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Check if the function $$ F\colon L^2[0,1]\to L^2[0,1], (F(x))(t)=\sin x(t) $$ is Gâteaux- resp. Fréchet-differentiable at $x=0$.

I started checking if the function is Gâteauch-differentiable at $x=0$ with

$$\lim\limits_{s\to 0}\frac{F(x+sh)(t)-F(x)(t)}{s}=\lim\limits_{s\to 0}\frac{\sin (sh)(t)}{s}$$

But now I do not know how to continue the calculation...

Could anabody pls help me?

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up vote 1 down vote accepted

By using the mean value theorem, you have that $$\frac{\sin(x+sh)(t)-\sin(x)(t)}{s}=\frac{\cos(u(t))sh(t)}{s}$$

where $u(t)\in[x(t),(x+sh)(t)]$ or $u(t)\in[(x+sh)(t),x(t)]$. Now if you let $s\rightarrow 0$ you have from the last equality that $$\lim_{s\rightarrow 0}\frac{\sin(x+sh)(t)-\sin(x)(t)}{s}=\cos(x(t))h(t)$$

Hence, if $D_GF(x)$ denotes you Gâteux derivative, you have $$D_GF(x)h=\cos(x)h$$

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Can you please say what is wrong with my solution – Tomás Jan 30 '13 at 11:28
In general this is an standard argument. Now try to veirfy if it is Frechet differentialbe and if you have trouble leave a comment here. – Tomás Jan 30 '13 at 12:08
There is a more tractable way to do this. Take a look in Theorem 1.9 page 14:… – Tomás Jan 30 '13 at 12:49
I try to solve the problem by myself. Assumption: $x_n\to 0$ a.e. $(\lvert\cos x_n h-h\rvert^2)_{n\in\mathbb{N}}$ is a sequence in $L^2[0,1]$, which converges a.e. against $\lvert\cos (0)h-h\rvert^2=0$, i.e. the nullfunction. Moreover, $\lvert\cos x_n h-h\rvert^2\leq (\lvert\cos x_n h\rvert+\lvert h\rvert)^2$. So one can apply Lebesgue's theorem, which says $\lim\limits_{n\to\infty}\int\limits_0^1\lvert\cos x_n(t)h(t)-h(t)\rvert^2\, dt=\int\limits_0^1\lim\limits_{n\to\infty}\lvert\cos x_n(t)h(t)-h(t)\rvert^2\, dt=0$ forall $h\in L^2[0,1]$. And so the supremum above converges against 0. Hope ok – math12 Jan 30 '13 at 16:01
That's correct, you did it. – Tomás Jan 30 '13 at 16:06

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