Example where Gâteaux differentiability does not imply Fréchet differentiability

Question

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?

Answer 1

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 $h \mapsto f'(x) \, h$.