A HASTY INTERPOLATION

TERENCE TAO

Abstract. A cautionary note about using interpolation incorrectly.

1. Introduction

Let f_{0}, …, f_{n−1} be functions on a measure space X which are orthogonal. From Pythagoras’ theorem we have

∥f_{0} + … + f_{n−1}∥_{2} = (∑_{k=0}^{n−1} ∥f_{k}∥_{2}^{2})^{1/2}

whereas from the triangle inequality we have

∥f_{0} + … + f_{n−1}∥_{1} ≤ ∑_{k=0}^{n−1} ∥f_{k}∥_{1}

One may thus be led to conjecture the interpolated inequality

∥f_{0} + … + f_{n−1}∥_{p} ≤ (∑_{k=0}^{n−1} ∥f_{k}∥_{p}^{p})^{1/p}  (1)

for all 1 < p < 2. However, such estimates are false, even when n = 2. An example is

f_{0} = χ_{[−N^{2},0]} + Nχ_{[0,1]}, f_{1} = χ_{[−N^{2},0]} − Nχ_{[0,1]};

we have

∥f_{0} + f_{1}∥_{p} = 2N^{2/p}, (∥f_{0}∥_{p}^{p} + ∥f_{1}∥_{p}^{p})^{1/p} = 2^{1/p}(N^{2} + N^{p})^{1/p},

and so the right-hand side of (1) is larger than the left for sufficiently large N. By iterating this example, e.g. using the four functions f_{i} ⊗ f_{j} for i, j = 0, 1, we can make the ratio between the two sides of (1) arbitrarily large.

A similar naive interpolation would yield

∥f_{0} + … + f_{n−1}∥_{p} ≤ (∑_{k=0}^{n−1} ∥f_{k}∥_{p}^{p′})^{1/p′}  (2)

for 2 < p < ∞, but this is also false using the same example:

∥f_{0} + f_{1}∥_{p} = 2N^{2/p}, (∥f_{0}∥_{p}^{p} + ∥f_{1}∥_{p}^{p′})^{1/p′} = 2^{1/p′}(N^{2} + N^{p})^{1/p},

for sufficiently small N.

The problem with both of these “interpolations” is that the inequality

∥f_{0} + … + f_{n−1}∥_{2} ≤ (∑_{k=0}^{n−1} ∥f_{k}∥_{2}^{2})^{1/2}

does not hold for all f_{j} ranging over Banach spaces, but is subject to a constraint which is not amenable to interpolation.

However, substitutes exist when the orthogonality is made more robust. For instance, both (1) and (2) hold under the additional assumption that the |f_{j}| are constant multiples of characteristic functions (why?). Also, we have

Lemma 1.1. [1] Let R_{k} be a collection of rectangles in frequency space such that the dilates 2R_{k} are almost disjoint, and suppose that f_{k} are a collection of functions whose Fourier transforms are supported on R_{k}. Then for all 1 ≤ p ≤ ∞ we have

(∑_{k} ∥f_{k}∥_{p}^{p^{∗}})^{1/p^{∗}} ≲ ∥ ∑_{k} f_{k}∥_{p} ≲ (∑_{k} ∥f_{k}∥_{p}^{p_{∗}})^{1/p_{∗}},

where p_{∗} = min(p, p′), p^{∗} = max(p, p′).

References

[1] T. Tao, A. Vargas, A bilinear approach to cone multipliers II. Applications, to appear, GAFA.

Department of Mathematics, UCLA, Los Angeles, CA 90024

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