Let $f:E\rightarrow F$ be a map between Banach spaces E and F; E finite dimensional (>0) and F infinite dimensional. Let $F$ be equipped with its weak topology and suppose that $f$ is strong-weak continuous.


Under what additional conditions can we guarantee that $\operatorname{span}(f(E))$ is a *finite-dimensional* subspace of F?



Control on dimension of image
Just to give an example on how weird this can become: take $E = \mathbb{R}$ and $F = \ell^2$ with standard Hilbert basis $e\_0 e\_1, e\_2, \ldots$. Then take a smooth bump function $\chi \in C^\infty(\mathbb{R})$ with support in the unit interval and, say $\chi(1/2) = 1$ to make things non-trivial.


Define the highly non-linear but smooth as smooth can be map
\begin{equation}
f(t) = \sum\_{n = 0}^\infty \chi(t-n) e\_n
\end{equation}
Then the span of the image contains the Hilbert basis...


As a slight variation one can als consider a holomorphic example, e.g. by
\begin{equation}
g(z) = \sum\_{n=0}^\infty \frac{z^n}{n!} e\_n
\end{equation}
Since the $e\_n$ are unit vectors, this series is absolutely convergent for all $z \in \mathbb{C}$. It is holomorphic as a vector-valued function: here weakly coincides with strongly anyway, but this can also to be seen elementary. Now the derivatives at $0$ are the basis vectors again, so there is no finite-dimensional subspace and no open neighbourhood $U$ of $0$ containing $f(U)$: otherwise the derivatives would be in this subspace...


Maybe this example is a bit more illustrative since the first one has the
feature that *locally in $t$* the image is contained in finite-dimensional subspaces. This is not the case with $g$.


