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I'm trying to prove the following lemma for a vector space:

Suppose $X = span\{ e_1, ..., e_n \}$, with the $e_i$ linearly independent. (i.e. a basis, so you can uniquely write each $x\in X$ as a linear combination $\sum_j x^j e_j$ of the basis elements).

There exists $C > 0$ such that $|| x || \leq 1$ implies $| x^j | \leq C$ for all $j = 1, ..., n$.

At first I was thinking of using induction, since the base case is very simple : $||x^1e_1||\leq 1\implies |x^1|\leq 1/||e_1||$, but the inductive step looks much more difficult. We have $||\Sigma_{j=1}^n x^je_j|| \leq 1$, and using the triangle inequality $||x+y||\leq ||x||+||y||$, we have $||\Sigma_{j=1}^{n-1} x^je_j|| + ||x^ne_n||$, but I don't know how to show that $||\Sigma_{j=1}^{n-1} x^je_j|| \leq 1$ in order to make use of the inductive hypothesis. Any ideas?

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Assume that $X$ is a linear space over $\mathbb{F}$, where $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Define $$f:\mathbb{F}^n\to [0,\infty), \quad f(x^1,\dots,x^n)=\|\sum_{j=1}^nx^je_j\|,$$

and let $$K:=\{(x^1,\dots,x^n)\in\mathbb{F}^n:\max_{1\le j\le n}|x^j|=1\}.$$

By definition, $f$ is continuous and $K$ is compact. Moreover, $0\notin f(K)$. Therefore, there exists $C>0$, such that if $\max_{1\le j\le n}|x^j|=1$, then $\|\sum_{j=1}^nx^je_j\|\ge C^{-1}$. As a result, if $\max_{1\le j\le n}|x^j|=C$, then $\|\sum_{j=1}^nx^je_j\|\ge 1$. That is to say, if $\|\sum_{j=1}^nx^je_j\|\le 1$, then $\max_{1\le j\le n}|x^j|\le C$.

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