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

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For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$

Statement

Let $(x_{k})$ and $(y_{k})$ be bounded sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $x_{k} \geq 0$ and $y_{k} \geq 0$ for every $k \in N$ . Then ${lim sup}_{k} x_{k}$ , ${lim sup}_{k} y_{k}$ and ${lim sup}_{k} (x_{k} y_{k})$ are real numbers, all $\geq 0$ , and

${lim sup}_{k} (x_{k} y_{k}) \leq ({lim sup}_{k} x_{k}) ({lim sup}_{k} y_{k}) .$

Both hypotheses are doing work. Boundedness makes all three quantities real, so that the product on the right is a product in the field $R$ and no extended multiplication is involved; without it the right-hand side could be an undefined product $0 \cdot (+ \infty)$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , 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 $x_{k} \geq 0$ and $y_{k} \geq 0$ for every $k$ ; their termwise product $(x_{k} y_{k})$ ; and $Λ : = {lim sup}_{k} x_{k}$ , $M : = {lim sup}_{k} y_{k}$ , $P : = {lim sup}_{k} (x_{k} y_{k})$ (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]

Tail ranges $T_{n}$ , extended tail suprema $s_{n} = \sup T_{n}$ and ${lim sup}_{k} z_{k} = \inf {s_{n}}$ all exist for every sequence; $s_{n}$ is the least upper bound of $T_{n}$ and $lim sup$ the greatest lower bound of ${s_{n}}$ (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 $R$ , 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 $\overline{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on $\overline{R}$ is total and transitive, restricts on $R$ to the order of $R$ , and has $+ \infty$ greatest and $- \infty$ least; a member of $\overline{R}$ lying between two reals is itself real (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , 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 $ε > 0$ one has $z_{k} < {lim sup}_{k} z_{k} + ε$ eventually (For finite $L$ : $L = lim sup x_{k}$ iff for every $ε > 0$ one has $x_{k} < L + ε$ eventually and $x_{k} > L - ε$ frequently).

[L4]

Boundedness of a sequence of reals: there is a real $B$ with $∣ z_{k} ∣ \leq B$ for every $k$ ; and $z \leq ∣ z ∣$ 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: $0 \leq a \leq b$ and $0 \leq c \leq d$ give $a c \leq b d$ , 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 $a < b ⟺ a c < b c$ and the nonstrict form following by adjoining the case $a = b$ , where the two products are equal.

[L6]

Order arithmetic in $R$ : 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 $η > 0$ there is a natural $m \geq 1$ with $1 / m < η$ ; and $0 < a \leq b$ gives $0 < 1 / b \leq 1 / a$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , 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 $N$ being total (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Order on the natural numbers, $\leq$ is a linear order on $N$ ).

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 $∣ x_{k} ∣ \leq B$ and $∣ y_{k} ∣ \leq B$ for every $k$ , and $B \geq ∣ x_{0} ∣ \geq 0$ . With $x_{k} \geq 0$ and $y_{k} \geq 0$ this gives $0 \leq x_{k} \leq B$ and $0 \leq y_{k} \leq B$ for every $k$ , hence $0 \leq x_{k} y_{k} \leq B \cdot B$ by [L5].

givenL4L5L6

2.1

Each of $Λ$ , $M$ , $P$ is a real number $\geq 0$ . Indeed, for every $n$ the real $B$ is an upper bound of the $n$ -th tail range of $(x_{k})$ , so $s_{n} \leq B$ and hence $Λ \leq s_{0} \leq B$ ; and $s_{n} \geq x_{n} \geq 0$ for every $n$ , so $0$ is a lower bound of ${s_{n}}$ and $0 \leq Λ$ . Being between the reals $0$ and $B$ , the element $Λ$ is real. The same argument gives $0 \leq M \leq B$ , and, using the bound $B \cdot B$ from step 1.1, $0 \leq P \leq B \cdot B$ .

step 1.1L1L2

3.1

Let $δ > 0$ be an arbitrary real and put $C : = Λ + M + 1$ , a real with $C \geq 1 > 0$ . Take a natural $m_{1} \geq 1$ with $1 / m_{1} < 1$ and a natural $m_{2} \geq 1$ with $1 / m_{2} < δ / C$ , let $m$ be the larger of $m_{1}$ and $m_{2}$ , and set $ε : = 1 / m$ , so that $0 < ε < 1$ and $ε C < δ$ . By [L3] there are thresholds beyond which $x_{k} < Λ + ε$ and beyond which $y_{k} < M + ε$ ; let $N$ be the larger. For $k \geq N$ we have $0 \leq x_{k} \leq Λ + ε$ and $0 \leq y_{k} \leq M + ε$ , so $x_{k} y_{k} \leq (Λ + ε) (M + ε) = Λ M + ε (Λ + M + ε) \leq Λ M + ε C < Λ M + δ$ , the middle step because $Λ + M + ε \leq C$ and $ε > 0$ . Hence $Λ M + ε C$ is an upper bound of the $N$ -th tail range of $(x_{k} y_{k})$ , so $P \leq Λ M + ε C < Λ M + δ$ .

step 2.1L1L3L5L6L7L8algebra

4.1

Suppose $P > Λ M$ . Both are real by step 2.1, so $δ_{0} : = P - Λ M > 0$ , and step 3.1 applied with $δ = δ_{0}$ gives $P < Λ M + δ_{0} = P$ , which is impossible. By totality $P \leq Λ M$ , 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 $x_{k} \leq Λ + ε$ by $y_{k} \leq M + ε$ , which is legitimate only when all four quantities are $\geq 0$ (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 $ε$ with a fixed coefficient. Expanding gives $ε (Λ + M + ε)$ , and restricting to $ε < 1$ replaces the varying coefficient by the constant $C = Λ + M + 1$ , after which one choice of $ε$ makes the whole error smaller than the prescribed $δ$ . Both restrictions on $ε$ 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 $lim sup (x_{k} y_{k}) = 0 < 4$ ↗ 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 $Λ$ could be $+ \infty$ and $M$ could be $0$ , and then the right-hand side is not an element of $\overline{R}$ at all (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , 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 $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence ↗ shows.

Depends on

Used by

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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(xk​yk​)≤(limsupxk​)(limsupyk​)