup$ – MOMO Dec 6 '18 at 23:32
  • $\begingroup$ Moreover, I recommend you try and understand why the equality between integrals you wrote are incorrect, although it is not needed to the exercise. $\endgroup$ – MOMO Dec 6 '18 at 23:37
  • $\begingroup$ 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). $\endgroup$ – Will M. Dec 8 '18 at 6:09

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