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.

  • $\begingroup$ I posted the other cases too, because this was not so in my way edited. $\endgroup$ – 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 $DL(0)=0.$

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

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

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

  • $\begingroup$ Not sure why you posted this after I posted my answer. $\endgroup$ – zhw. Nov 7 '17 at 23:26
  • $\begingroup$ I didn't see your answer. Sorry. Would you like me to delete my answer? $\endgroup$ – Matematleta Nov 7 '17 at 23:28
  • $\begingroup$ Ah, okay, no there's no problem. It happens, no worries. $\endgroup$ – zhw. Nov 7 '17 at 23:33
  • 1
    $\begingroup$ 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 '17 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.