Let $\ell^2 \equiv \ell^2(\mathbb{N})$ denote the space of square-summable real sequences.
Let $f \colon \ell^2 \to \ell^2$. Let $f_v(x) := \langle v, f(x) \rangle$.
Suppose there exists a continuous linear operator $D_x \colon \ell^2 \to \ell^2$ such that that for each unit vector $v \in \ell^2$, $$ \big|f_v(x + h) - \{f_v(x) + \langle D_x h, v \rangle\} \big| = o\big(\|h\|\big), \quad \mbox{as}~h \to 0. $$
Does it then imply that $D_x$ is the Fréchet derivative of $f$ at $x$? If instead we restrict to basis vectors $v = e_j$, $j = 1, 2, \ldots$, does this still hold?
In general, it would also be useful to know of a reference where one can find such basic results regarding differentiability of functions on $\ell^2$ (or equivalently, over a separable Hilbert space).
At least the restriction to basis vectors seems to fail: Set $$ f:l^2 \to l^2, \quad f(x)=\left(\frac{\cos(2^kx_k)}{k}\right)_{k=1}^\infty, $$ consider differentiability at $0$ and set $D_0x:=0$ $(x \in l^2)$.
Now $f_{e_k}(x)= \cos(2^kx_k)/k$ and for $h \not=0$ we have $$ \frac{1}{\|h\|}|f_{e_k}(h)-f_{e_k}(0)-\langle D_0h,e_k \rangle|= \frac{1}{\|h\|}| \frac{\cos(2^kh_k)}{k}-\frac{1}{k}| $$ $$ \le \frac{1}{|h_k|}| \frac{\cos(2^kh_k)}{k}-\frac{1}{k}| \to 0 \quad (\|h\| \to 0). $$ Note for the inequality above that we have $\le 0$ if $h_k=0$.
On the other hand $f$ is not totally differentiable in $0$: For $h \not=0$ we have $$ \frac{1}{\|h\|}\|f(h)-f(0)- D_0h\| = \frac{1}{\|h\|} \left(\sum_{k=1}^\infty|\frac{\cos(2^kh_k)}{k}-\frac{1}{k}|^2\right)^{1/2} \quad =:g(h) $$ Inserting $h^{(n)}:=\frac{\pi}{2^n} e_n$ (note that $\|h^{(n)}\| \to 0$) yields $$ g(h^{(n)})=\frac{2^n}{\pi}|\frac{\cos(\pi)}{n}-\frac{1}{n}|=\frac{2^{n+1}}{n\pi} \to \infty \quad (n \to \infty). $$