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I was reading this: Proving an inequality with $\|x\|_p$ metrics?
Since it is a really old post, and I'm not sure that I'll get an answer, I hope asking this way, as an independent post, doesn't bother anyone. My question is about the answer that is given:

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 $$0\leq |x_i|\leq 1\implies 0\leq |x_i|^q \leq |x_i|^p \leq 1.$$ 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 $x=\mathbf{0}$.


And my comment/question is:
I don't quite understand why you need that $\|x\|_p=1$ to prove the first inequality, I thought that if $p \le q$ then $\frac pq \le 1$ so: $$\left( \sum_i |x_i|^q \right)^{(1/q)(p/p)} = \left( \sum_i |x_i|^q \right)^{p/(qp)} = \left( \left( \sum_i |x_i|^q \right)^{p/q} \right)^{1/p} \le \left(\sum_i |x_i|^{q(p/q)} \right)^{1/p}$$ Hence: $\|x\|_q\le \|x\|_p$
I got this idea from How do you show that $l_p \subset l_q$ for $p \leq q$? however it's likely that I didn't understood well.

Community
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Ana Galois
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2 Answers2

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I find it easier to use $\|x\|_\infty = \sup |x_i|$. It isn't hard to see that $\|x\|_\infty \le \|x\|_p$ for any $p > 0$. This leads to $$\|x\|_q^q = \sum_i |x_i|^q = \sum_i |x_i|^{q-p} |x_i|^p \le \|x\|_\infty^{q-p} \sum_i |x_i|^p \le \|x\|_p^{q-p} \sum_i |x_i|^p = \|x\|_p^{q-p} \|x\|_p^p = \|x||_p^q$$ provided that $q > p$. Now take the $q^{th}$ root on both sides.

Umberto P.
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Taking roots (power $<1$) pushes values towards $1$. Taking powers $>1$ pushes values away from $1$. In general if some of your coordinates are less than $1$ and some are larger than $1$, it becomes a mess. If $\|x\|_p = 1$, then all coordinates are less than $1$, so taking roots or powers will push all coordinates the same way. So, the idea is to take advantage of homogeneity (scaling) of the inequality to get this nice situtation where the proof becomes simple.

To add a little bit to address the question: While what you wrote is correct, using the inequality $\left( \sum_i |x_i| \right)^{\alpha} \leq \sum_i |x_i|^\alpha$, if you rearrange this inequality, you are really using that $\|x\|_1 \leq \|x\|_\alpha$ for $\alpha < 1$. Or if you like, replacing $|x_i|$ with $|x_i|^{1/\alpha}$, $\|x\|_{1/\alpha} \leq \|x\|_1$. So in some sense, the inequality you use is a specific case of what you proved. I'd say that's a bit too circular.

Evan
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