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Suppose $\{a_n\}_{n\geq 1}$ is a sequence of nonnegative real numbers. Define sequence \begin{align} b_n = \frac{\sum_{i = 1}^n a_i}{n}. \end{align}

Prove the following conjecture:

There exists a universal constant $C>0$ such that for any $n$ positive integer, any $\epsilon>0$ and any sequence $\{a_n\}$ of nonnegative real numbers, we have \begin{align} \sum_{i = 1}^n a_i \ln\left ( \frac{a_i}{b_i} \right ) \mathbf{1}(b_i\leq \epsilon) & \leq C n\epsilon \cdot \ln(n+1) . \end{align}

We adopt the convention that $0 \ln(0/0) = 0$.

Partial converse: it is clear that one cannot relax the RHS upper bound. Consider the sequence $\{a_n\}$ such that $a_i = 0$, $1\leq i\leq m$, and $a_i = 1, i\geq m+1$. Take $n = \sup\{j: b_j \leq \epsilon\}$.

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  • $\begingroup$ If $C$ is universal, and $\epsilon$ arbitrary, the inequality must hold as $\le 0$, which is absurd. The quantifiers need work here. $\endgroup$ – user357151 Aug 27 '17 at 21:17
  • $\begingroup$ @Michelle I do not understand what you mean here. One needs to prove this inequality for ALL sequences $\{a_n\}$. For certain sequence the LHS may be negative, but it does not matter. $\endgroup$ – user401582 Aug 27 '17 at 21:18
  • $\begingroup$ I think what @Michelle meant is that since you have $\forall \epsilon$ in your condition, naively any inequality that holds as $\ldots \leq C \epsilon$ can be improved to $\ldots \leq 0$. // This should then be answered by pointing out the fact that on the LHS you have the indicator function of $\{ b_i \leq \epsilon\}$. $\endgroup$ – Willie Wong Aug 27 '17 at 22:51
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    $\begingroup$ Possibly buggy idea - the sum is clearly upper bounded by $\sum a_i \log (na_i/\sum_{k\le i} a_i) \mathbf{1}\{\sum_{k \le i} a_k \le i\epsilon\}$. Break the sum into $\log n \sum a_i \mathbf{1}\{\sum_{k \le i} a_i \le i\epsilon\} + \sum a_i \log \left(a_i/\sum_{k\le i} a_k\right) $. The second sum is clearly $\le 0$. For the first sum, consider the partial sums $\sum_{k \le i} a_k$, and note that the indicator zeroes out every term that pushes a partial sum above $i\epsilon$, and so the sum up to the $i$th term must be below the same. But then the first sum is bounded by $n\epsilon$. $\endgroup$ – stochasticboy321 Aug 27 '17 at 23:02
  • $\begingroup$ @WillieWong Yes! The indicator guarantees that the scaling is correct. $\endgroup$ – user401582 Aug 27 '17 at 23:15

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