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Let $1 \le p \le q \le \infty$ and $\Vert x \Vert_p = \big( \sum\limits_{i=1}^\infty \vert x \vert^p \big)^{1/p}$ where $x = (x_1, x_2, \cdots, x_k, \cdots )$. $l^p = \{x: \Vert x \Vert_p < \infty\}$.

Then $\Vert x \Vert_q \le \Vert x \Vert_p$ for $x \in l^p$.

I can show $l^p \subset l^q$, but I am a little struggling to show this. Any help would be appreciated!

withgrace1040
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    First prove this when $|x|_p=1$. For the general case consider $\frac x {|x|}$. – Kavi Rama Murthy Feb 24 '20 at 09:06
  • @KaviRamaMurthy Oh, you mean consider $\frac{x}{\Vert x \Vert_p}? Yes, it is easy to show. Thank you very much. I don't know why I cannot come up with this kind of technique. Do you have any tips for coming up with this idea? – withgrace1040 Feb 24 '20 at 09:14
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    @MaximilianJanisch Yes, thank you very much for referring me to that post! – withgrace1040 Feb 24 '20 at 09:16
  • If you can show $\ell^p\subset \ell^q$ you should have no problem in showing the inequality in the box. The inclusion amounts to proving $|x|_q\le C|x|_p$ for some $C>0$. Now you need to prove that $C=1$. – Giuseppe Negro Feb 24 '20 at 10:18
  • The way I interpret this question, it is not a duplicate. Here we need to prove that the constant in the inequality $|x|_q\le C|x|_p$ is $C=1$. In the other question, we only need to prove the embedding $\ell^p\subset \ell^q$, which amounts to proving the previous inequality without worrying about the value of $C$. – Giuseppe Negro Feb 24 '20 at 13:22

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You mentioned that you can show $\ell^p\subset \ell^q$ for $p<q$. This means that you have no problem showing that there is a $C>0$ such that $$\tag{1} \lVert x\rVert_{\ell^q}\le C\lVert x\rVert_{\ell^p}, \qquad \forall x\in \ell^p.$$ Now you need to prove that, indeed, the constant $C$ can be taken to be $1$. This is a classical case in which you can use the tensor power trick, as Terry Tao calls it.

WARNING. What follows is not the simplest way of proving what you want. But I think that it is an interesting application of a not so well-known trick.

For $x\in \ell^p$, define the tensor power of $x$ with itself to be the matrix $$ x\otimes x := \begin{bmatrix} x_1^2 & x_1x_2 & x_1x_3 & \ldots \\ x_1x_2 & x_2^2 & x_2x_3 & \ldots \\ x_3x_1 & x_3x_2 & x_3^2 & \ldots\\ \ldots & \ldots & \ldots & \ldots \end{bmatrix}.$$ Define the $\ell^p$ norm of a matrix as $$ \lVert A\rVert_{\ell^p}:= \left(\sum_{ij} |a_{ij}|^p\right)^\frac1p,\quad \text{ where }A=\begin{bmatrix} a_{ij}\end{bmatrix}_{i, j},$$ and define the $\ell^q$ norm in the obvious way.

There are now two key observations to be made.

The first key observation is that $$\tag{i} \lVert x\otimes x\rVert_{\ell^p}=\lVert x\rVert_{\ell^p}^2\quad \text{ and }\quad \lVert x\otimes x\rVert_{\ell^q}=\lVert x\rVert_{\ell^q}^2.$$ The second key observation is that, for any matrix $A$, $$\tag{ii} \lVert A\rVert_{\ell^q}\le C\lVert A\rVert_{\ell^p}, $$ where $C$ is the same constant that appears in (1).

Now we can conclude our proof that $C$ can be $1$. Let us apply (ii) with $x\otimes x$ in place of $A$. By (i), $$ \lVert x\rVert_{\ell^q}\le C^{1/2} \lVert x\rVert_{\ell^p},$$ which means that (1) still holds if $C$ is replaced by its square root. Iterating, we see that (1) still holds if $C$ is replaced by $C^{\tfrac1{2^n}}$ for any $n\in\mathbb N$. Letting $n\to \infty$ we conclude that $C$ can be replaced by $$ \lim_{n\to \infty} C^{\tfrac{1}{2^n}}=1.\quad \Box$$

Giuseppe Negro
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