关于特定指数-对数和不等式成立性的技术咨询

Hey there! I get that this inequality looks deceptively simple but is stumping you right now—let's break it down step by step.

First, let's restate the inequality you're asking about clearly:
$$\exp\left(\sum_{j>s} \log(1+a_j) \right)-1\leq C\sum_{j>s}a_j$$
where $C>0$ is some constant.

You mentioned this comes from equation (12) on page 8 of a specific paper, and you're wondering if it holds even without the paper's stated conditions on the sequence ${a_j}$.

First, let's simplify the left-hand side (LHS) using logarithm and exponent properties: the sum of logs is the log of a product, so $\exp\left(\sum_{j>s} \log(1+a_j)\right) = \prod_{j>s} (1+a_j)$. So the inequality simplifies to:
$$\prod_{j>s} (1+a_j) - 1 \leq C \sum_{j>s} a_j$$

Now let's walk through when this inequality holds, and when it doesn't:

情况1：所有$a_j \geq 0$ #

If every term in the sequence is non-negative, we can lean on standard inequalities here. We know for $x \geq 0$, $\log(1+x) \leq x$, so $\sum_{j>s} \log(1+a_j) \leq \sum_{j>s} a_j$. This means $\exp\left(\sum \log(1+a_j)\right) \leq \exp\left(\sum a_j\right)$.

• If $\sum_{j>s} a_j$ is uniformly bounded across all $s$ (say, there's a fixed $M$ where $\sum_{j>s} a_j \leq M$ for any $s$), then we can use the convexity of $\exp(x)-1$. For $0 \leq x \leq M$, the function $\exp(x)-1$ lies below the line connecting $(0,0)$ and $(M, \exp(M)-1)$, so $\exp(x)-1 \leq \frac{\exp(M)-1}{M}x$. Substituting $x = \sum_{j>s}a_j$, we get our constant $C = \frac{\exp(M)-1}{M}$, which works for all $s$.
• If $\sum_{j>s} a_j$ can grow arbitrarily large as $s$ changes, the inequality fails. The LHS grows exponentially while the RHS is linear—there's no fixed constant $C$ that can bridge that gap forever.

情况2：序列包含负数项 #

If some $a_j$ are negative, things get more complicated:

• For example, if $a_j = -1 + \epsilon$ (small $\epsilon >0$), then $1+a_j = \epsilon$, so the product becomes $\epsilon^k$ (for $k$ terms), making the LHS $\epsilon^k -1$ (a negative number). The RHS would be $k(-1+\epsilon)$ (also negative), so the inequality might hold here—but this is a niche case.
• If negative terms cause the product to blow up or the sum to behave erratically, the inequality can easily fail without additional constraints on ${a_j}$.

关键结论 #

You suspected the paper's conditions on ${a_j}$ might not matter, but they're almost certainly critical here. Common conditions that would make this inequality hold include:

• Non-negative $a_j$ with uniformly bounded partial sums $\sum_{j>s}a_j$
• Each $a_j$ is small in magnitude (e.g., $|a_j| \leq \delta < 1$) so the product doesn't explode or vanish unexpectedly
• The sequence ${a_j}$ is absolutely summable

Without these sorts of constraints, the inequality doesn't hold in general.

Hope this helps you untangle the problem!

备注：内容来源于stack exchange，提问作者Byeong-Ho Bahn