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Can you tell me whether my (partial) proof that $\|v\|_B := \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \}$ where $\varnothing \neq B \subset \mathbb R^d$ is open, bounded, $B = -B$ and convex defines a norm is correct? It seems too long (especially (ii)) and is therefore either too complicated or wrong. Also, could someone complete it for me by showing me how to do (iv)? Thanks.


Claim: Let $V$ be a real vector space. Then $\|v\|_B := \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \}$ where $B \subset \mathbb R^d$ is open, bounded, $B = -B$ and convex defines a norm.

Proof:

(i) $\|v\|_B \geq 0$ by definition.

In the following let $B$ be bounded by $K$, i.e. for all $b$ in $B$, $\|b\|_2 \leq K$.

(ii) $\|v\|_B = 0 \iff v = 0$: First $\implies$: Let $\|v\|_B = 0$. Then $\inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = 0$, in particular, for every $\lambda = \frac1n \in \mathbb N$, $\frac{1}{\lambda}\|v\|_2 = n \|v\|_2 \leq K$. Hence $\|v\|_2 = 0$ and hence $v = 0$ since $\|\cdot\|_2$ is a norm.

$\Longleftarrow$: Let $v=0$. Then for every $n \in \mathbb N$, $n \|v\|_2 = 0$ and hence for every $\lambda = \frac{1}{n}$, $\frac{1}{\lambda} \|v\|_2 = 0 \leq K$. Hence $ \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = \lim_{n \to \infty} \frac{1}{n} = 0$.

(iii) $\| \alpha v \|_B = |\alpha| \|v\|_B$ for all $\alpha \in \mathbb R$: $$ \| \alpha v \|_B = \inf \{\lambda > 0 \mid \frac{1}{\lambda} \alpha v \in B \} \stackrel{B = -B}{=} \inf \{\lambda > 0 \mid \frac{1}{\lambda} |\alpha| v \in B \} = \inf \{\lambda |\alpha| > 0 \mid \frac{|\alpha|}{|\alpha|\lambda} v \in B \} = |\alpha| \inf \{\lambda > 0 \mid \frac{1}{\lambda} v \in B \} = |\alpha| \|v\|_B$$

(iv) $ \|v + w \|_B \leq \|v\|_B + \|w\|_B$:

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  • $\begingroup$ You've forgotten to assume $B$ is nonempty. $\endgroup$ Jul 5 '12 at 14:15
  • $\begingroup$ Sometimes the long way is the only correct way. (I'm not asserting that to be the case, here, but be cautious of concluding a proof is incorrect based on its length.) $\endgroup$ Jul 5 '12 at 15:53
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    $\begingroup$ You also need to show that $\|.\|_B$ takes values in $\mathbb{R}$: the infimum may a priori be $\pm\infty$. $\endgroup$ Jul 8 '12 at 20:38
  • $\begingroup$ @wildildildlife Yes, thanks for pointing that out! $\endgroup$ Jul 8 '12 at 21:21
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Your proof of (ii) seems fine to me.

For (iv), let $\epsilon>0$, and choose $\lambda_w >0$ such that $\frac{w}{\lambda_w} \in B$, and $||w||_B > \lambda_w - \epsilon$. Choose $\lambda_v$ in a similar manner for $v$. By convexity of $B$, you have $$\frac{\lambda_w}{\lambda_w+\lambda_v} \frac{w}{\lambda_w} + \frac{\lambda_v}{\lambda_w+\lambda_v} \frac{v}{\lambda_v} \in B,$$ so in particular, $\frac{w+v}{\lambda_w+\lambda_v} \in B$. It follows that $||w+v||_B \leq \lambda_w+\lambda_v \leq ||w||_B+||v||_B + 2 \epsilon$. Since $\epsilon>0$ was arbitrary, we are finished.

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