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Let $(E,\|\cdot\|)$ be an Euclidean vector space. Prove that if $f: E \to E$ is linear then $f$ is continuous.

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

Because $E$ is finite dimensional, all norms on $E$ are equivalent. Let $\{e_i \mid 1 \le i \le n\}$ be a basis of $E$. WLOG, we assume that $\|x \| = \sum |\lambda_i|$ for $x = \sum \lambda_i e_i \in E$.

We have $$\begin{aligned}\|f\| &= \sup_{\|x\|=1} \|f(x)\| &&= \sup_{\|x\|=1} \left \|f \left( \sum \lambda_i e_i \right) \right\| \\ &= \sup_{\|x\|=1} \left \|\sum \lambda_i f(e_i) \right\| &&\le \sup_{\|x\|=1} \sum |\lambda_i| \left \|f(e_i) \right\| \\ &\le \sup_{\|x\|=1} \left ( \sum |\lambda_i|\right) \sum \|f(e_i) \| && = \sup_{\|x\|=1} \|x\| \sum \|f(e_i) \| \\ &= \sum \|f(e_i) \end{aligned}$$

As such, $f$ is bounded and thus continuous.

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By finite dimensionality the image of a closed sphere is compact. This immediately implies continuity at $0$ because $E$ is locally compact.

Continuity elsewhere follows from continuity at $0$ and linearity.

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