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I apologize for the poor title.

Let $p\in 2\mathbb N$ and $a_1,\ldots , a_n\geq 0$.

Show that $$\left (\sum a_j\right )^p \leq n^{p-1}\sum a_j^p $$

So, firstly, by rearranging the expression, it's sufficient to show that $$\left (\sum\frac{a_j}{n}\right )^p \leq \sum\frac{a_j^p}{n}. $$ It's known from theory of integrals that for any convex function $g:\mathbb R\to\mathbb R$ $$g(EX)\leq E g(X) \quad\mbox{the Jensen inequality}$$ Suppose $(\Omega,\mathcal F,\mathbb P)$ is a probability space with $X_1,\ldots, X_n$ some $\mathcal F$-measurable functions such that $EX_j \equiv a_j$. Since $p\in2\mathbb N$, among others, the mapping $x\mapsto x^p$ is convex. Define $$X := \sum \frac{X_j}{n}, $$ The inequality yields $$g(EX) = g\left (\int\sum\frac{X_j}{n}\mbox{d}\mathbb P\right ) = \left (\sum\frac{a_j}{n}\right )^p\leq \int\left (\sum\frac{X_j}{n}\right )^p\mbox{d}\mathbb P = Eg(X) $$ ..and I don't see any way to make progress or I'm missing something very trivial or is the choice of $g$ or $X$ a poor one? Are there any suggestions?

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Alternatively, you can also use Holder's inequality. It states that, if $q=\frac{p}{p-1}$, then

\begin{align*} \|f g\|_{1} &\leq \|f\|_{p}\|g\|_{q} \\ \|f g \|_{1}^{p} &\leq \|f\|_{p}^{p} \|g\|_{q}^{p} \end{align*}

By comparing the expressions in the problem, we note that we want $f_{i} \cdot g_{i} = a_{i}$. In this case,

\begin{align*} \|f g\|_{1}^{p} &= \left(\sum_{i=1}^{n}{|a_i|}\right)^{p} \\ &= \left(\sum_{i=1}^{n}{a_i}\right)^{p} & a_i \geq 0 \end{align*}

Next, if we take $g_{i}=1$, obtain \begin{align*} \|g\|_{q}^{p} &= \left(\sum_{i=1}^{n}{1^{q}}\right)^{\frac{p}{q}} \\ &= n^{\frac{p}{q}} = n^{p-1} & q = \frac{p}{p-1} \end{align*}

Finally, since $f_ig_i=a_i$ and $g_i=1$, obtain $f_i=a_i$. That is, \begin{align*} \|f\|_{p}^{p} &= \sum_{i=1}^{n}{a_i^p} \end{align*} Finally, plugging in the expressions, into Holder's inequality \begin{align*} \|f g \|_{1}^{p} &\leq \|f\|_{p}^{p} \|g\|_{q}^{p} \\ \left(\sum_{i=1}^{n}{a_i}\right)^{p} &\leq n^{p-1} \sum_{i=1}^{n}{a_i^p} \end{align*}

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  • $\begingroup$ Oh wow, I like this one even more :) $\endgroup$ – Alvin Lepik Nov 7 '17 at 20:28
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Although your line of reasoning is good, the choice of $X$ is poor. A common trick to transform empirical means into expectations is to take the empirical distribution. For this, let $\Omega=\{a_1,\ldots,a_{n}\}$ be such that $P(\{a_i\})=n^{-1}$ and $X(\omega)=\omega$. We say that $X$ follows the empirical distribution because it takes with probability $n^{-1}$ each value in the sample.

Next, let $g(x)=x^{ p}$, for $p \geq 1$. It follows from your reasoning that

\begin{align*} \left(\sum_{i=1}^{n}{a_i \cdot n^{-1}}\right)^{p} &= g(E[X]) \\ &\leq E[g(X)] = \sum_{i=1}^{n}{a_{i}^{p} \cdot n^{-1}} \end{align*}

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  • $\begingroup$ PS: There seems to be a side case when $p=0$. You can solve this one by hand, since it simplifies to $1 \leq 1$. $\endgroup$ – madprob Nov 7 '17 at 19:59
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    $\begingroup$ Yes, I understand that. I didn't consider $E[g(X)]^p \leq E[g(X)^p]$, though. Thanks for your help! $\endgroup$ – Alvin Lepik Nov 7 '17 at 20:01
  • $\begingroup$ Ops, that was a typo! I fixed now. $\endgroup$ – madprob Nov 7 '17 at 20:08

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