Proof: $\|X\|_2\leq \sqrt{n}\|X\|_\infty$

Question

(Euclidean Norm) Proof that $\|X\|_2\leq \sqrt{n}\|X\|_\infty$, if $x \in \mathbb{R^n}$.

With that, proof $\|X\|_\infty \to 0 \implies \|X\|_2\to 0$

I'm stuck:

$\|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{n}\sqrt{x^2_1+x^2_2+...+x^2_n}$

I don't know how to use the fact: $\|X\|_\infty=\max(|x_1|,|x_2|,...,|x_n|)$

Answer 1

Hint: $a_1^2+a_2^2+\ldots+a_k^2\leq k\cdot\max(a_1,a_2,\ldots,a_k)^2$.

Edit: this is valid for positive $a_i$, if you want it valid for all $a_i\in\mathbb{R}$ then you only have to add some absolute values: $a_1^2+a_2^2+\ldots+a_k^2\leq k\cdot\max(|a_1|,|a_2|,\ldots,|a_k|)^2$

Answer 2

Hint: Try using the fact that $x_i^2 \leq \| X \|_\infty^2$ in the definition of $\| X \|_2$

Answer 3

Let $a=\max(|x_1|,|x_2|,\ldots, |x_n|)$ $$ \|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{a^2+a^2+...+a^2}=\sqrt{n}a $$