# Frechet differential in $L^\infty$ spaces

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.

• I posted the other cases too, because this was not so in my way edited. – user160069 Nov 8 '17 at 10:39

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 $DL(0)=0.$

• what would be the frechet derivative for all $t$? – user160069 Nov 8 '17 at 9:43
• @user160069 It looks to me like $DL(f)(h) = (-\sin f)\cdot h.$ – zhw. Nov 8 '17 at 20:30

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

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

• Not sure why you posted this after I posted my answer. – zhw. Nov 7 '17 at 23:26
• I didn't see your answer. Sorry. Would you like me to delete my answer? – Matematleta Nov 7 '17 at 23:28
• Ah, okay, no there's no problem. It happens, no worries. – zhw. Nov 7 '17 at 23:33
• I type latex very slowly. I really should learn to use Mathjax. In the future, I will try to read new posts before posting mine! At least I can upvote you for beating me to the punch on the answer. – Matematleta Nov 7 '17 at 23:40