If $T$ is a linear map then $\|T(x)\|<\infty$ Let $T:V\to W$ be a linear map, $V, W$ be vector spaces over $\mathbb{R}$ or $\mathbb{C}$. Then if $T$ is continuous at $x_0\in V$ then $\nu:=\sup\{\|Tx\|:x\in V, \|x\|\leq 1\}<\infty$.
I'm stuck with the proof. First of all, if $T$ is continuous at $x_0$, this implies that $T$ is continuous on $V$, correct? Do I need to prove that? I can prove this for the case when $T$ is continuous at $0$, but I'm not sure how to prove for a general $x_0$.
Also, can someone please give me a hint what I should pay my attention to in order to realize that $\nu<\infty$? In my understanding, $\nu$ is the maximum norm of $T(x)$.
 A: Assume that $T$ is continuous at $x_0$. If $x_n\to0$, then $x_n+x_0\to x_0$. By the continuity at $x_0$, 
$$
Tx_0+ Tx_n=T(x_0+x_n)\to Tx_0.
$$
It follows that $Tx_n\to0$, so $T$ is continuous at $0$. We can use the same idea to deduce continuity at any other point. 
A: By Martin's argument, $T$ is continuous at 0. Thus if $x_n \to x$, then $x - x_n \to 0$, so $T(x - x_n) \to T(0) = 0$, so $T(x) - T(x_n) \to 0$, which implies $T(x_n) \to T(x)$.
To show boundedness in the operator norm $\nu$, suppose that $T$ is unbounded. Then there exists a sequence $\{x_n\}$ with $\|x_n\| = 1$, such that $\|T(x_n)\| \geq n$ for each $n$. Define
\begin{align*}
y_n \doteq \frac{x_n}{n}
\end{align*}
Then $\|y_n\| = 1/n \to 0$, so $y_n \to 0$, but
\begin{align*}
T(y_n) = \frac{1}{n}T(x_n) \Rightarrow \|T(y_n)\| = \frac{1}{n}\|T(x_n)\| \geq 1
\end{align*}
for all $n$. But this is a contradiction since $T$ is continuous at $0$.
A: Hint
Continuity at $x_0$ implies continuity at $0$. (Use linearity.)
Continuity at $0$ implies boundedness. (Use this to construct a bound for $\| Tv \|$.)
