13

I'm having difficulty with the following problem. Any help would be appreciated.

Problem: Consider the sequence spaces $l_p$ with the usual norm. If $1\le p\le q\le \infty$, I want to show the following inequality for any sequence $a$.

$$\|a\|_q\le \|a\|_p$$

If we restrict to $\mathbb{R}^n$ but still use the $l_p$ norms, I also want to show this: $$\|a\|_q\le \|a\|_p\le n^{\frac{1}{p}-\frac{1}{q}}\|a\|_q$$

Work so far: I strongly suspect that a clever application of Hölder is needed here, but I tried the following for the first inequality:

First, we consider the case where a finite number of elements in the sequence are nonzero. We want to prove

$$||x||_q\le ||x||_p \Leftrightarrow \left(\sum_1^n |x_j|^q\right)^{\frac{1}{q}} \le \left(\sum_1^n |x_j|^p\right)^{\frac{1}{p}}.$$

We induct on $n$. The base case is clear. Because we can multiply all of the variables by a constant without affecting the inequality, we assume $x_n=1$. Assume we have proven the inequality for $n-1$. Then

$$\left(\sum_1^{n-1} |x_j|^q\right) \le \left(\sum_1^{n-1} |x_j|^p\right)^{\frac{q}{p}}$$

It suffices to show that

$$\left(\sum_1^{n-1} |x_j|^q\right) + 1 \le \left(\sum_1^{n-1} |x_j|^p+1\right)^{\frac{q}{p}}$$

This is equivalent to

$$\left(\sum_1^{n-1} |x_j|^q\right)\le \left(\sum_1^{n-1} |x_j|^p+1\right)^{\frac{q}{p}}-1$$

So we need to show that if $f(x)=x^{q/p}$, then $f(x+1)\ge f(x)+1$. But this is clear, as $q\ge p$. Now I think it should be an easy matter to pass to the $l_p$ spaces by taking limits.

I'm not sure what to do about the second inequality yet.

Martin Sleziak
  • 53,687
Potato
  • 40,171
  • A suggestion for the first inequality: assume without loss off generality that $|x|_p = 1$ and use that for $p \leq q$ and $|t| \leq 1$ we have $|t|^q \leq |t|^p$. For the second inequality, you want to assume that $a$ only has $n$ non-zero entries, otherwise it is wrong. If you make this assumption, consider the sequence having $n$ entries $1$ and apply Hölder to a modification of $a$, as you suspected. – t.b. Jun 01 '12 at 09:37
  • Is there a way to use Hölder to get the first inequality, the one that shows the $p$ norm dominates the $q$ norm? – Potato Jun 01 '12 at 19:45

1 Answers1

14

For any $x,y\in\mathbb{R}^n,$ let us define $x\ast y=(x_iy_i)_{i=1,\ldots,n}\in\mathbb{R}^n.$ For any $p,q,r\in[1,\infty]$ such that $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$, we have a generalization of Hoelder inequality $$||x\ast y||_r\leq ||x||_p||y||_q\tag{*}.$$

By applying (*) taking $y=(1,\ldots,1),$ we get $$||x||_r\leq n^{\frac{1}{r}-\frac{1}{p}}||x||_p.$$

Edit About the first inequality $||x||_p\geq ||x||_q,\textrm{ when }1\leq p\leq q\leq\infty.$ Apart from the trivial case $q=\infty,$ a possible derivation is as follows $$||x||_p^q=\left(\Sigma_{i}|x_i|^p\right)^{q/p}\geq \Sigma_{i}|x_i|^q=||x||_q^q.$$

Here we have used the majoration $\left(\Sigma_{i}|x_i|^p\right)^{q/p}\geq \Sigma_{i}|x_i|^q$ which is justifed by the remark that, for any $\alpha\in [1,\infty[,$ the function $f(t)=(1+t)^\alpha- 1-t^\alpha$ is nonnegative.

agt
  • 4,772
  • 2
    The generalization is obtained by using Hoelder inequality for the pair of conjugate exponents $(\frac{p}{r},\frac{q}{r}).$ – agt Jun 01 '12 at 10:22
  • @Potato It is not possible because $p,q,r\in[1,\infty]$ and $\frac{1}{p}+\frac{1}{q}=\frac{1}{r},$ so $\frac{r}{p},$ $\frac{r}{q}$ are non-negative and their sum is $1$, consequently they have to belong to $[0,1].$ Now where are they reciprocals? in $[1,\infty].$ – agt Jun 01 '12 at 10:31
  • Do you see a way to use Hölder to prove the first inequality? I am curious if this is possible. – Potato Jun 01 '12 at 19:46
  • @Potato I don't know if it is possible, but I have reported an alternative derivation. – agt Jun 02 '12 at 17:29