# Approximating compactly supported $L^2$ functions with Schwartz functions “from within”?

Crossposted from MathOverflow.

It is well known that the class of Schwartz functions $$\mathcal{S}$$ in dense in all $$L^p$$ spaces therefore for each $$f \in L^2$$ there exists a sequence of Schwartz functions $$(f_k)$$ such that $$\lVert f - f_k \rVert_{L^2} \to 0$$ as $$k \to \infty$$.

If we suppose further that $$f$$ has compact support can we find a sequence of Schwartz functions $$(f_k)$$ such that $$\lVert f - f_k \rVert_{L^2} \to 0$$ as $$k \to \infty$$ (as above) and additionally $$\operatorname{supp}(f_k) \subseteq \operatorname{supp}(f)$$ for all $$k$$?

If so, what argument can I appeal to?

If $$E$$ is a compact set with positive measure and empty interior, and if $$f$$ is the characteristic function of $$E$$, then $$f$$ is nonzero in $$L^2$$ and there is no nonzero continuous function whose support is contained in the support of $$f$$.

To construct such a set $$E$$, find an enumeration of $$[0,1]\cap\mathbb Q$$, say $$\{q_n\}_{n=1}^\infty$$, and for each $$n$$ choose an open interval $$I_n$$ around $$q_n$$ with length $$\varepsilon /2^n$$.

Then $$U:= \bigcup I _n$$ is an open set whose measure is at most $$\varepsilon$$, and $$E:= [0,1]\setminus U$$ is a compact set with empty interior and measure at least $$1-\varepsilon$$.

• What was the need to repeat my answer ? – reuns Dec 6 '20 at 3:14

Not if $$f = \lim_{N\to \infty} f_N, \qquad f_N=1_{[0,1]} \prod_{n=1}^N \prod_{k=0}^{2^n-1} (1-1_{\displaystyle [\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}}]})$$ $$\lim_{N\to \infty} f_N$$ converges in $$L^p(\Bbb{R})$$, the limit is non-zero since $$\int_0^1 f_N(x)dx\ge 1-\sum_{n=1}^N 2^{-n-1}$$

The support of $$f$$ doesn't contain any $$k/2^n$$ so it doesn't contain any open set.

• This is what I call obscuring a very simple idea. – Black Dec 6 '20 at 4:47