I want to prove the following statement. Let $$ f,\,g\in L_1(\mathbb{R}). $$ Also $$ \forall x<0 \quad f(x)=g(x)=0 $$ and $$ \forall x\in\mathbb{R}\quad f*g=0, $$ where $*$ denotes a convolution. Then $f\equiv0$ or $g\equiv0$ almost everywhere.

It is advised to prove it using Fourier transform. I have read through this question: convolution of non-zero functions, but I can't figure it out if I can apply this techinque to my problem, because I don't have compact support.

As advised in the comments, I note that this variation should be proved without the application of Titchsmarsh theorem.

  • $\begingroup$ reading en.wikipedia.org/wiki/Titchmarsh_convolution_theorem it seems like your variation follows from it, have you tried that? $\endgroup$ – supinf Jun 1 '17 at 12:17
  • $\begingroup$ @supinf yes, but then I have to prove the theorem itself. $\endgroup$ – Michael Freimann Jun 1 '17 at 12:18
  • $\begingroup$ ok i understand. Maybe you should add to your question that you are not allowed to use that theorem itself $\endgroup$ – supinf Jun 1 '17 at 12:19
  • $\begingroup$ @supinf edited the question $\endgroup$ – Michael Freimann Jun 1 '17 at 12:21

Lemma If convolution of two finite functions is zero, then one of these functions is zero almost everywhere.

Proof. Can be found here https://math.stackexchange.com/questions/1024340/convolution-of-non-zero-functions?noredirect=1&lq=1

Now take an arbitrary $A>0$ and assume $$ f_A=f\cdot\chi_{[0;A]} $$ and $$ g_A=g\cdot\chi_{[0;A]} $$ Observe that for $y\leq A$ $$ 0=f*g(y)=\int_{-\infty}^{+\infty}f(x)g(y-x)\;dx=\int_{0}^{y}f(x)g(y-x)\;dx=\int_{0}^{y}f_A(x)g_A(y-x)\;dx=\int_{-\infty}^{+\infty}f_A(x)g_A(y-x)\;dx=f_A*g_A(y). $$ Thus, $f_A$ or $g_A$ is zero almost everywhere (apply the Lemma), meaning that $f$ or $g$ is almost everywhere zero on every interval, therefore, it vanishes almost everywhere on $\mathbb{R}$.


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