Clarification on proof that for a linear map T, being continuous and being Lipschitz are equivalent statements

Question

I know that in general for any function $f:V \longrightarrow W $ from a normed vector space $V$ to a second normed vector space $W$, being a Lipschitz function implies being continuous, is the second implication on my question the one I'm having troubles to understand.

Answer 1

Hint: Consider an open ball around $0$ within $f^{-1}(B_1(0))$ to obtain that $f$ is bounded and use linearity.

Answer 2

If $\dim V$ and $\dim W$ are finite, here it is a proof for the sake of curiosity.

Based on such assumption, let us consider a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ for $V$ and a basis $\mathcal{B}' = \{b'_{1},b'_{2},\ldots,b'_{m}\}$ for $W$, where $n = \dim V$ and $m = \dim W$.

Then one has the following matricial representation for $T$: \begin{align*} [T]_{\mathcal{B}}^{\mathcal{B}'} = [[T(b_{1})]_{\mathcal{B}'}\, [T(b_{2})]_{\mathcal{B}'} \ldots\,[T(b_{n})]_{\mathcal{B}'}] \end{align*}

Consequently, if $v = a_{1}b_{1} + a_{2}b_{2} + \ldots + a_{n}b_{n}$, one has that \begin{align*} T(v) = [T]_{\mathcal{B}}^{\mathcal{B}'}[v]_{\mathcal{B}} = a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}} \end{align*}

Finally, according to the triangle inequality as well as the Cauchy-Schwarz inequality, we have that

\begin{align*} \|T(v)\| & = \|a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}} \|\\\\ & \leq \|a_{1}[T(b_{1})]_{\mathcal{B'}}\| + \|a_{2}[T(b_{2})]_{\mathcal{B'}}\| + \ldots + \|a_{n}[T(b_{n})]_{\mathcal{B'}}\|\\\\ & = |a_{1}|\|[T(b_{1})]_{\mathcal{B'}}\| + |a_{2}|\|[T(b_{2})]_{\mathcal{B'}}\| + \ldots + |a_{n}|\|[T(b_{n})]_{\mathcal{B'}}\|\\\\ & = \langle(|a_{1}|,|a_{2}|,\ldots,|a_{n}|),(\|[T(b_{1})]_{\mathcal{B'}}\|,\|[T(b_{2})]_{\mathcal{B'}}\|,\ldots,\|[T(b_{n})]_{\mathcal{B'}}\|)\rangle \leq K\|v\| \end{align*}

Finally, due to the linearity of $T$, the desired result holds: \begin{align*} \|T(v) - T(w)\| = \|T(v - w)\| \leq K\|v - w\| \end{align*}

Hopefully this helps!