$T$ continuous in $x_0$ then $T$ is continuous 
Let $T:V\to W$ be a linear operator, with $V, W$ normed spaces. Show that if there exist $x_0 \in V$ such that $T$ is continuous in $x_0$ then $T$ is continuous.

I'm thinking that given an arbitrary $x_1 \in V$, to prove continuity in $x_1$ I could in someway, try to make a sort of translation from $x_1$ to $x_0$, and using both linearity of $T$ and the continuity of $T$ in $x_0$ I will be able to prove continuity of $T$ in $x_1$. However I'm not being able to do that. Can you give me a hint, if possible, without defining new norms?
 A: Note that $T$ being continuous at $x_0$ means, by definition: for any $\epsilon > 0$, there exists a $\delta > 0$ such that
$$
\|y\|< \delta \implies \|T(x_0 + y) - T(x_0)\|  < \epsilon
$$
By linearity, we can rewrite this as
$$
\|y\|< \delta \implies \|T(0 + y) - T(0)\| <\epsilon
$$
So, $T$ is continuous at $x_0$ (an arbitrary point) if and only if it is continuous at $0$.
It follows that if $T$ is continuous at any single point, it is continuous everywhere, which is to say that $T$ is continuous.
A: Let $x \in V$ be arbitrary and define a sequence $\{p_n\}$ such that $p_n \to x$.  Now define for each $n \in \mathbb{N}$ the sequence $\{a_n\}$ where $a_n = p_n + x_0 - x$, certainly we have that $a_n \to x_0$.  Now by continuity $T(a_n) \to T(x_0)$ but observe that we have 
$$
T(a_n) \;\; =\;\; T(p_n) + T(x_0) - T(x) \;\; \to \;\; T(x_0).
$$  
A: Prove that the operator $T$ is continuous if and only if 
$$
T':x\mapsto T(x-x_0)
$$
is continuous. Assume $T$ is continuous. Notice the translation operator is continuous. So $T'$ is the composition of continuous operators. The same works to prove the other direction.
