Frechet differential in $L^\infty$ spaces

Question

define $L :L^\infty([0,1]) \to L^\infty([0,1])$, $f \to \cos f$.

Show that this operator is not Frechet differentiable at $f = 0$.

My idea was just to use the taylor expansion: $$\cos (f+h) = \cos f + h \sin f + \mathcal{o}(\|h\|) $$ to conclude that the derivative is given by $L'(f)(h)=h \sin f$. Is this correct? Thanks for any hints.

Answer 1

I might be missing something, but since

$$\cos h(t) = 1 - h(t)^2/2 + O(\|h\|_\infty^4)$$

and $1 = \cos 0,$ it appears to me that $DL(0)=0.$

Answer 2

$L$ is differentiable at $f=0$ because, taking $A=0$ we have

$\lim_{\space \|h\| \rightarrow 0} \frac{\|\cos(h)-1-A(h)\|}{\|h\|}=\lim_{\space \|h\| \rightarrow 0} \frac{\|\cos(h)-1\|}{\|h\|}=0.$