Proving that inequalities $\|a\|_{\infty} \leq \|a\|_2 \leq \sqrt{n} \|a\|_{\infty}$ are true and sharp.

Question

For all $a=(a_1,\cdots,a_n)\in \mathbb{R}^n$ we consider

$$\|a\|_{\infty}= \displaystyle\max_{i\in \{1,\dots,n\}} |a_i|~~and~~~~\|a\|_2=\left(\displaystyle\sum_{i=1}^n|a_i|^2\right)^{1/2}.$$

Why we have $$\|a\|_{\infty} \leq \|a\|_2 \leq \sqrt{n} \|a\|_{\infty},\;\forall\,a=(a_1,\cdots,a_n)\in \mathbb{R}^n?$$ And why the following inequalities are sharp?

Answer 1

Let $i_0\in\{1,2,....n\}$ such that $|a_{i_0}|= \max\limits_{1\le i\le n}(|a_i|)$ then for all $ i\in\{1,2,....n\}$ we have $|a_{i}|\le |a_{i_0}|$

$$|a_{i_0}|^2\le \sum_{i=1}^n|a_i|^2 \le \sum_{i=1}^n|a_{i_0}|^2 = n|a_{i_0}|^2 $$

taking the square root you get $$\|a\|_{\infty} \leq \|a\|_2 \leq \sqrt{n} \|a\|_{\infty}$$

The inequalities are sharp indeed taking $a= (1,0,0\cdots0)$ and $b= (1,1,1\cdots1)$

we get $$\color{red}{\|a\|_{\infty} = \|a\|_2 = 1~~~~and ~~~ = \|b\|_2 = \sqrt{n}\|b\|_{\infty}=\sqrt{n}}$$

Answer 2

Because $$\sqrt{\sum_{k=1}^n|a_k|^2}\le\sqrt{n\max\limits_{k}|a_k|^2}=\sqrt{n}\max\limits_{k}|a_k|$$