# Problem regarding vanishing set of convolution

Let $$f$$ vanishes on an open set containing 0. So there exists $$l>0$$ such that $$f$$ vanishes on $$B(0,2l).$$ So we can choose $$g\in C_c^\infty (\mathbb{R}^n)$$ (supported on $$B(0,l)$$) such that $$f*g$$ vanishes on an open set ( vanishes on $$B(0,l)$$).

My question: Is it remains true if we replace open set by set of positive Lebesgue measure i.e. if $$f$$ vanishes on an positive Lebesgue measure set around 0, then can we find a nonzero $$g\in C_c^\infty (\mathbb{R}^n)$$ such that $$f*g$$ vanishes on a set of positive Lebesgue measure?

• Yes, e.g. we can take $g$ identically 0 Commented Oct 30, 2021 at 19:27
• I have edited the question, adding the condition for $g$ to be nonzero. Commented Nov 2, 2021 at 12:41

## 1 Answer

As noted in Pietro Majer's comment, you can trivially take $$g=0$$ to get the "yes" answer.

To avoid this and make the question less trivial, one may additionally require that the function $$g$$ be nonnegative and nonzero. Then the answer becomes "no".

Indeed, e.g. let $$f:=1_{D},$$ where $$D:=\mathbb R\setminus C$$ and $$C$$ is a fat Cantor subset of the interval $$[0,1]$$, so that $$C$$ is a closed nowhere dense set of positive (Lebesgue) measure and hence $$D$$ is an everywhere dense open set.

So, $$f=0$$ on the set $$C$$ of measure $$|C|>0$$.

Now, take any nonnegative nonzero continuous function $$g\colon\mathbb R\to\mathbb R$$, so that for some real $$c>0$$ we have $$g\ge c$$ on a nonempty open interval $$I$$. Then, for any real $$x$$, $$(f*g)(x)=\int_{\mathbb R}f(x-y)g(y)\,dy \\ \ge c\int_I f(x-y)\,dy=c|D\cap(x-I)|>0,$$ because $$D$$ is an everywhere dense open set and hence $$D\cap(x-I)$$ is a nonempty open interval.

Thus, $$f>0$$ on $$\mathbb R$$.

• Good point, but what if all that is required of $g$ is that it is smooth, compactly supported, and not identically zero, which, most likely, was the original intention? Commented Nov 1, 2021 at 1:49
• @fedja : I think this is a good question. I did think about it, but at this point don't know an answer to it. Commented Nov 1, 2021 at 2:31
• Yes, I was looking for the answer for any non-zero $C_c^\infty$ function. But any way thanks for your efforts. Please update your answer if you able to make some progress. Commented Nov 2, 2021 at 8:11
• @Wilderness : This is indeed a good question, and I will have it in mind. Commented Nov 2, 2021 at 12:39