NOTES ON HW 3

TERENCE TAO

1. Comments on Q2.15

The quickest way to deduce Q2.15 from Q2.14 is to consider the operator T : l^{2}({1, …, N}) → L^{2}(X) defined by T ((c_{n})_{n=1}^{N}) := ∑_{n} c_{n}f_{n}. There are also several direct proofs of the Radamacher-Menshov inequality, which are essentially adapting the proof of Q2.14 rather than the result.

There is also a more general version of Radamacher-Menshov for nested projections: if π_{1}, …, π_{N} are a nested sequence of orthonormal projections on L^{2}(X) for N > 1, then

∥ sup_{1≤n≤N} |π_{n}f|∥_{L^{2}(X)} ≲ log N∥f∥_{L^{2}(X)}.

The proof is left to the reader.

2. Comment on Q3.1

Note that the monotone convergence theorem is applicable to Lorentz norms such as L^{1,∞}, which facilitates the reduction to the case of finitely many σ-algebras.

The need to keep things σ-finite is somewhat annoying. One can first truncate f by finite measure sets in B_{1} and take monotone limits to reduce to the case when the entire space has finite measure, at which point σ-finiteness is automatic no matter how coarse the σ-algebras become.

It is possible to establish the inequality (also known as Doob’s inequality) without performing most of the reductions suggested. Suppose one wants to show that

μ({sup_{n} |E(f|B_{n})| ≥ λ}) ≤ ∥f∥_{L^{1}(X)}/λ.

Let E_{n} be the set where |E(f|B_{n})| ≥ λ, but |E(f|B_{n′})| ≤ λ for all n′ < n; clearly we have

μ({sup_{n} |E(f|B_{n})| ≥ λ}) ≤ ∑_{n} μ(E_{n}).

On the other hand, E_{n} is clearly B_{n} measurable, and hence

∫_{X} 1_{E_{n}}|f| dμ = ∫_{X} 1_{E_{n}}E(|f||B_{n}) dμ ≥ ∫_{X} 1_{E_{n}}|E(f|B_{n})| dμ ≥ λμ(E_{n}).

Summing this in n (and using monotone convergence) we obtain the claim.

2 TERENCE TAO

One easy consequence of Doob’s inequality is that E(f|B_{n}) converges pointwise almost everywhere to E(f| ⋁_{n} B_{n}).

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

E-mail address: tao@math.ucla.edu