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Let $1<p<\infty$ and $x=(x_1,x_2,\ldots,x_N)\in\mathbb{R}^N$, $N\geq 2$. Then does the following inequality holds: $$ (|x_1|^{2(p-1)}+\ldots+|x_N|^{2(p-1)})^\frac{1}{2}\leq(|x_1|^p+\ldots+|x_N|^p)^\frac{p-1}{p}. $$

I can only see it holds for $p=2$. Can anyone please help to prove or disprove it for $p\neq 2$. Thanks.

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2 Answers 2

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Let the $p$-norm $\|x\|_p$ of $x\in\mathbb{R}^N$ be defined as $\left(\sum_{n=1}^N |x_n|^p\right)^{1/p}$. Notice then that the left hand side of the inequality is $(\|x\|_{2(p-1)})^{p-1}$, and the right hand side is $(\|x\|_p)^{p-1}$. Since we take $p\gt 1$, this implies that we can remove the powers and the inequality will be the same inequality as the one for $\|x\|_{2(p-1)}$ and $\|x\|_p$. Now since $p\gt 1$, there is a fact that the higher order p-norm is lesser than or equal to the lower order $p$-norm (intuition: higher order $p$-norms define a “stricter” sense of distance). Proof: let $b\gt a\gt 1$. Now, think about the expansion of the multinomial, when all values in the brackets are positive: $(\|x\|_a)^b\ge\sum_{n=1}^N |x_n|^{a\cdot (b/a)}=(\|x\|_b)^b$ which shows that the $a$-norm of $x$, the lower order norm, was greater than or equal to than the $b$-norm of $x$ - key here is the fact that both $a,b\gt 1$. This proves your inequality for all $p\ge2$ since $2(p-1)\ge p$ if $p\ge2$, so the $p$-norm is a lower order norm than the $2(p-1)$-norm, therefore the $p$-norm of $x$ is greater than or equal to, therefore the right hand side is greater than or equal to the left hand side.

It is false for $p$ in the range $(1,2)$ since in those ranges $p\gt 2(p-1)$, and therefore the norm in the LHS is of lower order, not higher, making the left hand side larger.

For example, take $x=(0.5,0.5)\in\mathbb{R}^2$, with $p=1.5$. The LHS is $(0.5^1+0.5^1)^{1/2}=1$, but the RHS is $(0.5^{3/2}+0.5^{3/2})^{1/3}=0.89089871813\lt1\lt LHS$.

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In $p$-Norm notation, we have $\|x\|_{2(p-1)}^{p-1}\leq \|x\|_p^{p-1}$. This is equivalent to $\|x\|_{2(p-1)}\leq \|x\|_p$, which is true, because $2(p-1)\geq p$ for $p\geq 2$. This is true by the monotonicity of the $p$-norms.

We prove $\|x\|_p\leq \|x\|_r$ for $r<p$. W.l.o.g. suppose that $\|x\|_p=1$. Then $\|x\|_p^p=1$ and $|x_i|\le 1$ for all $i$. Since $p>r$, we obtain $|x_i|^p \le |x_i|^r$ for all $i$ and we get $$1=\|x\|_p^p = \sum_{i} |x_i|^p \le \sum_{i} |x_i|^r = \|x\|_r^r.$$ The proof of the monociticy is not my own proof. You can find it on several places in the internet.

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  • $\begingroup$ Amusing how we gave almost the same answer within two minutes of each other $\endgroup$
    – FShrike
    Commented Jul 26, 2021 at 10:16
  • $\begingroup$ Ha ha, yeah, i didn't realized, that there is another answer :D $\endgroup$
    – Jochen
    Commented Jul 26, 2021 at 10:20
  • $\begingroup$ @Jochen and Idiotic Shrike Thank you very much. If I understood correctly, the inequality in the question holds for $p\geq 2$ and it is false for $1<p<2$. $\endgroup$
    – Mathguide
    Commented Jul 26, 2021 at 10:21
  • $\begingroup$ Well, it is only provably true for all $x$ if $p\ge 2$. Let me see if it is provably false for all $1\lt p\lt 2$ $\endgroup$
    – FShrike
    Commented Jul 26, 2021 at 10:25

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