I am trying to solve Ch. 7, Exercise 17 (a) from Rudin's Functional Analysis. Let $u$ be a distribution on $\mathbb{R}^n$ with compact support $K$, and suppose the Fourier transform $\hat{u}$ is a bounded function. I am asked to show that if $n = 1$ or $n = 2$, then $\psi u = 0$ for all $\psi \in C^{\infty}(\mathbb{R}^n)$ which vanishes on $K$. A prior exercise shows this is false for $n = 3$.
I have proceeded by the hint, but I'm stuck on the final step where it appears I have to use the fact that $n \in \{1,2\}$.
Here's what I have so far. For $\varepsilon > 0$, let $H_\varepsilon$ be the set of points outside of $K$ with distance less than $\varepsilon$ to $K$. Let $\{h_\varepsilon\}_{\varepsilon > 0}$ be the standard approximation to the identity, where $h_\varepsilon$ is supported on $B_{\varepsilon}(0)$. Each $u \ast h_\varepsilon$ is a function, so we can take their norms. Now see that by the Plancherel theorem,
\begin{align*} ||u \ast h_\varepsilon||_2 &= ||\hat{(u \ast h_\varepsilon)} ||_2 \\ &= || \hat{u} \hat{h_\varepsilon} ||_2 \\ &\leq ||\hat{u}||_{\infty} ||h_\varepsilon||_2 \\ &= ||\hat{u}||_{\infty} \varepsilon^{-\frac{n}{2}}||h_1||_2. \end{align*}
Now let $\psi \in C^\infty(\mathbb{R}^n)$ which vanishes on $K$. Note that the support of $(u \ast h_\varepsilon)\psi$ is contained in $H_\varepsilon$. Hence,
\begin{align*} \left|\int_{\mathbb{R}^n} (u \ast h_\varepsilon)(x)\psi(x)dx\right| &= \left|\int_{H_\varepsilon} (u \ast h_\varepsilon)(x)\psi(x)dx\right| \\ &\leq ||u \ast h_\varepsilon||_2 \left( \int_{H_\varepsilon} |\psi(x)|^2dx \right)^{\frac{1}{2}} \\ &\leq ||\hat{u}||_{\infty} ||h_1||_2 \left( \varepsilon^{-n} \int_{H_\varepsilon} |\psi(x)|^2dx \right)^{\frac{1}{2}}. \end{align*}
Taking $\varepsilon \to 0$ on the outermost sides of this inequality yields \begin{align*} |u(\psi)| \leq |\hat{u}||_{\infty} ||h_1||_2 \liminf_{\varepsilon \to 0}\left( \varepsilon^{-n} \int_{H_\varepsilon} |\psi(x)|^2dx \right)^{\frac{1}{2}}. \end{align*}
To complete the exercise, it suffices to show that the righthand quantity is zero, when $n = 1$ or $n = 2$, but not necessarily when $n \geq 3$. This boils down to estimating the measure of $H_\varepsilon$, which is on the order of $O(\varepsilon^{n-1})$. But this leaves me with a quantity on the order of $O(\frac{1}{\varepsilon})$, which doesn't tend to zero. I need to find a finer estimate which is somehow based on $n$, but I'm not sure where to proceed from here.
If $x\in H_\epsilon$, there is $y\in K$ such that $|x-y|\le\epsilon$. Since $\psi=0$ on $K$, you can use Lipschitz continuity in a big ball to estimate $|\psi(x)|=|\psi(x)-\psi(y)|\le M|x-y|\le M\epsilon$, so now you can estimate the expression inside parentheses by $\epsilon^{-n}M^2\epsilon^2|H_\epsilon|$. This goes to zero for $n=1,2$. Note that if $K$ is a ball of radius 1 then $|H_\epsilon|$ is up to a constant $(1+\epsilon)^n-1$ which goes to zero like $\epsilon$ and not $\epsilon^{n-1}$. This is why the proof fails for $n$ bigger than $2$.