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Let $\lVert\cdot\lVert$ be any norm on the vector space $E$ and let $\rho\left(\sum^n_{j=1}\lambda_je_j\right)=\left(\sum^n_{j=1}|\lambda_j|^2\right)^{1/2}$ where $(e_j)$ is a basis for $E$.

Now Young argues that $f(\lambda_1,\ldots,\lambda_n)=\left\lVert\sum^n_{j=1} \lambda_j e_j\right\lVert$ reaches its infimum (it is unclear whether this is a global og local infimun) $m\geq 0$ on the compact set $\left\{ (\lambda_1, \ldots,\lambda_n) \,|\, \sum^n_{j=1}|\lambda_j|^2=1 \right\}$

At the end of the proof, Young argues that whenever $\rho(x)=1$ then $\lVert x\lVert\geq m>0$, and this implies "by homogeneity" that $\lVert x\lVert \geq m\rho(x)$, for all $x\in E$. Can someone explain this? I find it mystifying.

Thanks in advance!

msx
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  • Can you please explain better the contest? – Onil90 Feb 16 '16 at 22:01
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    Note that $\rho(c x)=|c|, \rho(x)$ for any number $c\in \mathbb C$. This follows directly from the definition of $\rho$.

    Let $x\in E$ and $x\neq 0$, then $\rho(\frac x{\rho(x)})=1$ and $|\frac x{\rho(x)}|≥m$, which is the infimum of the norm on the set for which $\rho(x)=1$. Since $|\cdot|$ is also homogenous, this imples $|x|≥m,\rho(x)$.

     – s.harp Feb 17 '16 at 15:55
  • Fantastic, thanks! – msx Feb 17 '16 at 15:56

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