real analysis - Limit of $|x|_p$ as $prightarrowinfty$

I am not sure how to start this hw problem. Here it is:

Let $x$ be a given vector in $\ell^p$ with $1\le p\lt\infty$. Show that $x \in \ell^\infty$ and
$$\lim_{p \to \infty} \|x\|_p = \|x\|_\infty.$$ This seems like it would be obvious, but trying to prove it is different. I am not sure how to even start this one. Thanks.

EDIT: Would the triangle inequality be useful here?

Answer

Let $x\in\ell^{p_0}$ for some $1\le p_0<\infty$. In order for that limit to even have a chance to converge to a finite limit, we first need to prove $x\in \ell^{p}$ for all $p\ge p_0$.

Claim: $\|x\|_p\le \|x\|_{p_0}$ for all $p_0\le p\le\infty$.

Proof: First consider the case $p<\infty$. Let us assume $\|x\|_{p_0}=1$. It follows that $|x_n|\le 1$ for all $n$. Now

$$\|x\|_p^p =\sum_n |x_n|^p \le \sum_n |x_n|^{p_0}=1$$

so $\|x\|_p\le 1=\|x\|_{p_0}$. The claim for all $x\not=0$ follows by exploiting homogeneity of the inequality. Set $\lambda=\|x\|_{p_0}$. Then $\|x/\lambda\|_{p_0}=1$. Therefore

$$\|x\|_p=\lambda \left\|\frac{x}{\lambda}\right\|_p\le \lambda=\|x\|_{p_0}$$

Now we do the case $p=\infty$. By assumption, the sum $\sum_n |x_n|^{p_0}$ converges. Therefore $|x_n|^{p_0}\rightarrow 0$ as $n\rightarrow\infty$ and $|x_n|$ must be bounded. $\square$

This shows in particular $x\in \ell^{p}$ for all $p\ge p_0$ (including $p=\infty$). Now we show

$$\lim_{p\rightarrow\infty}\|x\|_p = \|x\|_\infty$$

Again, first assume $\|x\|_\infty=1$. Then $|x_n|\le 1$ for all $n$. Therefore

$$\limsup_{p\rightarrow\infty} \|x\|_p \le 1\tag{1}$$

On the other hand $1=\|x\|_\infty=\sup_n |x_n|$ implies that for every $\epsilon>0$ there is an $n_0$ such that $|x_{n_0}|>1-\epsilon$. Thus

$$\|x\|_p = \left(\sum_{n} |x_n|^p\right)^{1/p} \ge |x_{n_0}|>1-\epsilon$$

Letting $\epsilon\rightarrow 0$ gives

$$\liminf_{p\rightarrow\infty} \|x\|_p \ge 1\tag{2}$$

Now $(1)$ and $(2)$ imply together that $\lim_{p\rightarrow\infty} \|x\|_p$ exists and equals $1$.

In general we conclude again by homogeneity as follows: write $\lambda=\|x\|_\infty\not=0$. Then $\|x\|_p=\lambda \|x/\lambda\|_p$ and since $\|x/\lambda\|_\infty=1$ the above gives

$$\lim_{p\rightarrow\infty} \|x\|_p = \lambda \lim_{p\rightarrow\infty} \left\|\frac{x}{\lambda}\right\|_p = \lambda = \|x\|_\infty$$