TITLE: $C^1$-functions on Banach spaces
QUESTION [7 upvotes]: For Banach spaces $X,Y$ and an open subset $U$ of $X$  a function $f:U\to Y$ is $C^1$ if $U\to L(X,Y)$, $x\to f'(x)$ is continuous where, by definition, the derivative $f'(x)$ is a continuous linear operator from $X$ to $Y$ satisfying $$\|f(x+h)-f(x)-f'(x)(h)\|_Y = o(\|h\|_X)$$ and $L(X,Y)$ is the Banach space of continuous linear operators endowed with the operator norm. Then $f'$ is again a map between Banach spaces and one can define higher derivatives.
For Frechet instead of Banach spaces there are many different topologies on $L(X,Y)$ but none of them is Frechet. To stay in the category one thus needs a different definition. In his exposition of the Nash-Moser theorem
Hamilton therefore tells $f$ continuously differentiable if all directional derivatives $Df(x)(v)=\lim\limits_{t\to 0}(f(x+tv)-f(x))/t$ exist in $Y$ and $Df: U\times X \to Y$ is continuous. He then writes:
Note well that this definition of a $C^1$ map does not agree with the usual
definition for a Banach space. The derivative will of course be the same map (i.e. $Df(x)(v)=f'(x)(v)$),
but our continuity requirement is weaker.
What is an example (in Banach spaces) showing that the latter definition is strictly weaker than the former?

REPLY [4 votes]: Here is a different and very natural example from the forthcoming book Infinite-dimensional Lie Groups. General Theory and Main Examples by H. Glöckner and K.-H. Neeb. I describe it here with the permission of H. Glöckner.
The function $f:L^1[0,1]\to L^1[0,1]$  is the superposition $\xi\mapsto \sin\circ\xi=\sin(\xi)$. The directional derivatives are $Df(\xi)(\eta)=\cos(\xi)\eta$ -- which looks very plausible but the proof needs some care. On can show that $Df:L^1[0,1] \times L^1[0,1]\to L^1[0,1]$ is continuous but $Df:L^1[0,1] \to \mathcal L_{op}(L^1[0,1])$ is not: Because of $Df(0)=id$ the inverse mapping theorem would imply that the range $f(L^1[0,1])$ contains a $0$-neighbourhood in $L^1[0,1]$ which is not true because this range is contained in the unit ball of $L^\infty[0,1]$ (which is of first category). If one prefers a Hilbert space example one can replace $L^1$ by $L^2$.