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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.

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  • $\begingroup$ I posted the other cases too, because this was not so in my way edited. $\endgroup$
    – user160069
    Nov 8, 2017 at 10:39

2 Answers 2

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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 $DL(0)=0.$

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  • $\begingroup$ what would be the frechet derivative for all $t$? $\endgroup$
    – user160069
    Nov 8, 2017 at 9:43
  • $\begingroup$ @user160069 It looks to me like $DL(f)(h) = (-\sin f)\cdot h.$ $\endgroup$
    – zhw.
    Nov 8, 2017 at 20:30
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$L$ is differentiable at $f=0$ because, taking $A=0$ we have

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

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  • $\begingroup$ Not sure why you posted this after I posted my answer. $\endgroup$
    – zhw.
    Nov 7, 2017 at 23:26
  • $\begingroup$ I didn't see your answer. Sorry. Would you like me to delete my answer? $\endgroup$
    – Matematleta
    Nov 7, 2017 at 23:28
  • $\begingroup$ Ah, okay, no there's no problem. It happens, no worries. $\endgroup$
    – zhw.
    Nov 7, 2017 at 23:33
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    $\begingroup$ 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. $\endgroup$
    – Matematleta
    Nov 7, 2017 at 23:40

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