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.
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 $\mathbb{R}$.