# Showing that a given function is Frechet differentiable

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 '18 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 '18 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 '18 at 6:09