Continuous function in vector space Let $E$ a normed vector spaces. Let $B$ a basis (finite or infinite) of $E$ and $e\in B$ fixed. We know each $x$ in $E$ can be write as $x=w_x+\alpha_x e$ (where $w_x$ is in the generated space by $B\backslash\{e\}$ and $\alpha_x\in \mathbb{K}$. 
Define $T: E\to \mathbb{K} $, where $T(x)=\alpha_x$.
I proved $T$ is linear. Is $T$ continuous?
 A: 
Proposition:
Let $X$ be a normed vector space and $x_1,x_2...x_n$ linearly independent elements of $X$.Then exists  $C>0$(which depends on the norm and $x_1,x_2...x_n$)such that $\forall a_1,a_2...a_n \in F$ we have $C(|a_1|+|a_2|+...+|a_n|) \leq ||a_1x_1+...+a_nx_n||$

Now let $x_n,x \in X$ such that $||x_n-x|| \to 0$
Then $$x_n=a^1_ne_1+...+a^{m-1}_ne_{m-1}+a_n^me \to x=x_1e_1+...+x_{m-1}e^{m-1}+x_me$$
where $x_1,x_2...x_m$ are linearly independent.
From the proposition we have that: $$C|a_n^i-a_i| \leq C(|a_n^1-a_1|+...+|a_n^m-a_m|) \leq ||x_n-x|| \to 0$$
Thus $a_n^i \to a_i,\forall i \in \{1,2,3...m\}$
So for $a_x=a_m$ and $a_{x_n}=a_n^m$ we have that $$T(x_n) \to T(x)$$
Thus $T$ is continuous.
A: If you mean a HAMEL basis (i.e. every $x\in E$ can be uniquely written as a FINITE linear combination of basis elements) and if you believe in the axiom of choice the answer is negative: Consider e.g. $E=\ell^1$, let $L$ be the subspace of sequences with only finitely many non-zero entries, choose a basis $B_0$ of $L$ and extend it to a basis $B$ of $E$. For every $e\in B\setminus B_0$ the linear functional $T: X\to \mathbb K$ defined by $T(e)=1$, $T(b)=0$ for $b\in B\setminus \{e\}$ and linear extension to $E$ is discontinuous because it vanishes on the dense subspace $L$ but not on the whole space.
