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Let $|x|$ be the standard Euclidean norm, $|x|_1 = |x_1|+\dots + |x_n|$ and $|x|_\infty = \max\{|x_1|,\dots,|x_n|\}$. Show that $|x|_\infty \leqslant |x| \leqslant|x|_1 \leqslant n|x|_\infty.$

For $|x| \leqslant|x|_1$ if I write $x = \sum_{i=1}^n x_ie_i$ I have that

$|\sum_{i=1}^n x_ie_i| \leqslant \sum_{i=1}^n |x_i||e_i| = \sum_{i=1}^n |x_i| = |x|_1.$

But I'm not sure how to show the first inequalities with $|x|_\infty$... What should I do with these?

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Just consider $$ \lvert x \rvert_{\infty} = \max_{i = 1, ..., n} \lvert x_i \rvert = \lvert x_j \rvert $$ for an appropriate $j \in \lbrace 1, ..., n \rbrace$. Now, because $x \mapsto \sqrt{x}$ is increasing: $$ \lvert x_j \rvert = \sqrt{\lvert x_j \rvert^2} \leq \sqrt{\sum_{i = 1}^n \lvert x_i \rvert^2 } = \lvert x \rvert $$ The inequality $\lvert x \rvert \leq \lvert x \rvert_1$ follows immediateley from $\sqrt{a+b} \leq \sqrt{a}+\sqrt{b}$ for all $a, b \in \mathbb{R}$ (you should know this but in case you do not one can prove it by squaring both sides). Note that $$ \lvert x \rvert_1 = \sum_{i = 1}^n \lvert x_i \rvert \leq \sum_{i= 1}^n \left(\max_{k = 1, ..., n} \lvert x_k \rvert\right) = n \cdot \max_{k = 1, ..., n} \lvert x_k \rvert = n \lvert x \rvert_{\infty}. $$

Hyperbolic PDE friend
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  • What I don't see is how does $\sqrt{\lvert x_j \rvert^2} \leq \sqrt{\sum_{i = 1}^n \lvert x_i \rvert^2 }$ hold? –  Oct 06 '20 at 15:16
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    Because $\lvert x_j \rvert^2 \leq \displaystyle \sum_{i = 1}^n \lvert x_i \rvert^2$. – Hyperbolic PDE friend Oct 06 '20 at 15:19
  • $|x_j|$ is now the maximum of $x_1, \dots x_i$? I'm a bit confused about the indexes here... –  Oct 06 '20 at 15:21
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    It is the maximum of $x_1, ..., x_n$ but I think you mean the correct thing – Hyperbolic PDE friend Oct 06 '20 at 15:22
  • I see, however if it is the maximum why do you define $j \in \lbrace 1, ..., n \rbrace$? I guess there's only one maximum? –  Oct 06 '20 at 15:23
  • There could be more than one maximum, but because $\lbrace 1, ..., n \rbrace$ is finite, the maximum has to be attained at at least one $j \in \lbrace 1, ..., n \rbrace$. I mean it is obvious: If you have $n$ real numbers, at least one of it will be at least as great as all the other ones. – Hyperbolic PDE friend Oct 06 '20 at 15:25
  • Oh now that becomes obvious. So there's no way that the maximum $|x_j|^2$ could be larger than the sum of all $|x_i|^2$? –  Oct 06 '20 at 16:39
  • Of course. $\lvert x_j \rvert^2$ is among the all the $\lvert x_i \rvert^2$... – Hyperbolic PDE friend Oct 06 '20 at 17:56