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Let $f : L^2([0, \pi])\to L^2([0, \pi]); u\mapsto \sin(u)$. We can see that $f$ is everywhere Gâteaux differentiable, as $$\lim_{\tau\to 0} \frac{f(u+\tau v)-f(u)}{\tau} = v\cos(u)$$ But it's not hard to construct a counterexample to the Fréchet differentiability of $f$. Consider that for $u = 0$, the Gâteaux derivative in the direction of $v$ is $v$. Therefore, if $f$ is Fréchet differentiable at $u = 0$, its Fréchet derivative is $A = \mathrm{Id}$. Then, let $v = 1_{[0, s^2]}$ for $0\leq s\leq \sqrt{\pi}$. Notice that $$\|f(u+v)-f(u)-Av\|_{L^2([0, \pi])} = \sqrt{\int_0^{\pi} |\sin(v(t))-v(t)|^2\,\mathrm{d}t} = (1-\sin(1))s$$ and $\|v\|_{L^2([0, \pi])} = s$, so $$\lim_{s\to 0} \frac{\|f(u+v)-f(u)-Av\|_{L^2([0, \pi])}}{\|v\|_{L^2([0, \pi])}} = 1-\sin(1)\neq 0$$ and therefore, $f$ is not Fréchet differentiable at $u = 0$. However, a lot of sources that I see seem to imply that if the Gâteaux derivative of $f$ is a bounded linear functional at $u$ (and therefore continuous at $u$), then the Fréchet derivative of $f$ ought to exist at $u$ (see, for example, page 5 here). But, isn't the Gâteaux derivative at $u = 0$ a bounded linear functional? Why doesn't this imply existence of the Fréchet derivative?

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A sufficient condition for Fréchet differentiability is that the Gâteaux derivative depends continuously on the position of differentiation, that is if $$X \ni x \mapsto f'(x) \in \mathcal L(X,Y)$$ is continuous. This is not to be confused with the continuity of $h \mapsto f'(x) \, h$.

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  • $\begingroup$ I understand. I'm having trouble showing that this map is discontinuous at $u = 0$, though. $\endgroup$ – Michael Lee Feb 27 '17 at 20:08
  • $\begingroup$ That is, the map $u\mapsto \cos(u)$ from $L^2([0, \pi])$ to $L^2([0, \pi])$. $\endgroup$ – Michael Lee Feb 28 '17 at 1:40
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    $\begingroup$ No, you need to check continuity of the map $u \mapsto F(u)$, where $F(u) \in \mathcal L(L^2, L^2)$ is the linear operator $F(u) \, h = \cos(u) \, h$. This (more or less) boils down to the continuity of $u \mapsto \cos(u)$ from $L^2$ into $L^\infty$. $\endgroup$ – gerw Feb 28 '17 at 7:23
  • $\begingroup$ Ah, I see. Okay, that's easy, because we can just use $u = 1_{[0, s^2]}$ as above, and then $\|0-u\|_{L^2} = s$, but $\|\cos(0)-\cos(u)\|_{L^{\infty}} = 1$, so for $u$ arbitrarily small in $L^2$ we don't get $\cos(u)$ arbitrarily close to $\cos(0)$ in $L^{\infty}$. I'm not sure why you've identified $\mathcal{L}(L^2, L^2)$ with $L^{\infty}$, though. $\endgroup$ – Michael Lee Feb 28 '17 at 8:42
  • $\begingroup$ Although, I see that we can also choose $u$ such that the operator norm $\|\cos(0)-\cos(u)\|_{\mathcal{L}(L^2, L^2)}$ does not go to $0$ as $u$ goes to $0$. For $u$ as above, we let $v = s^{-1}1_{[0, s^2]}$. Then, $\|v\|_{L^2} = 1$, but $\|(\cos(0)-\cos(u))v\|_{L^2} = 1-\cos(1)$, so $\|\cos(0)-\cos(u)\|_{\mathcal{L}(L^2, L^2)}\geq 1-\cos(1)$ for $u$ arbitrarily close to $0$ in $L^2$. $\endgroup$ – Michael Lee Feb 28 '17 at 9:06

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