非负实序列下，对所有t∈ℕ，∂(liminfₙ∑ₜ=1ⁿaₜ)/∂aₜ=1是否恒成立？

Question 1: All real sequences #

The answer is no. We can construct a real sequence where adjusting (a_t) has no impact on the liminf of the partial sums, proving the partial derivative (or subgradient) isn’t necessarily 1.

Take (t = 1) for simplicity, and define the sequence as:

• (a_1 = x) (a real variable)
• For (k \geq 1), (a_{2k} = -x) and (a_{2k+1} = 0)

The partial sums work out to:

• (S_1 = x)
• (S_2 = x + (-x) = 0)
• All (S_n = 0) for (n \geq 2)

The liminf of these partial sums is always (0), no matter what value (x) takes. Changing (a_1) doesn’t shift the liminf at all, so the partial derivative with respect to (a_1) here is (0), not (1). This counterexample shows the statement fails for arbitrary real sequences.

Question 2: Non-negative sequences #

The answer is yes. When every (a_t \geq 0), the sequence of partial sums (S_n = \sum_{t=1}^n a_t) is non-decreasing (each term adds a non-negative value). For non-decreasing sequences, the liminf equals the limit of the sequence (which is just the infinite sum (\sum_{t=1}^\infty a_t), whether finite or infinite).

For any fixed (t \in \mathbb{N}):

• If you increase (a_t) by (\epsilon > 0), every partial sum (S_n) for (n \geq t) goes up by (\epsilon). The liminf (which is the limit here) increases by exactly (\epsilon).
• If you decrease (a_t) by (\delta > 0) (keeping (a_t - \delta \geq 0)), every partial sum (S_n) for (n \geq t) drops by (\delta), so the liminf decreases by exactly (\delta).

Even if the infinite sum is infinite, adjusting (a_t) maintains the infinite value but the directional derivative (rate of change) in the direction of (a_t) remains (1). For all non-negative sequences, (\frac{\partial(\liminf_n \sum_{t=1}^n a_t)}{\partial a_t} = 1) holds for every (t \in \mathbb{N}).

内容的提问来源于stack exchange，提问作者Yes