Is it possible to extend a continuous linear function from a basis to an entire Banach space? Let $X$ and $Y$ be two Banach spaces over $\mathbb{C}$.
Let $B \subset X$ be a subset of $X$ such that it's linearly independent and $X=\overline{\operatorname{span}(B)}$.
Let $f : B \to Y$ be a continuous function.
Is it possible to extend $f$ to $F : X \to Y$ such that $F$ is a continuous linear function and $\forall b \in B : f(b)=F(b)$?
 A: No, not necessarily. Here's an example (with $B$ compact and connected):
Let $K$ be a metric space with diameter at most 1.
Let $B = \{\delta_x \ :\ x \in K\} \subset \mathrm{Prob}(K)$ be the set of point masses, and let $X_0$ be the vector space of complex-valued measures on $K$ with finite total variation. We can give $X_0$ the bounded Lipschitz norm $$\|\mu\| = \sup \left\{ \big\lvert \int f\, d\mu \big\rvert \ : \ f \in \mathrm{Lip}^1(K) , \ \|f\|_\infty \leq 1 \right\}$$ where $\mathrm{Lip}^1(K)$ is the set of 1-Lipschitz functions $K \to \mathbb{C}$. This norm restricts to the Wasserstein/transportation metric $\bar{d}$ on $\mathrm{Prob}(K)$, so $B$ is compact and connected if $K$ is (the map $x \mapsto \delta_x$ is an isometry $K \to B$). $B$ is also linearly independent.
The span of $B$ is dense in $X_0$: any $\mu \in X_0$ can be written as a linear combination of four probability measures on $K$ (first decompose into real and imaginary parts, then decompose these into their positive and negative parts). Each of these probability measures can then be approximated in transportation distance to arbitrary precision by measures with finite support.
$X_0$ is actually not complete. 
But the span of $B$ is still dense in the completion $X$ of $X_0$, and $B$ is still compact and connected when considered as a subset of $X$, so we can work in $X$ instead. The important thing is that we have a (complete, if you want) normed space containing $\mathrm{Prob}(K)$ such that the norm gives the transportation metric on $\mathrm{Prob}(K)$.
Let $\tilde{f} \colon K \to \mathbb{C}$ be an arbitrary continuous function, and let $f \colon B \to \mathbb{C}$ be the continuous function given by $f(\delta_x) = \tilde{f}(x)$. 
Suppose there were a linear, continuous extension $F \colon X \to \mathbb{C}$. Since $F$ is a continuous functional on a normed space it must be bounded, say $|F(\mu)| \leq C \|\mu\|$. But this means that
$$|\tilde{f}(x) - \tilde{f}(y)| = |F(\delta_x - \delta_y)| \leq C \|\delta_x - \delta_y\| = C d(x,y),$$
i.e. $\tilde{f}$ is Lipschitz. So as long as $K$ admits non-Lipschitz functions, not every continuous function on $B$ admits a continuous linear extension to $X$.
A: For me continuous function on the basis $B$ means continuous on $Span(B)$ since $f$ is assumed to be inear, it has to be defined on a vector space.
Let $x\in X$, there exists $(x_n), x_n\in B, lim_nx_n=x$. Since the restriction of $f$ on $B$ is continuous, $\|f(x_n)-f(x_m)\|\leq \|f\|\|x_n-x_m\|$. This implies that $f(x_n)$ is a Cauchy sequence. Since $Y$ is complete, there exists $f(x)=im_nf(x_n)$.
Let $y_n$ be another sequence such that $lim_ny_n=x$. $lim_n(x_n-y_n)=0$. Since $f$ is continuous, $lim_nf(x_n-y_n)=lim_nf(x_n)-lim_nf(y_n)=0$. Implies that $f$ is well-defined on $X$.
$lim_n\|f(x_n)\|\leq lim_n\|f\|\|x_n\|=\|f\|\|x\|$ implies that $f$ is continuous.
