TITLE: Frechet differential in $L^\infty$ spaces
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: $L$ is differentiable at $f=0$ because, taking $A=0$ we have 
$\lim_{\space \|h\| \rightarrow 0} \frac{\|\cos(h)-1-A(h)\|}{\|h\|}=\lim_{\space \|h\| \rightarrow 0} \frac{\|\cos(h)-1\|}{\|h\|}=0.$