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I would just like to know if my following proof is correct:

Claim: If $T:\mathbb{R^n} \to\mathbb{R^m}$ is a linear map, then there exists $C > 0$ such that for every $x \in \mathbb{R^n}$$\|T(x)\| \le C\|x\|$.

Proof: We have $T(x) = \sum_{i=1}^{n}x_iT(e_i)$, so let $C = n\max(\|T(e_i\|)$. Then,

$$\|\sum_{i=1}^{n}x_iT(e_i)\| \le \sum_{i=1}^{n}|x_i|\|T(e_i)\|$$

by the triangle inequality, and

$$\sum_{i=1}^{n}|x_i|\|T(e_i)\| \le \sum_{k=1}^{n} \max|x_i|\max\|T(e_i)\| \le C\|x\|$$

  • 0
    If you use 2-norms for $\mathbb R^n$ and $\mathbb R^m$, your solution looks fine for me.2012-05-29
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    All norms are equivalent in $\mathbb{R}^n$. So, I think, it doesn't matter which norm he considers in this scenario.2012-05-29
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    @Ashok: indeed, but it does matter if inequalities applied are true or not, say in the latter row it is used that $$ \sum\limits_{k=1}^n\leq n\|x\| $$ which does not hold for any norm, but it does hold for the $2$-norm2012-05-29
  • 0
    (typo correction:) $\sum\limits_{k=1}^n|x_k|\leq n\|x\|$2012-05-29

1 Answers 1