Maximum of the limiting average of a sequence's square roots

Question

I'm looking for a proof (or counterexample) to the following: if $a_n \in [0,1]$ is such that $\lim_{n \rightarrow \infty} (a_1 + ... + a_n)/n = 1/4$, then $\lim_{n \rightarrow \infty} (\sqrt{a_1} + ... + \sqrt{a_n})/n$ is at most 1/2.

Thanks!

Answer 1

Since $f(x)=\sqrt{x}$ is concave,

$f(\sum_{i=1}^n\frac{1}{n}a_i)\geq \sum_{i=1}^n\frac{1}{n}f(a_i).$

It follows that if the limit of the square root sum exists, it's at most $\sqrt{\frac{1}{4}}=1/2$. There are cases when the Cesaro limit doesn't exist: Cesaro summability of nonnegative bounded sequence

Answer 2

Here's a straightforward proof:

Let $a_i = \frac14 + x_i$, then $\sum_i x_i = 0$. So

$$ \frac{1}{n } \sum_i \sqrt{a_i} = \frac{1}{n } \sum_i \sqrt{\frac14 + x_i} $$ Use that $\sqrt{\frac14 + x_i} \leq \frac12 + x_i$, with equality for $x_i = 0$.

So you have $$ \frac{1}{n } \sum_i \sqrt{a_i} \leq \frac{1}{n } \sum_i (\frac12 + x_i)= \frac12 $$

and again equality only if all $a_i = \frac14$.