** Translation Invariant Operators Fourier Multipliers Convolution Operators **

Theorem (Hardy–Littlewood–Sobolev inequality, Fractional Integration, Weak Young)Let and and . ThenIn particular,

for and .

Remark

- Convolution (generically) makes functions shorter and fatter. Consequently, if a convolution operator maps , then .
- If , then is not even a distribution!

*Personal Comment:* The Hedberg proof of this result was presented in a participating analysis seminar here at UCLA, and thus was not given in lecture. The transcription of that proof will be given shortly.

*Proof:* By Marcinkiewicz, if we prove then the result will still follow (we are necessarily dealing with an open set of exponents). (Indeed this weak-type statement is even true when -this is the -inequality in .) Given we split into

the relation between and will be given later.

Now, and

Therefore

Recall that we are trying to show . But

From the computation above, if we choose wisely we can make the second set empty, say

for some constant . We’re stuck trying to show

Well,

To conclude we note

As , we are done.

What is the Fourier multiplier associated to ? I.e., what is the (distributional) Fourier transform of ? Note that it must be spherically symmetric, and homogeneous of degree .

The left hand side of the last expression converges, distributionally, as to

and the right hand side converges in the same manner to

Thus we indeed have

where

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