I'm studying the norm $ℓ_∞$. It leads me to prove the following inequality

$$\left( \sum_{i=1}^m (x_{i})^{p} \right)^{q} \ge \left( \sum_{i=1}^m (x_{i})^{q} \right)^{p}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad p,q \in \mathbb R^+, \quad p \le q$$

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

Lemma: $x + y \ge (x^r + y^r)^{1/r}, \quad (x,y) \in {(\mathbb R^+)}^2, \quad 1 \le r \in \mathbb R$

It is easy to check that our desired inequality is equivalent to

$$\sum_{i=1}^m x_{i} \ge \left( \sum_{i=1}^m (x_{i})^{r} \right)^{1/r}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad 1 \le r \in \mathbb R$$

We proceed to prove this one by induction on $m$. Clearly, it holds for $m=1$. Assume that it holds for some $m \ge 1$. Then we have

$$\begin{aligned} \sum_{i=1}^{m+1} x_{i} &= \left (\sum_{i=1}^{m} x_{i} \right ) + x_{m+1} \\ &\ge \left( \sum_{i=1}^m (x_{i})^{r} \right)^{1/r} + x_{m+1} \quad \text{by inductive hypothesis} \\ &\ge \left (\sum_{i=1}^m (x_{i})^{r} + (x_{m+1})^r \right)^{1/r} \quad \text{by Lemma} \\ &= \left( \sum_{i=1}^{m+1} (x_{i})^{r} \right)^{1/r} \end{aligned}$$

This completes the proof.

  • 1
    $\begingroup$ It's correct of course. $\endgroup$ – Jakobian Aug 8 at 12:48
  • $\begingroup$ Thank you so much for your verification @Jakobian :) $\endgroup$ – Abstract Analysis Aug 8 at 13:15

I like to use Karamata here.

Indeed, let $\frac{q}{p}=r$, $x_i^{p}=a_i$ and $f(x)=x^{r}.$

Thus, $f$ is a convex function and we need to prove that $$f\left(\sum_{i=1}^na_i\right)\geq\sum_{i=1}^nf\left(a_i\right)$$ or $$f\left(\sum_{i=1}^na_i\right)+f(0)+...+f(0)\geq\sum_{i=1}^nf\left(a_i\right),$$ which is true by Karamata because $$(a_1+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n),$$ where $a_1\geq a_2\geq...\geq a_n.$


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