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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

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

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If you use 2-norms for $\mathbb R^n$ and $\mathbb R^m$, your solution looks fine for me. –  Ilya May 29 '12 at 7:52
    
All norms are equivalent in $\mathbb{R}^n$. So, I think, it doesn't matter which norm he considers in this scenario. –  Ashok May 29 '12 at 7:56
1  
@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 $2$-norm –  Ilya May 29 '12 at 8:01
    
(typo correction:) $\sum\limits_{k=1}^n|x_k|\leq n\|x\|$ –  Ilya May 29 '12 at 8:14

1 Answer 1

Your proof looks fine. $%And now I avoid the black box bad answer detector just a bit.$

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+1 for answering the question. Often when the OP provides a proof himself, the question will go unanswered. Ideally, one might try to convince the OP to post his solution as an answer and accept it, but for some reason this rarely happens. –  Eric Naslund May 29 '12 at 8:38

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