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Denote supnorm by $|x| = \max\{|x_1|,...,|x_n|\}$ where $x \in \mathbb{R}^n$. How can we show that this norm and Euclidean norm satisfies the following inequality?

$$ |x| \leq ||x|| \leq \sqrt{n}|x|$$

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up vote 4 down vote accepted

From the definition, we have that $\vert x \vert = \vert x_k \vert$ for some $k \in \{1,2,3,\ldots,n\}$. Hence, $$\Vert x \Vert^2 = \sum_{j=1}^n \vert x_j \vert^2 = \sum_{j=1}^{k-1} \vert x_j \vert^2 + \vert x_k \vert^2 + \sum_{j=k+1}^n \vert x_j \vert^2 \geq \vert x_k \vert^2 = \vert x \vert^2$$ Further, since $\vert x_k \vert$ is the maximum, we have $\vert x_j \vert \leq \vert x_k \vert$ forall $j$. Hence, we have $$\Vert x \Vert^2 = \sum_{j=1}^n \vert x_j \vert^2 \leq \sum_{j=1}^n \vert x_k \vert^2 = n\vert x_k \vert^2 = n \vert x \vert^2$$ Hence, we get that $$\vert x \vert \leq \Vert x \Vert \leq \sqrt{n} \vert x \vert$$

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