CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$x_{k} = 1 + (- 1)^{k}$ , $y_{k} = 1 + (- 1)^{k + 1}$ give $lim sup (x_{k} y_{k}) = 0 < 4$

Statement refuted

That the submultiplicativity of For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ can be improved to an equality: that for all bounded nonnegative sequences $(x_{k})$ , $(y_{k})$ of reals, ${lim sup}_{k} (x_{k} y_{k}) = ({lim sup}_{k} x_{k}) ({lim sup}_{k} y_{k}) .$

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ and the index maps $e, o$ ; the sequences $x_{k} : = 1 + s_{k}$ and $y_{k} : = 1 - s_{k}$ , which are the families usually written $1 + (- 1)^{k}$ and $1 + (- 1)^{k + 1}$ ; and the tail ranges $T_{n}$ of Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ .

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ , and $e$ , $o$ are strictly increasing with $n_{j} \geq j$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior in $\overline{R}$ : existence for every sequence, and the least-upper-bound and greatest-lower-bound descriptions of the tail bounds (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set, The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

Absolute value: $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Order and field arithmetic: $0 < 1$ , so $0 < 2 = 1 + 1$ and $0 < 4 = 2 \cdot 2$ ; $1 + 1 = 2$ , $1 - 1 = 0$ , $1 + (- 1) = 0$ and $1 - (- 1) = 2$ ; a product with a zero factor is $0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)).

[L5]

Submultiplicativity: for bounded nonnegative sequences, ${lim sup}_{k} (x_{k} y_{k}) \leq ({lim sup}_{k} x_{k}) ({lim sup}_{k} y_{k})$ , all three quantities being real (For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ ).

Counterexample

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ . When $s_{k} = 1$ the pair $(x_{k}, y_{k})$ is $(2, 0)$ , and when $s_{k} = - 1$ it is $(0, 2)$ . In either case $0 \leq x_{k} \leq 2$ and $0 \leq y_{k} \leq 2$ , so both sequences are bounded and nonnegative, and $x_{k} y_{k} = 0$ because one of the two factors is $0$ .

givenL1L3L4

1.2

For every $n$ both cases occur at an index $\geq n$ : $s_{e_{n}} = 1$ with $e_{n} \geq n$ and $s_{o_{n}} = - 1$ with $o_{n} \geq n$ .

givenL1

2.1

Hence $T_{n} (x) = {0, 2}$ and $T_{n} (y) = {0, 2}$ for every $n$ , each with least upper bound $2$ in $\overline{R}$ , since $2$ bounds both elements and belongs to the set; so ${lim sup}_{k} x_{k} = {lim sup}_{k} y_{k} = 2$ . The product sequence is constantly $0$ , so $T_{n} (x y) = {0}$ and ${lim sup}_{k} (x_{k} y_{k}) = 0$ .

step 1.1step 1.2L2L4

3.1

The hypotheses of [L5] are met by step 1.1, and the inequality it gives reads $0 \leq 2 \cdot 2 = 4$ . Since $0 < 4$ , it is strict, so the equality asserted above fails for this pair and the refuted claim is false.

step 2.1step 1.1L4L5∎

Remarks

The two sequences vanish at complementary indices. Wherever $x_{k}$ attains its maximum $2$ , its partner $y_{k}$ is $0$ , so the product is $0$ everywhere and the two limit superiors are attained along disjoint sets of indices. This is the multiplicative form of the phase mismatch behind $x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$ .
Nonnegativity and boundedness are satisfied, so the failure is not degenerate. Both hypotheses of For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ hold, and the right-hand side is an honest product of real numbers; the inequality simply cannot be reversed.
The gap can be made total. Here the product sequence is identically zero while the bound is $4$ , so no fraction of the bound is achieved. Equality does hold when one of the two sequences converges, for the same reason as in the additive case.

Depends on

Used by

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Sources

Limit superior and limit inferior (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)

xk=1+(−1)kxk​=1+(−1)k, yk=1+(−1)k+1yk​=1+(−1)k+1 give lim sup⁡(xkyk)=0<4limsup(xk​yk​)=0<4