Proof that a linear map between normed spaces is continuous if and only if it is bounded

Proof

We prove both directions. The metrics are $\|x - y\|_V$ and $\|x - y\|_W$.

1. Suppose $T$ is bounded, with $\|Tx\|_W \le C \|x\|_V$ for all $x$. By linearity, $\|Tx - Ty\|_W = \|T(x - y)\|_W \le C \|x - y\|_V$. So $T$ is Lipschitz with constant $C$.
2. Let $a \in V$ and $\varepsilon > 0$, and let $\delta = \varepsilon / (C + 1)$. If $\|x - a\|_V < \delta$, then $\|Tx - Ta\|_W \le C \delta < \varepsilon$.
3. By continuity via open sets, $T$ is continuous. So $T$ is a continuous linear map.
4. Conversely, suppose $T$ is continuous. The open ball $B = \{w \in W : \|w\|_W < 1\}$ is open, so $T^{-1}(B)$ is open in $V$.
5. Since $T0 = 0 \in B$, we have $0 \in T^{-1}(B)$. By the ball criterion, there is $\eta > 0$ with $\|Tx\|_W < 1$ whenever $\|x\|_V < \eta$.
6. Let $\delta = \eta / 2$, and let $x \in V$ with $x \neq 0$. Put $y = \delta x / \|x\|_V$. Then $\|y\|_V = \delta < \eta$, so $\|Ty\|_W < 1$.
7. By linearity, $Ty = (\delta / \|x\|_V) \, Tx$. So $\|Tx\|_W = (\|x\|_V / \delta) \|Ty\|_W < \|x\|_V / \delta$.
8. For $x = 0$, $Tx = 0$. So $\|Tx\|_W \le C \|x\|_V$ for all $x \in V$, with $C = 1 / \delta$, and $T$ is bounded.