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Let $1 \leq p < q \leq \infty$ and $x \in\mathbb{R}^n$. Show that $\|x\|_q \leq \|x\|_p \leq n^\frac{1}{p}\|x\|_q$, where $\|x\|_p$ is the metric $\left(\sum_{j=1}^n{|x_j|^p}\right)^\frac{1}{p}$.

A hint is given: "For the left-hand inequality do first the case where $\|x\|_p = 1$, and for the right-hand inequality do first the case $\|x\|_q = 1$." So first I set $\|x\|_p = 1$ and got $\sum_{k = 1}^n|x_k|^q \leq 1 \leq n^\frac{q}{p}\sum_{k = 1}^n|x_k|^q$. This makes sense since $\frac{q}{p} > 1$.

I left this and used the other half of the hint; I considered the inequality again and set $\|x\|_q = 1$ to obtain $1 \leq \sum_{k = 1}^n|x_k|^p \leq n$. I'm not quite as sure what this inequality means.

My question is, how to extend these two inequalities to the more general cases where $\|x\|_p, \|x\|_q \neq 1$? And how can I make my two results relate to each other, since they seem to be two completely different cases? Perhaps this question is simpler than I am making it out to be and I might not be wording myself clearly but once I understand the general principle behind inequalities like this I will be able to do more complex ones on my own.

james h
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  • Looking at my second inequality, I can see that it naturally follows that $\left(\sum_{k=1}^n|x_k|^p\right)^\frac{1}{p} = |x|_p \leq n^\frac{1}{p}$. Which seems to be just a step or two away from what I need - what I'm confused about is how I can extend the inequality for a specific case to the general case. – james h Oct 28 '11 at 18:43

1 Answers1

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Note: The hint is that you should consider the special case $\lVert x\rVert_p = 1$ in order to prove that in that case, you have $$\lVert x\rVert_q \leq \lVert x\rVert_p,$$ ignoring the second inequality; and then, separately, that you consider the case $\lVert x\rVert_q=1$ to prove that in that case $$\lVert x\rVert_p \leq n^{1/p}\lVert x\rVert_q,$$ while ignoring the first inequality

First: How do you prove these special cases?

Assume first that $\lVert x\rVert_p = 1$, so that $$\sum_{j=1}^n|x_i|^p = 1.$$ That means that $0\leq |x_i|\leq 1$ for all $i$, and since $q\gt p$, then $$0\leq |x_i|\leq 1\Rightarrow 0\leq |x_i|^q \leq |x_i|^p \leq 1.$$ Hence $\sum |x_i|^q \leq \sum |x_i|^p =1$, and taking $q$th roots you conclude that $\lVert x \rVert_q \leq 1 = \lVert x\rVert_p$.

Try something along those lines for the second inequality, starting from the assumption that $\lVert x \rVert_q =1$.

Second: how do you extend the special case to the general case?

Given an arbitrary $x\neq \mathbf{0}$, let $\lambda=\frac{1}{\lVert x\rVert_p}$. If you already know the inequality when the $p$-norm is $1$, then you know that $$\lVert \lambda x \rVert_q \leq \lVert \lambda x \rVert_p,$$ since the $p$-norm of $\lambda x$ is $1$. But since $\lambda$ is a positive scalar, this is equivalent to $$\lambda\lVert x \rVert _q \leq \lambda \lVert x\rVert_p,$$ and cancelling $\lambda$ gives the desired inequality for arbitrary $x\neq\mathbf{0}$.

A similar trick works for the second inequality. And of course, the inequality trivially holds if $x=\mathbf{0}$.

Arturo Magidin
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  • This is very helpful. I can probably do the rest on my own. Thanks! – james h Oct 28 '11 at 19:03
  • I don't quite understand why you need that $||x||_p=1$ to prove the first inequality, I thougth that if $p \le q$ then $\frac pq \le 1$ so: $$\left( \sum_i |x_i|^q \right)^{p/qp} \le \left(\sum_i |x_i|^{q(p/q)} \right)^{1/p}$$ Hence: $||x||_q\le ||x||_p$
    I got this idea from http://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q however it's likely that I didn't understood well     – Ana Galois Aug 27 '13 at 22:30