Take $f$ and $g$ $\in L^2(\mathbb R)$, then I want to show that $\lim_{h \to 0} \int f(x-y-h)g(y)dy = f \ast g(x)$. My idea is to first take $f$ to be a continuous function with a compact support, then $f$ has to be uniformly continuous. Then we know that $f \ast g$ has to be continuous by exchanging the order of the limit and integration. Then use the fact that $C_{c}(\mathbb R)$ is dense in $L^2(\mathbb R)$ to conclude the proof. Is this the right idea? Is there any simpler arguments?
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Let $f_n\to f$ in $L^2$. With $$ |((f_n-f)*g)(x)|\le\int|f_n(x-y)-f(x-y)||g(y)|\,dy $$ and Cauchy-Schwarz you see that $(f_n*g)(x)\to (f*g)(x)$.
For $h\in\Bbb R$ define the operator $T_h : L^2\to L^2$ by $(T_hf)(x) := f(x-h)$ and show that $T_hf\to f$ in $L^2$ as $h\to 0$. Then you are done because, by the above, $(T_hf*g)(x)\to (f*g)(x)$ as $h\to 0$.
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