NOTES ON HW 1

TERENCE TAO

1. Comments on Q1,Q2

It was pointed out to me by Shaunglin Shao that Q1 and Q2 in fact follow from a stronger statement:

Proposition 1.1 (Quasi-triangle inequality). Let ∥∥ be a quasinorm with quasitriangle inequality constant C_{0} ≥ 1, thus

∥f + g∥ ≤ C_{0}(∥f∥ + ∥g∥)

and let 0 < p ≤ 1 be the exponent such that (2C_{0})^{p} = 2, i.e. p = \frac{log 2}{log 2C_{0}} . Then for any f_{1}, …, f_{N} we have

∥f_{1} + … + f_{N}∥ ≤ 2C_{0}(∑_{n=1}^{N} ∥f_{n}∥^{p})^{1/p}.

To prove this proposition, we first observe that for any f, g

∥f + g∥^{p} ≤ C_{0}^{p}(∥f∥ + ∥g∥)^{p} ≤ C_{0}^{p}2^{p}max(∥f∥, ∥g∥)^{p} ≤ 2max(∥f∥^{p}, ∥g∥^{p}).

In particular, if we have ∥f∥^{p} ≤ 2^{k} and ∥g∥^{p} ≤ 2^{k} for some k, then ∥f +g∥^{p} ≤ 2^{k+1}. We can iterate this to get

Lemma 1.2. Suppose we have ∥f_{n}∥^{p} ≤ 2^{k_{n}} for n = 1, …, N and some integers k_{1}, …, k_{N}, and suppose that 2^{k_{1}} + … + 2^{k_{N}} ≤ 2^{k}. Then ∥f_{1} + … + f_{N}∥^{p} ≤ 2^{k}.

Proof We induct on N. The case N = 1 is trivial. Now suppose N > 1 and the claim has already been proven for N − 1. Then we order k_{1} ≥ … ≥ k_{N}, and normalise k_{N−1} = −1. Then ∥f_{N−1}∥^{p}, ∥f_{N}∥^{p} ≤ 1/2 and hence by the preceding discussion ∥f_{N−1} + f_{N}∥^{p} ≤ 1. Now observe that 2^{k_{1}} + … + 2^{k_{N−2}} + 1 is equal to 2^{k_{1}} + … + 2^{k_{N}} rounded up to the nearest integer and is thus also less than 2^{k}. The claim now follows from the induction hypothesis.

To prove the proposition, we can of course assume the f_{n} are all non-zero. We normalise

∑_{n=1}^{N} ∥f_{n}∥^{p} = 1

and then let k_{n} be the integer such that

2^{k_{n}−1} < ∥f_{n}∥^{p} ≤ 2^{k_{n}}

which implies that

2^{k_{1}} + … + 2^{k_{N}} ≤ 2 ∑_{n=1}^{N} ∥f_{n}∥^{p} = 2

and so by the lemma

∥ ∑_{n=1}^{N} f_{n}∥^{p} ≤ 2 = (2C_{0})^{p}

and the claim follows.

It is not hard to deduce both Q1 and Q2 from the proposition. Basically the proposition is pursuing a nearly optimal “divide and conquer” strategy, of which the specific strategies in the hints for Q1 and Q2 are essentially special cases.

2. Comments on Q3

One approach to Q3 that is shorter than that suggested in the notes is to observe that for any set E of finite measure there is an exceptional set E_{n} ⊂ E of measure at most μ(E_{n}) ≤ μ(E)/2N such that

∫_{E\E_{n}} |f_{n}| dμ ≲ ∥f_{n}∥_{L^{1,∞}} log N.

(This is a modification of the usual dual characterisation of weak L^{1,∞}.) Letting E′ := E\(E_{1}∪…∪E_{n}) then μ(E′) ≥ μ(E)/2, and on summing the above inequality we conclude

∫_{E′} | ∑_{n=1}^{N} f_{n}| dμ ≲ ∑_{n=1}^{N} ∥f_{n}∥_{L^{1,∞}} log N.

Using the dual characterisation of weak L^{1,∞} we obtain the claim again. (This is close to the original argument of Stein and Weiss.)

It was pointed out by Thai Hoang Le that there is in fact a sharper inequality

∥f_{1} + … + f_{N}∥_{L^{1,∞}} ≤ 2 ∑_{n=1}^{N} ∥f_{n}∥_{L^{1,∞}} log \frac{2 ∑_{m=1}^{N} ∥f_{m}∥_{L^{1,∞}}}{∥f_{n}∥_{L^{1,∞}}} ,

where to avoid degeneracies we take the f_{n} to all be non-zero. This can be done by optimising the cutoffs more carefully (so instead of truncating each f_{n} to be larger than 1/2N, one chooses a cutoff α_{n} adapted to each f_{n} and then optimises in α_{n} using Jensen’s inequality).

The simplest counterexample in Q3 that shows that the logarithm is necessary is to take X = Z/NZ with counting measure, to set f_{1}(n) = 1/n for n = 1, …, N, and f_{j}(n) = f_{1}(n−j+1) for j = 2, …, N. Then one easily verifies that ∥f_{i}∥_{L^{1,∞}} = 1 but ∑_{i=1}^{N} f_{i} is the constant function ∑_{i=1}^{N} \frac{1}{i} and thus ∥ ∑_{i=1}^{N} f_{i}∥_{L^{1,∞}} = N ∑_{i=1}^{N} \frac{1}{i} ∼ N log N. The basic strategy is of course to make each f_{i} “multiply-scaled” (in

NOTES ON HW 1 3

order to make the L^{1} norm much larger than the L^{1,∞} norm) while keeping ∑_{i=1}^{N} f_{i} “singly scaled”.

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

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