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?
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$\begingroup$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.
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