# Infinitely differentiable functions with compact support are dense in $L^p$

This problem is in Stein.

In $$L^p$$, $$1\leq p<\infty$$ on $$\mathbb{R}^d$$ with Lebesgue measure.

(a) Continuous function with compact support are dense in $$L^p$$. I have already proves this :)
(b) Infinitely differentiable functions with compact support are dense in $$L^p$$.

How proves (b)? I read that this hard...

The argument I know depends on a few facts.

• Consider the convolution $$(f*g)(x)=\int_{\mathbb R^d} f(t)\,g(x-t)\, dt.$$

• One can show that if $$f\in L^p$$ and $$g\in L^1$$, then $$\|f*g\|_p\leq\|f\|_p\|g\|_1.$$ so $$f*g\in L^p$$.

• If $$f\in L^p$$, $$g\in C^r(\mathbb R^d)$$ with compact support and $$D$$ is a mixed partial derivative of order $$m$$, then $$D(f*g)(x)=(f*Dg)(x).$$ In particular, $$f*g\in C^r(\mathbb R^d)$$. So if $$g\in C^\infty$$, so does $$f*g$$.

• Let $$f\in L^p(\mathbb R^d)$$, $$g\in L^1(\mathbb R^d)$$ with $$\int_{\mathbb R^d} g=1$$. Let $$g_n(x)=n^d\,g(nx)$$. Then $$\lim_{n\to\infty}\|f-f*g_n\|_p=0.$$

• Since the compactly supported functions are dense in $$L^p$$, we may assume that $$f$$ is compactly supported. Then the functions $$f*g_n$$ from above are compactly supported.

• Let $$f\in L^p(\mathbb R^d)$$. Take $$g$$ to be any compactly supported $$C^\infty$$ function with $$\int_{\mathbb R^d} g=1$$. Fix $$\varepsilon>0$$. There exists $$f_0\in L^p(\mathbb R^d)$$, with compact support, such that $$\|f-f_0\|_p<\varepsilon/2$$. By the above steps, we have that $$f_0*g_n\in C^\infty_c(\mathbb R^d)$$ and $$\|f_0-f_0*g_n\|_p<\varepsilon/2$$ for big enough $$n$$. Then $$\|f-f_0*g_n\|_p\leq\|f-f_0\|_p+\|f_0-f_0*g_n\|_p<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$ Thus we get that $$f$$ is a limit of compactly supported infinitely differentiable functions.

• Finally, it remains to construct a nonzero infinitely differentiable, compactly supported function. Start with $$h(t)=\begin{cases} e^{-1/x^2},&\ x>0\\ 0,&\ x\leq0\end{cases}$$ Then $$h\in C^\infty(\mathbb R)$$. Now, given any interval $$[a,b]$$, form $$h_{a,b}(x)=h(x-a)h(b-x).$$ Then $$h_{a,b}\in C^\infty(\mathbb R)$$ with support in $$[a,b]$$. Now given any box $$B=\prod_{j=1}^d[a_j,b_j]\subset \mathbb R^d$$, the function $$g_B(x)=h_{a_1,b_1}(x_1)\cdots h_{a_d,b_d}(x_d)$$ is nonzero, non-negative, $$C^\infty$$, and with support in $$B$$.

• I got stuck trying to prove the following... Could you help please? "Let $f\in L^p(\mathbb{R}^d), g\in L^1(\mathbb{R}^d)$ with integral 1, and $g_n(x):=n^dg(nx)$. Then $\lim ||f-f*g_n||_p=0$." Commented Mar 25, 2020 at 1:10
• I posted the details here. Commented Mar 25, 2020 at 3:22
• Very Clear and Helpful! Also for some who want to know more about this, you can check Folland's Real Analysis. Commented Oct 12, 2023 at 17:39