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ID: past-exam-of-the-functional-analytic-methods-for-partial-differential-equations-course-of-the-university-of-oxford/2023/3/b/ii/solution

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Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 3 / b / ii / Solution by

Codex  0 2026-10-25

The Lp norm of the forcing at the critical exponent is constant:

∥ f_{m} ∥_{L^{6/5} (B)}^{6/5} = ∣ E_{m} ∣^{- 1} \int_{B} χ_{E_{m}} d x = 1.

To prove weak convergence in

L^{6/5} (B)

, the duality of Lp spaces says that we must test against an arbitrary

g \in L^{6} (B)

. Holder inequality, restricted to the shrinking measurable set

E_{m}

, gives

\int_{B} f_{m} g d x \leq ∣ E_{m} ∣^{- 5/6} \int_{E_{m}} ∣ g ∣ d x \leq ∣ E_{m} ∣^{- 5/6} ∣ E_{m} ∣^{5/6} (\int_{E_{m}} ∣ g ∣^{6} d x)^{1/6} = (\int_{E_{m}} ∣ g ∣^{6} d x)^{1/6} ⟶ 0.

The last step is absolute continuity of the Lebesgue integral, since

∣ E_{m} ∣ = r_{m}^{3} ∣ E ∣ \to 0

and

∣ g ∣^{6}

is integrable. Thus

f_{m} ⇀ 0 in L^{6/5} (B), ∥ f_{m} ∥_{6/5} = 1.

Testing only bounded or smooth functions would not by itself establish this weak convergence; the restricted Holder inequality handles every member of the full dual space.

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