CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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Strong convergence of subcritical powers

Statement

Assume Countable Choice. Let $Ω \subset R^{n}$ have finite Lebesgue measure, let $1 \leq q < r < \infty$ and $m \geq 1$ , and let $(u_{j})$ be a sequence in $L^{q} (Ω; R)$ with $u_{j} \to u$ in $L^{q} (Ω; R)$ and $\sup_{j} ∥ u_{j} ∥_{L^{r} (Ω)} < \infty$ . Then $u \in L^{r} (Ω)$ , and for every $1 \leq s < r / m$ the nonlinear maps converge: $∣ u_{j} ∣^{m - 1} u_{j} ⟶ ∣ u ∣^{m - 1} u in L^{s} (Ω) .$ The range is nonempty only when $m < r$ ; the endpoint $s = r / m$ is not asserted.

Facts & Assumptions

Given: Countable Choice, a finite-measure set $Ω \subseteq R^{n}$ , exponents $1 \leq q < r < \infty$ , a real number $m \geq 1$ , and real-valued measurable classes $u_{j}, u$ on $Ω$ with $u_{j} \to u$ in $L^{q} (Ω)$ and $M : = \sup_{j} ∥ u_{j} ∥_{L^{r} (Ω)} < \infty$ .

[F1]

Hölder inclusion on a finite-measure space. If $1 \leq a < b < \infty$ and $g$ is measurable on the finite-measure space $Ω$ , then $g \in L^{a}$ whenever $g \in L^{b}$ , with $∥ g ∥_{L^{a}} \leq ∣ Ω ∣^{1 / a - 1 / b} ∥ g ∥_{L^{b}}$ ; this is Hölder applied to $∣ g ∣^{a}$ and the constant function $1$ . (Holder's inequality for integrals, including the endpoint cases, The space $L^{p} (μ)$ as the quotient by null functions)

[F2]

Lyapunov interpolation. If $1 \leq p_{0} < p < p_{1} < \infty$ , $θ \in (0, 1)$ and $1 / p = θ / p_{0} + (1 - θ) / p_{1}$ , then every $f \in L^{p_{0}} \cap L^{p_{1}}$ lies in $L^{p}$ with $∥ f ∥_{L^{p}} \leq ∥ f ∥_{L^{p_{0}}}^{θ} ∥ f ∥_{L^{p_{1}}}^{1 - θ}$ . (Lyapunov interpolation inequality for $L^{p}$ norms)

[F3]

Almost-everywhere subsequences. Every sequence converging in $L^{q}$ , $1 \leq q \leq \infty$ , has a subsequence whose representatives converge almost everywhere to a representative of the limit. (Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences)

[F4]

Fatou's lemma. For nonnegative measurable functions $f_{k}$ , $\int {lim inf}_{k} f_{k} \leq {lim inf}_{k} \int f_{k}$ . (Fatou's lemma)

[F5]

Mean value theorem. If $f : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then $f (b) - f (a) = f^{'} (c) (b - a)$ for some $c \in (a, b)$ . (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ )

[F6]

Hölder's inequality for products. For conjugate exponents $1 \leq p, q \leq \infty$ and measurable $f, g$ , $\int ∣ f g ∣ \leq ∥ f ∥_{L^{p}} ∥ g ∥_{L^{q}}$ . (Holder's inequality for integrals, including the endpoint cases)

Proof

Proof technique: Extract an almost-everywhere subsequence to obtain the $L^{r}$ -bound of the limit, transfer convergence to the exponent $m s < r$ by interpolation, and apply the pointwise mean value bound followed by Hölder.

1.1F3F4given

By [F3] fix a subsequence $(u_{j_{k}})$ convergent almost everywhere to $u$ . Then ${lim inf}_{k} ∣ u_{j_{k}} ∣^{r} = ∣ u ∣^{r}$ pointwise almost everywhere, so [F4] gives $∥ u ∥_{L^{r}}^{r} \leq {lim inf}_{k} ∥ u_{j_{k}} ∥_{L^{r}}^{r} \leq M^{r} < \infty$ , hence $u \in L^{r} (Ω)$ with $∥ u ∥_{L^{r}} \leq M$ . Consequently $∥ u_{j} - u ∥_{L^{r}} \leq ∥ u_{j} ∥_{L^{r}} + ∥ u ∥_{L^{r}} \leq 2 M$ for every $j$ .

2.1F1F2step 1.1

Fix $1 \leq s < r / m$ and put $b : = m s < r$ . If $b = q$ then $∥ u_{j} - u ∥_{L^{b}} = ∥ u_{j} - u ∥_{L^{q}} \to 0$ ; if $b < q$ , then [F1] gives $∥ u_{j} - u ∥_{L^{b}} \leq ∣ Ω ∣^{1 / b - 1 / q} ∥ u_{j} - u ∥_{L^{q}} \to 0$ . If $q < b < r$ , choose $θ \in (0, 1)$ with $1 / b = θ / q + (1 - θ) / r$ ; [F2] applied to the classes $u_{j} - u \in L^{q} \cap L^{r}$ gives $∥ u_{j} - u ∥_{L^{b}} \leq ∥ u_{j} - u ∥_{L^{q}}^{θ} ∥ u_{j} - u ∥_{L^{r}}^{1 - θ} \leq ∥ u_{j} - u ∥_{L^{q}}^{θ} (2 M)^{1 - θ} \to 0$ by step 1.1. In both cases $∥ u_{j} - u ∥_{L^{m s}} \to 0$ , and $\sup_{j} ∥ u_{j} ∥_{L^{m s}} \leq ∣ Ω ∣^{1 / (m s) - 1 / r} \sup_{j} ∥ u_{j} ∥_{L^{r}} < \infty$ by [F1], while $∥ u ∥_{L^{m s}} \leq ∣ Ω ∣^{1 / (m s) - 1 / r} M$ by step 1.1.

3.1F5F6step 2.1given∎

If $m = 1$ then $∣ u_{j} ∣^{m - 1} u_{j} = u_{j}$ and $∣ u ∣^{m - 1} u = u$ , so the claim is step 2.1 itself with $b = s < r$ . If $m > 1$ , consider $N (t) : = ∣ t ∣^{m - 1} t$ on $R$ ; $N$ is differentiable with $N^{'} (t) = m ∣ t ∣^{m - 1}$ , and for real $a \neq b$ every point $t$ of the closed interval between them satisfies $∣ t ∣^{m - 1} \leq ∣ a ∣^{m - 1} + ∣ b ∣^{m - 1}$ . By [F5] applied to $N$ on that interval, $∣ N (a) - N (b) ∣ \leq m (∣ a ∣^{m - 1} + ∣ b ∣^{m - 1}) ∣ a - b ∣ .$ Writing $a = u_{j} (x)$ , $b = u (x)$ and applying [F6] with exponents $m / (m - 1)$ and $m$ to the product $(∣ u_{j} ∣^{m - 1} + ∣ u ∣^{m - 1}) ∣ u_{j} - u ∣$ gives $∥ N (u_{j}) - N (u) ∥_{L^{s}} \leq m (∥ u_{j} ∥_{L^{m s}}^{m - 1} + ∥ u ∥_{L^{m s}}^{m - 1}) ∥ u_{j} - u ∥_{L^{m s}},$ and the right-hand side tends to $0$ by the bounds and convergence of step 2.1. Hence $∣ u_{j} ∣^{m - 1} u_{j} \to ∣ u ∣^{m - 1} u$ in $L^{s} (Ω)$ .

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Sources

Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes) (standard reference, not scraped)