$l^1 \subseteq l^2 \subseteq l^\infty$ proof

Question

I am trying to prove $l^1 \subseteq l^2 \subseteq l^\infty$. I was able to prove $$\Vert x\Vert_\infty \le \Vert x\Vert _2 \le \Vert x\Vert_1$$ in finite-dimensional space $\mathbb R^n$. However, I have been struggling to extend the above inequalities into the infinite-dimensional space $\mathbb R^{\mathbb N}$. Can anyone help me? Thank you!

Answer 1

If $(x_n)\in \ell_1$, there is an integer $N$ such that $n\ge N\Rightarrow |x_n|<1$. Then, $\sum^{\infty}_{n=N}|x_n|^2<\sum^{\infty}_{n=N}|x_n|<\infty$, which gives the first inclusion. And the same reasoning shows that if $(x_n)\in \ell_2$ then $\sup_{n\in\mathbb N}(|x_n|)$ is bounded, from which the second inclusion follows.

Let $x=(x_n)$ and $\epsilon>0$. For the first norm inequality, find an integer $N$ such that $\|x\|_{\infty}<|x_N|-\epsilon.$ Then, $$\|x\|_{\infty}<|x_N|-\epsilon\le \sqrt{\sum^{\infty}_{n=1}|x_n|^2}-\epsilon=\|x\|_2-\epsilon$$ and the result follows.

For the second inequality, use the fact that $f(t)=\sqrt t$ is concave and $f(0)=0$, so that that for fixed $N$, hence for all integers $n$,

$$\|\ x \|_2 = \sqrt{ \sum_{n=1}^{N} | x_n |^{2} } \leq \sum_{n=1}^{N} \sqrt{ | x_ n |^{2} } = \sum_{n=1}^{N} |x_n| = \|\ x \|_1$$

Answer 2

Here is a proof of the bounds. Consider

$$ p_n := \frac{|x_n| }{\sum_{n=1}^\infty |x_n|} $$

which defines a discrete probability: $0\le p_n \le 1$ and $\sum_n p_n =1$. Clearly

$$ p_n^2 \le p_n $$

summing both sides we get

$$\parallel x \parallel_2 \, \le \, \parallel x \parallel_1$$

For the other inequality use

$$ \sup_n |x_n|^2 \le \sum_{n=1}^\infty |x_n|^2 $$

but $\sup_n |x_n|^2 = \left ( \sup_n |x_n| \right )^2$. Hence we get

$$ \parallel x \parallel_\infty \, \le \, \parallel x \parallel_2 $$