Need help with Exercise 14, Chapter 6 of Rudin's Functional Analysis

Question

The exercise goes:

Suppose $K$ is the closed unit ball in $\mathbb{R}^n$, $\Lambda\in\mathcal{D}'(\mathbb{R}^n)$ has its support in K, and $f\in C^\infty(\mathbb{R}^n)$ vanishes on $K$. Prove that $f\Lambda = 0$. Find other sets $K$ for which this is true. (Compare with Exercise 12.) (Exercise 12 basically says that the statement isn't true when $K=\{0\}$ and $\Lambda=\delta'$)

I can solve the first half of the question by showing that $\Lambda\phi=0$ whenever $\phi\in\mathcal{D}(\mathbb{R}^n)$ and $\phi$ vanishes on $K$, and that every such $\phi$ is the limit of a sequence of test functions $\{\phi_i(x)=\phi((1-|\epsilon_i|)x)\}$ when $\epsilon_i\rightarrow0$, whose supports don't intersect $K$.

I have difficulty with the second half. Here are my thoughts: I'm guessing the statement is true when $K=\overline{K^\mathrm{o}}$. In this case $\phi=0$ on $K$ implies $D^\alpha\phi=0$ on $K$ for every multi index $\alpha$, so I can find a similar sequence of test functions $\{\phi_i=g_i\phi\}$ as in the first part, where $\{g_i\}$ are smooth functions taking $0$ near $K$ and taking $1$ far from $K$. The existence of such $\{g_i\}$ can be shown by a smooth version of Urysohn's lemma. I can't prove this is a necessary condition, though.

Answer 1

Necessary and Sufficient condition: $K$ has no isolated point and $|K|>1$.

For simplicity of notation I will do for 1-dimension which we can easily extend.

For sufficient condition, let $f\in C_c^\infty(\mathbb R)$ with $f\equiv 0$ on $K$ then for any $x\in K$ there exists a sequence $x_n\in K/\{x\}$ such that $x_n\to x.$ Now, using Mean value theorem or Rolle's theorem on the line joining $x,x_n$, we will get $\theta_n$ such that $$ f'(\theta_n)=0$$ and $\theta_n\to x$. Now by continuity of $f'$, $f'(x)=0$ hence, $f'\equiv 0$ on $K$. Like that we can show, $f^n\equiv 0$ on $K$ for every $n$. Now, using Exercise 19 of Rudin's Functional Analysis, we got $\Lambda(f)=0$. That means $f\Lambda=0.$

To prove to the necessary condition, if possible let $K$ has an isolated point say $p\in K$. Then, there exists $r>0$ such that $B_r(p)$ does not intersect $K$. Using $C^\infty$ Urysohn lemma, consider an $\phi\in C_c^\infty(\mathbb R)$ such that $$\phi(x)=\begin{cases}1:\qquad x\in B_{\frac{r}{2}}(p)\\0:\qquad x\in\mathbb R-B_r(p).\end{cases}$$Now, take $f(x)=(x-p)\phi(x)$. Consider, $\Lambda=\delta'_p$ which has support $\{p\}\subseteq K$. Now, $f\equiv 0$ on $K$ but $\delta_p'(f)=1$ that is $f\delta_p'\ne 0.$ So, $K$ should not have any isolated point.

Edit: I will need little modification of the set for higher dimension. The above description of the set only works for 1-dimension. For $n=2,$ take, $K= [-1,1]\times \{0\}$ and $\Lambda=\partial _y\delta_0$ and $f\equiv y$ on a small nbd around $[-1,1]$ and supported on a bigger disk. Then, $\partial_y\delta_0(f)\ne 0$ but $f=0$ on $k=[-1,1]\times \{0\}$.