0
$\begingroup$

I'm having some trouble understanding Terence Tao's proof of the Hausdorff-Young theorem in his lectures notes 1. The theorem states

Proposition 3.1. Let $B$ denote the open unit ball in $\mathbb R^n$ and let $1\leq p,q\leq\infty$ such that $\|\hat f\|_{L^q(B)}\leq C\|f\|_{L^p(\mathbb R^n)}$ for all test functions $f\in\mathcal S(\mathbb R^n)$ with support in $B$. Then we have $p\in[1,2]$ and $q\leq p'$, where $p':=p/(p-1)$ denotes the conjugate index of $p$.

In the proof of the proposition prior to this one (where $B=\mathbb R^n$), we showed that the inequality $\|\hat f\|_{L^q(\mathbb R^n)}\leq C\|f\|_{L^p(\mathbb R^n)}$ gives us $$\lambda^n\lambda^{-n/q}\leq\widetilde C\lambda^{n/p}\qquad\text{for all }\lambda>0.$$ By rearranging this inequality we obtain $\lambda^{1-1/p-1/p}\leq\widetilde C^{1/n}$ for all $\lambda>0$, which is only possible, if $\widetilde C\geq1$ and $1-1/p-1/q=0$, where the latter is equivalent to $q=p'$.

Now, we want to utilize the same trick in Proposition 3.1. to show $q\leq p'$. Since we are restricted to the unit ball $B$, we can make $\lambda$ only so large before the support of $\psi(\cdot/\lambda)$ leaves $B$. Thus, we only have $$\lambda^n\lambda^{-n/q}\leq\widetilde C\lambda^{n/p}\qquad\text{for all }\lambda\in(0,b]$$ for some $b\geq1$ (we can choose $\psi$ such that $b>1$). Rearranging this inequality now gives us $\lambda^{n-q/n-p/n}\leq\widetilde C$ for all $\lambda\in(0,b]$. And if $\widetilde C\geq1$, it follows that $n-\frac nq-\frac np\geq0$. But using this inequality I always get $p'\leq q$ and not $q\leq p'$. Am I missing something or is this approach flawed? If so, do you know of another proof of this theorem (or the $q\leq p'$ part at least)?

$\endgroup$
  • $\begingroup$ $\widetilde{C}\geq 1$ doesn't imply that $n-\frac{n}{q}-\frac{n}{p}\geq 0$. However if $\widetilde{C}\leq 1$ then you get the inequality in the other direction which is what you need. $\endgroup$ – Yanko May 11 '19 at 19:42
0
$\begingroup$

As it turns out it is completely irrelevant which values $\widetilde C$ can attain, it is far more important which values $\lambda$ attains.

Proposition: Let $U\subseteq\mathbb R^n$ be an open set and let $\Lambda:=\{\lambda>0|\lambda U\subseteq U\}$ have a limiting point at $\lambda=0$. If $1<p,q<\infty$ such that $\|\hat f\|_{L^q(U)}\leq C\|f\|_{L^p(\mathbb R^n)}$ for all $f\in\mathcal S(\mathbb R^n)$, we have $q\leq p'$ (and $p\in[1,2]$, which I'm not gonna prove).

Proof: Choose some $\psi\in\mathcal S(\mathbb R^n)$ such that $\|\hat\psi\|_{L^q(U)}>0$ and any $\lambda\in1/\Lambda$. For $f:=\psi(\cdot~/~\lambda)$ we have $\hat f(\xi)=\lambda^n\hat\psi(\lambda\xi)$, and thus $$ \lambda^n\lambda^{-n/q}\|\hat\psi\|_{L^q(U)} =\|\hat f\|_{L^p(\lambda^{-1}U)} \overset{\lambda^{-1}\in\Lambda}\leq\|\hat f\|_{L^q(U)} \leq C\|f\|_{L^p(\mathbb R^n)} =C\lambda^{n/p}\|\psi\|_{L^p(\mathbb R^n)} $$ which is equivalent to $$ \lambda^{n-\frac nq-\frac np}\leq\underbrace{C\|\hat\psi\|_{L^q(U)}\|\psi\|_{L^p(\mathbb R^n)}}_{\text{independent of }\lambda}\in[1,\infty[. $$ Since $\lambda\in1/\Lambda$ can be chosen arbitrarily large, we have $n-\frac nq-\frac np\leq0$ giving us $q\leq p'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.