Understanding the proof of the Hausdorff-Young theorem

Question

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)?

Answer 1

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 $q\leq p'$.