So I've already proven why $\left\lVert x\right\rVert_2\geq \left\lVert x\right\rVert_\infty$. I'm having trouble proving that $\sqrt{m}{\left\lVert x\right\rVert_\infty}\geq \left\lVert x\right\rVert_2$.

I've tried looking at the individual elements, but I'm not getting anywhere. I think I may have to use Holder's inequality, but I'm not sure if that's applicable, or how I would use it.

How should I do this?


I assume you are talking about norms on $\mathbb{R}^m$. Then, for $x\in\mathbb{R}^m$, $$ \lVert x\rVert_2^2 = \sum_{i=1}^m x_i^2 \leq m\cdot\max_{1\leq i\leq m} x_i^2 = m(\max_{1\leq i\leq m} |x_i|)^2 = m\lVert x\rVert_\infty^2 $$ so that $$ \lVert x\rVert_2 \leq \sqrt{m}\lVert x\rVert_\infty $$ as claimed.

  • $\begingroup$ could you please define what is $||x||_2$ and $||x||_{\infty}$ $\endgroup$ – Marso May 15 '13 at 12:09
  • 2
    $\begingroup$ For a vector $x\in\mathbb{R}^n$, $\lVert x\rVert_2=\sqrt{\sum_{i=1}^n x_i^2} = \langle x,x\rangle$ and $\lVert x\rVert_\infty=\displaystyle\max_{1\leq i\leq n} |x_i|$. $\endgroup$ – Clement C. May 15 '13 at 12:14

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