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How can we prove:

$$|\alpha|*||x|| = ||\alpha x|| $$ With the norm defined as: $$ ||x|| := \inf\left\{ \lambda > 0 \mid x/\lambda\in B \right\} $$

Where B is convexe, open, symmetric and bounded and $ 0\in B$.

I thought about to proof this by contradiction, but I never encountered a situation where I wanted to pull out a constant from a definition of a set.

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    What is $x$? Is it just a fixed value? – 2017-02-22

2 Answers 2

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Sketch of one direction:

Observe that $\|\alpha x\|=\inf\{\lambda>0:\alpha x/\lambda\in B\}$. Since this is an infimum, by the definition of an infimum, for all $\varepsilon>0$, there exists a $\lambda_\varepsilon$ such that

  1. $\frac{\alpha x}{\lambda_\varepsilon}\in B$

  2. $\|\alpha x\|\leq \lambda_\varepsilon<\|\alpha x\|+\varepsilon$.

Consider $\lambda'_\varepsilon:=\frac{\lambda_\varepsilon}{|\alpha|}$. This lambda satisfies $\frac{x}{\lambda'_\varepsilon}\in B$ since $\frac{x}{\lambda'_\varepsilon}=\frac{|\alpha| x}{\lambda_\varepsilon}$, which we know is in $B$ since $\frac{\alpha x}{\lambda_\varepsilon}\in B$ and $B$ is symmetric about the origin. Therefore, we know that $\lambda'_\varepsilon$ is one of the elements of $\{\lambda>0:x/\lambda\in B\}$. Hence, for the infimum, $\|x\|\leq\lambda'_\varepsilon$.

From the inequalities above, we know that $$ \frac{1}{|\alpha|}\|\alpha x\|\leq \lambda'_\varepsilon<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}. $$ Combining inequalities, we know that $$ \|x\|\leq\lambda'_\varepsilon<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}. $$ In other words, $$ \|x\|<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}. $$ Since $\varepsilon$ was arbitrary, we can let it be as small as possible, and, in the limit, we get $$ \|x\|\leq \frac{1}{|\alpha|}\|\alpha x\|. $$

This gives the proof of one side, for the other direction, mimic this proof, but start with $\|x\|$. You can, alternatively, replace $\alpha x$ by $x$ and $\alpha$ by $\frac{1}{\alpha}$ to use this proof as a lemma.

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    I'm sorry my question was ambiguous, I'd appreciate if you could have another look on it. – 2017-02-22
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    This is really nice, I'll try this! – 2017-02-22
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I would suggest letting $m = |\alpha| \mathrm{inf} \{ \lambda > 0 \mid x/\lambda \in B \}$ and verifying that $m$ satisfies the requirements to be the infimum of the set $\{ \lambda > 0 \mid \alpha x / \lambda \in B \}$. That is, prove that

  • If $\lambda > 0$ satisfies $\alpha x/ \lambda \in B$, then $m \le \lambda$; and
  • If $m'$ is another value such that $m' \le \lambda$ for all $\lambda > 0$ such that $\alpha x/\lambda \in B$, then $m' \le m$.

You'll use the fact that $m$ is $|\alpha|$ times an infimum of another set in your proof.

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    I'm not clear about what the variables $x$ or $B$ refer to or how they are quantified, but, as a general principle, this proof technique is a good way of proving that an expression involving one infimum (or supremum) is equal to another infimum (or supremum). – 2017-02-22
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    I'm sorry my question was ambiguous, I'd appreciate if you could have another look on it. – 2017-02-22
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    @Felix.C: Ah, OK, in that case my answer goes through just fine with $x$ taken to be an arbitrary, fixed element of the space. – 2017-02-22
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    I tried to do so, perhaps I'm too tired, I would be grateful about a hint for #1. – 2017-02-22
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    The $x/\alpha \lambda$ bits should have been $\alpha x / \lambda$... that might help. For the first bit, suppose $\lambda > 0$ is such that $\alpha x / \lambda \in B$. By symmetry, we have $|\alpha| x / \lambda \in B$. Letting $\lambda' = \lambda/|\alpha|$ gives $x/\lambda' \in B$, and hence $\lVert x \rVert \le \lambda'$. But then $m = \alpha \lVert x \rVert \le \lambda$. – 2017-02-22