# Understanding the proof of the Hausdorff-Young theorem

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)?

• $\widetilde{C}\geq 1$ 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. – Yanko May 11 '19 at 19:42

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 $$1 such that $$\|\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'$$.