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How does one show that for $1\leq p<q<\infty$, and $x_i\geq 0$, $(\sum_{i=1}^n x_i^p)^{1/p}\leq n^{1/p-1/q}(\sum_{i=1}^n x_i^q)^{1/q}$?

cyc
  • 2,973

2 Answers2

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The inequality is equivalent to $$\left(\frac{1}{n}\sum_{i = 1}^n x_i^p\right)^{1/p} \leq \left(\frac{1}{n}\sum_{i = 1}^n x_i^q\right)^{1/q}. \tag{1}$$ If we define a random variable $X$ by $P(X = x_i) = 1/n$, for $i = 1, 2, \ldots, n$, then $(1)$ reads $$[E(X^p)]^{1/p} \leq [E(X^q)]^{1/q}. \tag{2}$$ It is well known that when $1 \leq p < q$, the $L_p$ norm of $X$ is not greater than the $L_q$ norm of $X$, hence the inequality follows.

Details: For a nonnegative random variable $X$, let $r = \dfrac{q}{p} > 1$, by H$ \ddot{\text{o}}$lder's inequality $$E(X^p) = E(X^p\cdot 1) \leq \{E[(X^p)^r]\}^{1/r} \times 1 = [E(X^q)]^{p/q}.$$ Rearrangement gives $(2)$.

Zhanxiong
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Hint: Use Jensen's inequality.

$f(x)=x^{q/p}$ is convex when $p<q$.

Let me know if you need more information.

Sisi
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