Approximation of compactly supported continuous functions that conserve the integral value

Question

Say $f\in L^p((0,1))$ is a function satisfying $$ \int_0^1 f = 0 $$ Is it possible to find a sequence $(f_k)$ of compactly supported continuous function also satisfying $$ \int_0^1 f_k = 0 $$ such that $f_k \to f$ in $L^p((0,1))$?

I was wondering if this is possible but I don't know how one could go about proving it. Any suggestions are appreciated!

EDIT: If it helps, I am willing to add some assumptions on $f$. For instance, that $f$ is continuous.

Answer 1

First make a sequence $f_n$ approximating $f$ (say $\lVert f-f_n\rVert_p<\frac1n$) such that

• $\int_0^1 f_n=0$, and
• $f_n$ is an $L^p(I_n)$ function extended by 0 to $L^p(\mathbb{R})$, where $I_n$ is a compact subset of $(0,1)$.

This is possible because $\displaystyle\int_0^1 f=\varinjlim_{K}\int_K f$ over compact subsets $K\subset(0,1)$, so for large enough compact subset $\int_K f$ is approximately $0$, so $f1_K-\frac{1}{\lvert K\rvert}(\int_K f)$ is close to $f$ with total integral $0$.

Now $I_n\subset[\delta_n,1-\delta_n]$ for some $\delta_n>0$, so you can mollify $f_n$ to get a smooth function $g_n=f_n\ast\varphi_{\varepsilon_n}$ supported on $[\frac12\delta_n,1-\frac12\delta_n]$ which satisfies $\lVert g_n-f_n\rVert_p<\frac1n$. So $g_n$ is a compactly supported function approximating $f$ in $L^p$.

Finally, we need to check $g_n$ has total integral $0$, but that is a property of mollification: $$\require{cancel} \begin{align*} \int_0^1 g_n(x)\,\mathrm{d}x &=\int_\mathbb{R} (f_n\ast\varphi_{\varepsilon_n})(x)\,\mathrm{d}x\\ &= \int_\mathbb{R}\int_\mathbb{R}\varphi_{\varepsilon_n}(y)f_n(x-y)\,\mathrm{d}y\,\mathrm{d}x\\ &=\int_\mathbb{R}\varphi_{\varepsilon_n}(y)\cancel{\color{red}{\int_\mathbb{R}f_n(x-y)\,\mathrm{d}x}}\,\mathrm{d}y\\ &=0 \end{align*} $$ as desired.