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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?

davyjones
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homegrown
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1 Answers1

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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

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

J.R.
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  • Why is it that we are allowed to assume that $|x|_{p_0} = 1$? – homegrown Feb 09 '14 at 16:34
  • Homogeneity. First we prove it under the assumption $|x|{p_0}=1$, then we use that to prove it for general $x$. This works because any general $x$ can be written as $|x|{p_0} \cdot \frac{x}{|x|_{p_0}}$ and norms are homogenuous, i.e. for any constant $\lambda$ we have $|\lambda x|=|\lambda|\cdot |x|$. – J.R. Feb 09 '14 at 16:43
  • Ok. I'm still a little fuzzy about you showed that $x\in l^\infty$. Is it just because you showed that $\Sigma_n |x_n|^{p_0}$ is bounded? – homegrown Feb 09 '14 at 16:58
  • Yes. If $|x_n|^{p_0}$ is bounded, then also $|x_n|$ is bounded, so $|x_n|\le C$. That means $|x|_\infty=\sup_n |x_n|\le C$. – J.R. Feb 09 '14 at 17:03
  • Another thing: It was already given in the problem that $x\in l^p$. Why did we have to prove that? – homegrown Feb 09 '14 at 17:13
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    It was not clear in the problem statement whether $x\in\ell^p$ for some chosen $p$ or for all $p$, i.e. $x\in \bigcap_p \ell^p$. So I just assumed the weaker of both. – J.R. Feb 09 '14 at 17:17
  • OK. That makes sense. Thank you for being precise with that. – homegrown Feb 09 '14 at 17:23