Proving an inequality with $\|x\|_p$ metrics?

Solution 1:

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 $x=\mathbf{0}$.