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 $\forall b \in B : f(b)=F(b)$?

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.
