Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 5 / 1 / b

Solution

The relevant Banach-Alaoglu theorem says that for

1 < p < \infty

the closed bounded balls of

L^{p} (R^{n})

are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of

L^{p} = (L^{q})^{*}

; its weak and weak-star test spaces are both

L^{q}

Here is a direct proof of both compactness and the subsequence assertion. Let

∥ f_{j} ∥_{p} \leq M

and choose a countable dense set

{g_{m}}

in the separable Lp space

L^{q}

\int f_{j} g_{m}

is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives

f_{j_{k}}

for which these pairings converge for every

m

. For arbitrary

g \in L^{q}

\int (f_{j_{k}} - f_{j_{l}}) g \leq 2 M ∥ g - g_{m} ∥_{q} + \int (f_{j_{k}} - f_{j_{l}}) g_{m} .

First approximate

g

g_{m}

and then take

k, l

large. This proves convergence of every pairing, and the limit

Λ (g) = lim_{k} \int f_{j_{k}} g

is linear with

∣Λ (g) ∣ \leq M ∥ g ∥_{q}

. Apply the already proved duality of Lp spaces with

q

in place of

p

Λ (g) = \int f g

for an

f \in L^{p}

with

∥ f ∥_{p} \leq M

. Thus

f_{j_{k}} ⇀ f in L^{p}, ∥ f ∥_{p} \leq M .

To obtain compactness rather than only a sequential statement, on the radius-

M

d (f, h) = \sum_{m = 1}^{\infty} 2^{- m} \frac{∣ \int ( f - h ) g _{m} ∣}{1 + ∣ \int ( f - h ) g _{m} ∣} .

Density and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

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