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The $\implies$ part interests me. The proof given goes like this:

Let $A$ be continuous in 0 (because the 0 vector is in every vector space)

$B_y(0,r)=\{y \in Y | \| y\|<r \} \implies \exists B_x(0,s), A(B_s(0,s))\subset B_y(0,r), \sup_{\| x\|\subseteq S}\|Ax \| \leq r.$

Then it says for any

$$x \in X: \|Ax\|= S^{-1}\|x\|\|A(\frac{Sx}{\|x\|})\|\leq S^{-1}r\|x\| > \implies \|Ax \| \leq M \|x \|$$

Jonas Meyer
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Bozo Vulicevic
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  • what is your question? and $S$ is some positive real, but what means then $\sup_{| x|\subseteq S}|Ax |$ or should it be $\sup_{| x|\leq S}|Ax |$ ? – user190080 Jul 22 '15 at 01:48
  • My question is the clarification of the highlighted and what you wrote second should be correct i think – Bozo Vulicevic Jul 22 '15 at 01:50
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    You are asking about normed spaces, not just metric spaces, right? (Otherwise this doesn't make sense or is missing other assumptions.) – Jonas Meyer Jul 22 '15 at 02:18

1 Answers1

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What happens is the following:

first we state that $\sup_{\| x\|\leq S}\|Ax \|=\sup_{\| x\|= S}\|Ax \|\leq r \tag 1 $

then we see that for $x\in X$ the following holds, since $A$ is linear $$ \|Ax\|= A\left (S^{-1}\|x\|\frac{Sx}{\|x\|}\right )=S^{-1}\|x\|\|A(\frac{Sx}{\|x\|})\| $$ then we see that $\forall x\in X: \|\frac{Sx}{\|x\|}\|=S$ which means we can apply $(1)$ which gives us $$ \|Ax\|=S^{-1}\|x\|\|A(\frac{Sx}{\|x\|})\|\leq S^{-1}\|x\|r=M\|x\| $$ with $M:=S^{-1}r$. This means we have $$ \|Ax\|\leq M\|x\| $$ and $A$ is bounded. You can find a very nice proof also here by Anthony Carapetis.

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user190080
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