TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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For bounded nonnegative sequences, $\limsup(x_k y_k) \le (\limsup x_k)(\limsup y_k)$

Statement

Let $(x_k)$ and $(y_k)$ be bounded sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $x_k \ge 0$ and $y_k \ge 0$ for every $k \in \mathbb{N}$ . Then $\limsup_k x_k$ , $\limsup_k y_k$ and $\limsup_k (x_k y_k)$ are real numbers, all $\ge 0$ , and

$\limsup_{k} (x_k y_k) \;\le\; \Big(\limsup_{k} x_k\Big)\Big(\limsup_{k} y_k\Big).$

Both hypotheses are doing work. Boundedness makes all three quantities real, so that the product on the right is a product in the field $\mathbb{R}$ and no extended multiplication is involved; without it the right-hand side could be an undefined product $0 \cdot (+\infty)$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined). Nonnegativity is what lets two upper estimates be multiplied: for sequences of mixed sign the inequality is false in the stated form, since a product of two negative numbers is positive and the estimate would point the wrong way. Strictness is possible, and a witness is recorded on the companion page.

Facts & Assumptions

Given: Bounded sequences $(x_k)$ , $(y_k)$ of reals with $x_k \ge 0$ and $y_k \ge 0$ for every $k$ ; their termwise product $(x_k y_k)$ ; and $\Lambda := \limsup_k x_k$ , $M := \limsup_k y_k$ , $P := \limsup_k(x_k y_k)$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ ).

[L1]

Tail ranges $T_n$ , extended tail suprema $s_n = \sup T_n$ and $\limsup_k z_k = \inf\{s_n\}$ all exist for every sequence; $s_n$ is the least upper bound of $T_n$ and $\limsup$ the greatest lower bound of $\{s_n\}$ (The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ , Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on $\overline{\mathbb{R}}$ is total and transitive, restricts on $\mathbb{R}$ to the order of $\mathbb{R}$ , and has $+\infty$ greatest and $-\infty$ least; a member of $\overline{\mathbb{R}}$ lying between two reals is itself real (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation for a real limit superior: for every real $\varepsilon > 0$ one has $z_k < \limsup_k z_k + \varepsilon$ eventually (For finite $L$ : $L = \limsup x_k$ iff for every $\varepsilon > 0$ one has $x_k < L + \varepsilon$ eventually and $x_k > L - \varepsilon$ frequently).

[L4]

Boundedness of a sequence of reals: there is a real $B$ with $|z_k| \le B$ for every $k$ ; and $z \le |z|$ always (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set, Basic properties of the absolute value).

[L5]

Products of inequalities: $0 \le a \le b$ and $0 \le c \le d$ give $ac \le bd$ , which Multiplying inequalities of positives states in exactly this nonstrict form; and multiplication by a positive element preserves the order, Sign rules for products and monotonicity of multiplication stating the strict form $a < b \iff ac < bc$ and the nonstrict form following by adjoining the case $a = b$ , where the two products are equal.

[L6]

Order arithmetic in $\mathbb{R}$ : inequalities may be added and translated, and the order is total, so exactly one of $a < b$ , $a = b$ , $b < a$ holds (Order is preserved by adding a constant and by adding inequalities).

[L7]

Reciprocal Archimedean property: for every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ ; and $0 < a \le b$ gives $0 < 1/b \le 1/a$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L8]

Two properties each holding eventually hold together from the larger of the two thresholds on, the order on $\mathbb{N}$ being total (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ ).

Proof

technique · direct

1.1

Both sequences are bounded, so there are reals bounding $|x_k|$ and $|y_k|$ ; let $B$ be the larger of the two, so that $|x_k| \le B$ and $|y_k| \le B$ for every $k$ , and $B \ge |x_0| \ge 0$ . With $x_k \ge 0$ and $y_k \ge 0$ this gives $0 \le x_k \le B$ and $0 \le y_k \le B$ for every $k$ , hence $0 \le x_k y_k \le B \cdot B$ by [L5].

givenL4L5L6

2.1

Each of $\Lambda$ , $M$ , $P$ is a real number $\ge 0$ . Indeed, for every $n$ the real $B$ is an upper bound of the $n$ -th tail range of $(x_k)$ , so $s_n \le B$ and hence $\Lambda \le s_0 \le B$ ; and $s_n \ge x_n \ge 0$ for every $n$ , so $0$ is a lower bound of $\{s_n\}$ and $0 \le \Lambda$ . Being between the reals $0$ and $B$ , the element $\Lambda$ is real. The same argument gives $0 \le M \le B$ , and, using the bound $B \cdot B$ from step 1.1, $0 \le P \le B \cdot B$ .

step 1.1L1L2

3.1

Let $\delta > 0$ be an arbitrary real and put $C := \Lambda + M + 1$ , a real with $C \ge 1 > 0$ . Take a natural $m_1 \ge 1$ with $1/m_1 < 1$ and a natural $m_2 \ge 1$ with $1/m_2 < \delta/C$ , let $m$ be the larger of $m_1$ and $m_2$ , and set $\varepsilon := 1/m$ , so that $0 < \varepsilon < 1$ and $\varepsilon C < \delta$ . By [L3] there are thresholds beyond which $x_k < \Lambda + \varepsilon$ and beyond which $y_k < M + \varepsilon$ ; let $N$ be the larger. For $k \ge N$ we have $0 \le x_k \le \Lambda + \varepsilon$ and $0 \le y_k \le M + \varepsilon$ , so $x_k y_k \le (\Lambda + \varepsilon)(M + \varepsilon) = \Lambda M + \varepsilon(\Lambda + M + \varepsilon) \le \Lambda M + \varepsilon C < \Lambda M + \delta$ , the middle step because $\Lambda + M + \varepsilon \le C$ and $\varepsilon > 0$ . Hence $\Lambda M + \varepsilon C$ is an upper bound of the $N$ -th tail range of $(x_k y_k)$ , so $P \le \Lambda M + \varepsilon C < \Lambda M + \delta$ .

step 2.1L1L3L5L6L7L8algebra

4.1

Suppose $P > \Lambda M$ . Both are real by step 2.1, so $\delta_0 := P - \Lambda M > 0$ , and step 3.1 applied with $\delta = \delta_0$ gives $P < \Lambda M + \delta_0 = P$ , which is impossible. By totality $P \le \Lambda M$ , which is the asserted inequality.

step 3.1step 2.1L2L6∎

Remarks

The estimate is the product of two one-sided estimates, and that is why nonnegativity is needed. Step 3.1 multiplies $x_k \le \Lambda + \varepsilon$ by $y_k \le M + \varepsilon$ , which is legitimate only when all four quantities are $\ge 0$ (Multiplying inequalities of positives). For sequences of mixed sign the same two estimates say nothing about the product; the correct general statement in that setting involves absolute values and is not needed on this page.
The error term is linear in $\varepsilon$ with a fixed coefficient. Expanding gives $\varepsilon(\Lambda + M + \varepsilon)$ , and restricting to $\varepsilon < 1$ replaces the varying coefficient by the constant $C = \Lambda + M + 1$ , after which one choice of $\varepsilon$ makes the whole error smaller than the prescribed $\delta$ . Both restrictions on $\varepsilon$ are met at once by taking the larger of two Archimedean choices.
The inequality is strict for some bounded nonnegative pairs, and $x_k = 1 + (-1)^k$ , $y_k = 1 + (-1)^{k+1}$ give $\limsup(x_k y_k) = 0 < 4$ ↗ on the companion page is the witness: there the product sequence is identically $0$ while the right-hand side is $4$ .
The bounded hypothesis is not merely for convenience. Without it $\Lambda$ could be $+\infty$ and $M$ could be $0$ , and then the right-hand side is not an element of $\overline{\mathbb{R}}$ at all (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined); the behaviour of the products in that situation really is unconstrained, as Null times divergent has no rule: $x_k = 1/k$ with $y_k = ck$ gives product limit $c$ , and with $y_k = k^2$ gives divergence ↗ shows.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)
J. Lebl, Basic Analysis I, §2.3 (standard reference, not scraped)
N. Donaldson, Math 140A: Real Analysis notes (standard reference, not scraped)

For bounded nonnegative sequences, lim sup⁡(xkyk)≤(lim sup⁡xk)(lim sup⁡yk)\limsup(x_k y_k) \le (\limsup x_k)(\limsup y_k)limsup(xk​yk​)≤(limsupxk​)(limsupyk​)