I am dealing with a question requiring me to prove that if $f,g\in L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|\rightarrow\infty$.
I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.
Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=\int f(x-y)g(y)dy$$ Then as $f\in L_{2}$ and $g\in L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fg\in L_{1}$, and $$|f*g(x)|\leq\int|f(x-y)g(y)|dy\leq\|f(x-y)\|_{2}\|g(y)\|_{2}=\|f(y)\|_{2}\|g(y)\|_{2}<\infty$$
The last equality is by translation invariance and reflection invariance.
Therefore, $|f*g(x)|$ is defined everywhere and bounded.
Now for continuity, let $x,z\in\mathbb{R}$, then consider $$|f*g(x)-f*g(z)|=\Big|\int f(x-y)g(y)dy-\int f(z-y)g(y)dy\Big|$$ $$\leq\int|g(y)(f(x-y)-f(z-y))|dy$$ $$\text{now by Holder and by translation invariance and reflection invariance}$$ $$\leq\|g\|_{2}\Big(\int|f(x-y)-f(z-y)|^{2}\Big)^{1/2}=\|g\|_{2}\Big(\int|f(-y)-f((z-x)-y)|^{2}\Big)^{1/2}$$ $$=\|g\|_{2}\Big(\int|f(y)-f(y-(z-x))|^{2}\Big)^{1/2}=\|g\|_{2}\|f_{h}(y)-f(y)\|_{2}\ \text{where}\ h=z-x$$
Therefore, by the translation continuity property, we know that $$\|f_{h}(y)-f(y)\|_{2}\rightarrow 0\ \text{as}\ z\rightarrow x$$
It follows immediately that $$|f*g(x)-f*g(z)|\rightarrow 0\ \text{as}\ x\rightarrow z$$.
I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.
Any idea will be really appreciated!
There exists a continuous function $h$ with compact support such that $\|f-h\|_2 <\epsilon$. We have $|(f*g)(x)| \leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $\leq \epsilon \|g\|_2$. Now consider $\int h(x-y)g(y)dy$. Choose a continuous function $\phi$ with compact suport such that $\|g-\phi\|_2 <\epsilon /M$ where $M=\|h\|_2$. Then $|\int h(x-y)g(y)dy|<\epsilon +|\int h(x-y)\phi(y)dy|$. Can you now show that $\int h(x-y)\phi(y)dy \to 0$ as $ x\to \infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].