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How could I show that: $$\lVert x\rVert_{\infty}\leqslant\lVert x\rVert_2\leqslant\lVert x\rVert_\infty\;.$$ I was able to show the right side but troubling with the left-hand side. What would be the best way to approach it?

Can I just write this without proving? $$\lVert x\rVert_{\infty}=\max\limits_{1\leqslant i\leqslant n}\big\{x_i\big\}\leqslant \sqrt{\sum_{i=1}^nx_i^2}=\lVert x\rVert_2\;.$$

Angelo
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Daniel Fu
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    Well, both inequalities can't possibly hold, because then we'd be able to deduce $|x|_\infty = |x|_2$, which is often not true. You're probably missing a constant factor. – Misha Lavrov Oct 20 '23 at 04:00
  • Right hand inequality is not even true. How did you manage to prove it? – geetha290krm Oct 20 '23 at 04:57
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    Hint: Try to prove that $\lVert x \rVert_\infty \leq \lVert x \rVert_2 \leq \sqrt{n} \ \lVert x \rVert_\infty $. – M.B. Oct 20 '23 at 05:06

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Your definition of equivalent norms is incorrect. One says that two norms on a Banach space $X$ are equivalent if there exist constants $C,D > 0$ such that $$C\Vert x \Vert_1 \leq \Vert x \Vert_2 \leq D \Vert x \Vert_1$$ for all $x \in X$. I assume you are working in $X = \mathbb R^n$. In this case, all norms are equivalent and this result can be found in many places online (including this site).

SescoMath
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