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This problem has given me some trouble. Let $F$ be the operator on $L^2[0,1]$ defined by $F(g)(t)=\sin g(t)$. I'm trying to determine whether or not $F$ is (Fréchet) differentiable in that space. I know that it is in $C[0,1]$ because I have seen this one before.

The notion if Fréchet derivative is a direct expansion of the Gâteaux derivative:

$f : U \to Y$ where $U\subset X$ and $X, Y$ normed spaces is called (Fréchet) differentiable at $u\in U$ if there exists a bounded linear operator $T$ from $X$ to $Y$ such that for $h\to 0$ we have $$ \frac{f(u+hv)-f(u)}{h} \to Tv $$ uniformly for all $v\in B_X$, e.g. in the closed unit ball in $X$.

So this is basically Gâteax differentiabilty with added uniformity of convergence. At first, I couldn't really make sense of the difference, but the example here helped me a great deal with that.

I checked for Gâteaux differentiability and after some contortions and the realization that the MVT should be applied I figured out the derivative in $g$ to be $T(f)(t)=\cos(g(t))f(t)$. However, I can't come up with an argument as to whether Fréchet differentiability holds true in this case. If the Fréchet derivative exists, it is equal to $T$. The point in question should be the null function. But how to proceed from here?

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  • $\begingroup$ what you wrote is not the definition of the Frechet derivative; the division by $h$ doesn't even make sense. It should be the existence of a bounded linear $T$ such that $\frac{\|f(u+h)-f(u)-T(h)\|}{\|h\|}\to 0$ as $h\to 0$. $\endgroup$
    – peek-a-boo
    May 24, 2021 at 21:09
  • $\begingroup$ @peek-a-boo: For them $h$ is a scalar, they're trading the limit among vector $h$, for the scalar limit being uniform in $v$ lying in the unit ball of the space. $\endgroup$
    – Jose27
    May 24, 2021 at 21:12
  • $\begingroup$ BTW, right now there's a mistake in the last step of my answer; the bound that we get should be $\|v\|_{L^2}$ instead of $|h|\| v\|_{L^2}$ (and it's not enough to get us what we need), I'll fix it in a bit. $\endgroup$
    – Jose27
    May 24, 2021 at 21:22
  • $\begingroup$ @Jose27 oh my bad, I missed the uniformity in the limit $\endgroup$
    – peek-a-boo
    May 24, 2021 at 21:25

1 Answer 1

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Well, turns out my mistake was not so simple after all: The map is NOT Frechet differentiable. Consider $g=0$, and $v_p(t)=\sqrt{1-p}t^{-p/2}=a_pt^{-p/2}$ so that $\| v_p\|_{L^2[0,1]}=1$ for all $0<p<1$ and the quotient we're looking at is $$ \dfrac{\| \sin(hv_p)-hv_p\|_{L^2[0,1]}}{|h|} = \dfrac{1}{|h|} \left(\int_0^1 (\sin ha_pt^{-p/2})-(ha_pt^{-p/2}))^2\, dt \right)^{1/2}. $$ Now let $b_p= \left(\frac{a_p h}{100}\right)^{2/p} <1$, and notice that for $0<t<b_p$ we have $$ |\sin ha_pt^{-p/2}-(ha_pt^{-p/2})| \geq ha_p t^{-p/2}-1 \geq \dfrac{ha_p}{2t^{p/2}}, $$ and so $$ \dfrac{1}{|h|} \left(\int_0^1 (\sin ha_pt^{-p/2})-(ha_pt^{-p/2}))^2\, dt \right)^{1/2} \geq \dfrac{1}{|h|} \left( \int_0^{b_p} \left(\dfrac{ha_p}{2t^{p/2}}\right)^2\, dt \right)^{1/2}= \dfrac{a_p}{2} \left( \int_0^{b_p} \dfrac{dt}{t^{p}}\, dt \right)^{1/2}= \dfrac{a_p}{2\sqrt{1-p}} b_p^{(1-p)/2}= \dfrac{1}{2}\left( \dfrac{a_ph}{100}\right)^{(1-p)/p}. $$

To finish the argument, choose $h=1/n$ and $p_n= \frac{n}{n+1}$, to see that the difference quotient tends to $1/2$.

I hope I didn't miss anything, be sure to double check!

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  • $\begingroup$ But then wouldn't this imply that it is differentiable? $\endgroup$
    – Hölderlin
    May 24, 2021 at 22:17
  • $\begingroup$ I've edited, you're right: it's not! $\endgroup$
    – Jose27
    May 24, 2021 at 22:43
  • $\begingroup$ Thanks, I'm currently trying to cook up an example that's a bit simpler, but I have a much better feeling for the problem now. $\endgroup$
    – Hölderlin
    May 25, 2021 at 11:13
  • $\begingroup$ I think I've got a good solution now, but I would be grateful for some control over what I did since I'm not quite sure. I will edit the question shortly. $\endgroup$
    – Hölderlin
    May 25, 2021 at 12:07

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