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 Incoming links: Power singularity integrability

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 22H / b / ii / Solution Created 2026-09-24 Updated 2026-10-03

No. Take

X = L^{1} (R^{n}) \cap L^{2} (R^{n}), ∥ f ∥_{X} = ∥ f ∥_{1} + ∥ f ∥_{2} .

This is a Banach space, and

X

-convergence implies

L^{2}

convergence, which has an almost-everywhere convergent subsequence. Thus the hypotheses hold.

However, let

f (x) = ∣ x ∣^{- 3 n /4} 1_{{∣ x ∣ < 1}} .

The power singularity integrability test gives

f \in L^{1}

but

f \in / L^{2}

. Its truncations

f_{m} = min (f, m)

belong to

X

and converge to

f

in

L^{1}

by dominated convergence. Hence

(f_{m})

is

L^{1}

-Cauchy in

X

but has no

L^{1}

limit belonging to

X

. Therefore

X

need not be complete in the

L^{1}

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