$C^1$-functions on Banach spaces 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?
 A: In Banach spaces, continuity of $Df:U\times X \to Y$ and linearity in the second variable implies that $Df: U\to L(X,Y)$ is continuous for the topology of uniform convergence on compact subsets of $X$, and not the operator norm in general.
If $f$ is $C^2_c$ then $Df: U\to L(X,Y)$ is actually continuous into the operator norm.
Since I am travelling, I am now quite incomplete. Sorry.
EDIT:
I had the opportunity to look up the literature now. An example on $\ell^2$, due to Smolyanov, is in
12.13 of:

*

*Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997.  (pdf)

Edit by Jochen Wengenroth
The example is the following: Take a smooth function $\varphi$ with support in $[-1/4,1/4]$ and $\varphi(0)=1$. Define $f_k:\ell^2\to \mathbb R$ by $x\mapsto \varphi(k(x_k-1)) \prod\limits_{j<k}\varphi(jx_j)$ and $f=\sum\limits_{k=1}^\infty \frac{1}{k^2} f_k$.
A: $\def\kp#1{\kern.#1mm}\def\bbNo{\mathbb N_0}\def\seq#1{\langle\kern.8mm{#1}\kern.6mm\rangle}\def\bbR{\mathbb R}\def\bbZp{\mathbb Z^+}\def\sp#1{\kern.#1mm}$On page 63 in Differential Calculus in Locally Convex Spaces H. H. Keller refers to
Y. P. Wong: Differential Calculus and Differentiable Partitions of Unity in Locally Convex Spaces, §1.2, University of Toronto, 1974
for the assertion that $C_{\rm c}^{\kp4 1}$ is strictly weaker than $C_{\rm b}^{\kp4 1}$ for maps between Banach spaces. I have not looked at this paper, and thus I do not know whether one can find there an explicit example or not. I have constructed such an example of a map $c_0(\bbNo)\to c_0(\bbNo)\kp7$, if I recall correctly, but it is unpublished and lies somewhere in my files. The construction and the associated proof are not quite trivial.
Added. (14.7.2016) I haven't found my original example but here is a similar one. Let $f:c_0(\bbZp)=E\to E$ be defined by $x\mapsto\seq{n^{-2\,}\varphi\,(\sp3 n^{\,2\,}x_{\kp5 n}):n\in\bbZp}$ where for example $\varphi:\bbR\to\bbR$ is given by $t\mapsto\frac 12\,t\,(\sp4 3 - t^{\,2}\sp2\big)$ for $-1\le t\le 1$ and $t\mapsto{\rm sgn\,}t$ otherwise. It is almost trivial to see that $f$ is continuous. Also the directional derivatives ${\rm D\,}f\,x\,u=\delta\,f\,(\sp3 x\sp3,u\sp3) = \lim_{\,t\,\to\,0\,}t^{-1\,}(\sp3 f\,(\sp3 x + t\,u\sp3) - f\,x\sp3) = \seq{{\rm D\,}\varphi\,(\sp3 n^{\,2\,}x_{\kp5 n}\sp3)\,u_{\kp5 n}:n\in\bbZp}$ are defined for all $x,u$ in $E$. One easily verifies that ${\rm D\,}f\,x$ and $\delta\,f\,(\,\cdot\,,u\sp3)$ are continuous for all fixed $x,u$ in $E\,$. This means that we have $f$ in $C_{\kp3\rm s}^{\,1}$ in Keller's sense, and by Keller's Theorem 1.0.2 (p. 61) we get that also $f$ is $C_{\rm c}^{\,1}\,$, i.e. once continuously differentiable in Hamilton's sense. That $f$ is not $C_{\rm b}^{\,1}\,$, i.e. not once continuously differentiable in the classical Banach space sense, is seen by taking $x = n^{-1\,}{\rm e}_{\kp5 n}$ and $u = {\rm e}_{\kp5 n}$ and letting $n\to +\infty$ since for all $n\in\bbZp$ we then have $((\kp6{\rm D\kp5}f\,x-{\rm D\kp5}f\kp9 0_E\sp3)\,u\sp3)_{\kp5 n} = -\frac 32\,$, if I computed correctly.
One observes that the same example also works for $E=\ell^{\kp9 p\kp5}(\bbZp)$ with $1\le p < +\infty\,$.
Proof of directional differentiability (added 15.7.2016)

A: 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 $L^2$. 
