Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$. I'm trying to prove the discrete version of this result, i.e.,
Let $x= (x_1, \ldots, x_m) \in \mathbb K^m$ and $\lambda=(\lambda_1, \ldots, \lambda_m) \in \mathbb R_{\ge 0}^m$. Then $$ \lim_{p \to \infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} = \max _{1 \leq i\leq m} |x_{i}|. $$
Could you confirm if my below attempt is correct?
Proof WLOG, we assume $x_1 = \max _{1 \leq i\leq m} |x_{i}|$. Then $$ (\lambda_1 |x_1|^p)^{1/p} \le \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p}. $$
We take the limit $p \to \infty$ and get $$ \begin{align} \liminf_{p\to\infty} (\lambda_1 |x_1|^p)^{1/p} &\le \liminf_{p\to\infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \liminf_{p\to\infty} \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p} \\ \limsup_{p\to\infty} (\lambda_1 |x_1|^p)^{1/p} &\le \limsup_{p\to\infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \limsup_{p\to\infty} \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p}. \end{align} $$
The claim then follows by noticing that
$$ \lim_{p\to \infty} (\lambda_1)^{1/p} =\lim_{p \to \infty}\bigg (\sum_{i=1}^m \lambda_i \bigg )^{1/p}=0. $$
This completes the proof.