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

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.

## 1 Answer

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.