real analysis - The $l^{infty}$-norm is equal to the limit of the $l^{p}$-norms.

If we are in a sequence space, then the $l^{p}$-norm of the sequence $\mathbf{x} = (x_{i})_{i \in \mathbb{N}}$ is $\displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p}$.

The $l^{\infty}$-norm of $\mathbf{x}$ is $\displaystyle \sup_{i \in \mathbb{N}} |x_{i}|$.

Prove that the limit of the $l^{p}$-norms is the $l^{\infty}$-norm.

I saw an answer for $L^{p}$-spaces, but I need one for $l^{p}$-spaces. Besides, I didn’t really understand the $L^{p}$-answer either.

Thanks for your help!

Answer

Let me state the result properly:

Let $x=(x_n)_{n \in \mathbb{N}} \in \ell^q$ for some $q \geq 1$. Then $$\|x\|_{\infty} = \lim_{p \to \infty} \|x\|_p. \tag{1}$$

Note that $(1)$ fails, in general, not hold if $x=(x_n)_{n \in \mathbb{N}} \notin \ell^q$ for all $q \geq 1$ (consider for instance $x_n := 1$ for all $n \in \mathbb{N}$.)

Proof of the result: Since $$|x_k| \leq \left(\sum_{j=1}^{\infty} |x_j|^p \right)^{\frac{1}{p}}=\|x\|_p$$ for all $k \in \mathbb{N}$, $p \geq 1$, we have $\|x\|_{\infty} \leq \|x\|_p$. Thus, in particular $$\|x\|_{\infty} \leq \liminf_{p \to \infty} \|x\|_p. \tag{1}$$

On the other hand, we know that $$\|x\|_p = \left( \sum_{j=1}^{\infty} |x_j|^{p-q} \cdot |x_j|^q \right)^{\frac{1}{p}} \leq \|x\|_{\infty}^{\frac{p-q}{p}} \cdot \left( \sum_{j=1}^{\infty} |x_j|^q \right)^{\frac{1}{p}} = \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}$$ for all $q

$$\limsup_{p \to \infty} \|x\|_p \leq \limsup_{p \to \infty} \left( \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}\right) = \|x\|_{\infty} \cdot 1. \tag{2}$$

Hence, $$\limsup_{p \to \infty} \|x\|_p \leq \|x\|_{\infty} \leq \liminf_{p \to \infty} \|x\|_p.$$ This shows that $\lim_{p \to \infty} \|x\|_p$ exists and equals $\|x\|_{\infty}$.