# Prove variation of Titchsmarsh convolution theorem

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.

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

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 $\mathbb{R}$.