Let $X$ be the vector space of continuous functions on the interval $[0,\pi]$ equipped with the uniform norm. Let $F: X\rightarrow X$ be a function given by
$$
[F(f)](x) = \sin(f(x)),\quad (f\in X).
$$
And I am trying to compute the Fréchet derivative $F'$. Here is my attempt:
Take some $f,h\in X$ and try to evaluate the difference $F(f+h) – F(f)$:
\begin{align*}
[F(f+h)-F(f)](x) &= \sin(f(x)+h(x)) – \sin(f(x))\\
&= \sin(f(x))\cos(h(x)) + \cos(f(x))\sin(h(x)) – \sin(f(x))\\
&= [\cos(h(x))-1]\sin(f(x)) – \cos(f(x))\sin(h(x))
\end{align*}
So when $h\rightarrow 0$, the terms $[\cos(h(x))-1]\sin(f(x))$ and $\cos(f(x))\sin(h(x))$ both tend to zero, so I am quite confused about how to choose a valid candidate for the derivative.
Could anyone please give me a hint about how to proceed? Any of your help will be highly appreciated! 🙂
Best Answer
Hint: Define $L_f: X\to X$ by $L_fh(x)=-h(x)\cos (f(x))$. Then $T$ is a continuous linear map on $X$. Show that $\frac {\|F(f+h)-F(f)-L_fh\|} {\|h\|} \to 0$ as $\|h\|\to 0$ using the following basic facts: $\frac {\cos a -1}a \to 0$ as $ a \to 0$ and $\frac {\sin a } a -1 \to 0$ as $ a\to 0$. Hence $F'(f)=L_f$.