Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let define $(a_n)_{n\geq1}$ as real series. Prove, that

$$ \sum_{n=1}^{\infty}\left|\frac{a_{1}+\cdots+a_{n}}{n}\right|^{p}\leq\left(\frac{p}{p-1}\right)^{p}\sum_{n=1}^{\infty}|a_{n}|^{p} $$

(*) Extended level question - is the constant $\left(\frac{p}{p-1}\right)^p$ optimal?

I've tried induction methods, getting the logarithm of each sides, but it seems to be not working...

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

The following argument can be found in Problems in Real Analysis: Advanced Calculus on the Real Axis.

Due to Mond and Pecaric: $$ \left[ \frac{1}{n} \sum_{k=1}^n \left( \frac{a_1+\cdots+a_k}{k} \right)^p \right]^{1/p} \leq \frac{1}{n} \sum_{k=1}^n \left( \frac{1}{k} \sum_{i=1}^k a_i^p\right)^p, $$

with equality if and only if $a_1= \cdots a_n$. The above inequality can be written, equivalently $$ \tag 1 \sum_{k=1}^n \left( \frac{a_1+\cdots+a_k}{k} \right)^p \leq n^{1-p} \left[ \sum_{k=1}^n \left( \frac{1}{k} \sum_{i=1}^k a_i^p\right)^{1/p} \right]^p. $$

One may also prove the inequality: $$ \tag 2 \sum_{k=1}^n \left( \frac{1}{k} \right)^{1/p} < \frac{p}{p-1} n^{1-1/p}, $$

for all integers $n\geq 1$ and any real number $p>1$.

Set $S_n := \sum_{j=1}^n a_j^p$. Thus, by $(1)$ and $(2)$ and the observation that $S_n\geq \sum_{j=1}^k a_j^p$ for all $1\leq k \leq n$, we obtain

\begin{align} \sum_{k=1}^n \left( \frac{a_1+\cdots+a_k}{k} \right)^p &\leq n^{1-p} S_n \left[ \sum_{k=1}^n \left( \frac{1}{k} \right)^{1/p} \right]^p \\ &\leq n^{1-p} S_n \frac{p^p}{(p-1)^p} n^{p-1} \\ &= \frac{p^p}{(1-p)^p} \sum_{k=1}^n a_k^p. \end{align}

Let $n \to \infty $ in the above inequality to get the result.

In order to show that $p^p (p-1)^{-p}$ is the best constant take the sequence $a_n= n^{-1/p}$ if $n\leq N$ and $0$ elsewhere where $N$ is a fixed positive integer. A straightforward computation shows that for every $\epsilon \in (0,1) $ there exists a positive integere $N(\epsilon)$ such that

$$ \sum_{n=1}^\infty \left( \frac{1}{n} \sum_{k=1}^{n} a_k \right)^p > (1-\epsilon) \frac{p^p}{(1-p)^p} \sum_{n=1}^\infty a_k^p, $$

for the above choice of $(a_n)_{n\geq 1}$ and for all $N \geq N(\epsilon)$. This justifies that $p^p (p-1)^{-p}$ cannot be replaced with a smaller one.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.