The $|\cdot|_{p}$ norm will become the maximum norm when $p \to \infty$ I'm trying to prove the $|\cdot|_{p}$ norm will become the maximum norm when $p \to \infty$.

Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$, and $x= (x_1, \ldots, x_m) \in \mathbb K^m$. Then $$\lim_{p \to \infty} \left ( \sum_{i=1}^m |x_i|^p \right )^{1/p} = \max _{1 \leq i\leq m} |x_{i}|$$

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:
It suffices to prove the statement in case $x \in \mathbb {(R^+)}^{m}$, where it becomes $$\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} = \max _{1 \leq i\leq m} x_{i}$$
Let $l:= \max _{1 \leq i\leq m} x_{i}$. We have $$l = (l^p)^{1/p} \le \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \le (ml^p)^{1/p} = m^{1/p}l$$
Then $$l = \lim_{p \to \infty} l \le \lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \le \lim_{p \to \infty} (m^{1/p}l) = l$$
and thus by squeeze theorem $$\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} = l$$
This completes the proof.
 A: On the basis of @mathworker21's suggestion, I added a proof that $\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p}$ exists here. It would be great if someone helps me verify my attempt.

My attempt:

Lemma: $$\left( \sum_{i=1}^m (x_{i})^{p} \right)^{q} \ge \left( \sum_{i=1}^m (x_{i})^{q} \right)^{p}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad p,q \in \mathbb R^+, \quad p \le q$$

$$\text{Let } l:= \max _{1 \leq i\leq m} x_{i}.\ \text{Then } \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \ge (l^p)^{1/p} = l.\ \text{As such, the sequence}$$
$$\left \langle \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \right \rangle_{p \in \mathbb N}$$ is bounded from below. Next we prove that this sequence is decreasing by showing $$\left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \ge \left ( \sum_{i=1}^m (x_i)^{p+1} \right )^{1/(p+1)}$$ or equivalently $$\left ( \sum_{i=1}^m (x_i)^p \right )^{p+1} \ge \left ( \sum_{i=1}^m (x_i)^{p+1} \right )^{p}$$, which is true by our Lemma. This completes the proof.
