# Is the sine operator on $L^2[0,1]$ Fréchet differentiable or not and why?

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?

• 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 $h\to 0$. May 24, 2021 at 21:09
• @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. May 24, 2021 at 21:12
• 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. May 24, 2021 at 21:22
• @Jose27 oh my bad, I missed the uniformity in the limit May 24, 2021 at 21:25

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 $$0 and the quotient we're looking at is $$\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 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 $$1/2$$.