CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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$L^{p}$ convergence need not imply $L^{q}$ convergence when $p < q$

Statement refuted

For $1 \leq p < q < \infty$ , $L^{p}$ convergence need not imply $L^{q}$ convergence.

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ , $1 \leq p < q < \infty$ , and $X_{n} = (n + 1)^{1 / q} 1_{(0, 1 / (n + 1))}$ for $n \in N$ .

[L1]

$L^{r}$ convergence is vanishing of $E ∣ X_{n} - X ∣^{r}$ for the relevant exponent $r$ ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1

Direct calculation gives [L1] $E ∣ X_{n} ∣^{p} = (n + 1)^{p / q - 1} \to 0$ , since $p / q - 1 < 0$ . Thus $X_{n} \to 0$ in $L^{p}$ by [L1].

2.1

But $E ∣ X_{n} ∣^{q} = (n + 1) / (n + 1) = 1$ for every $n$ , so the [step 1.1, L1] $L^{q}$ norm never tends to zero. Hence [L1] rules out $L^{q}$ convergence.

step 1.1L1∎

Depends on

$L^p$ convergence for random variables

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

S. Roch, Lecture 3: Modes of convergence, Example 3.11 (standard reference, not scraped)

LpLp convergence need not imply LqLq convergence when p<qp<q