Prove that a convolution product $(f\ast g)(x)$, with $f,g\in\mathcal{L}^2(\mathbb{R})$, goes to $0$ as $|x|$ goes to $\infty$.

Question

I want to proof that the convolution product $(f\ast g)(x)$ of two square-integrable functions $f,g\in\mathcal{L}^2(\mathbb{R})$ goes to $0$ as $|x|$ goes to $+\infty$. The convolution product here denotes

$$(f\ast g)(x)=\int_\mathbb{R}f(y)g(x-y)dy.$$

I know that, using Hölder's inequality,

$$|(f\ast g)(x)|\leq \int_\mathbb{R}|f(y)g(x-y)|dy\leq ||f||_2||g||_2$$

but I do not immediately see how this can help me to finalise my proof.

Answer 1

To do this question, you'd better know some density theorem of $L^p$ space. We will use a theorem which shows for an $L^2(\mathbb{R})$-function, there exists a sequence of simple measurable functions $f_k$ with bounded support such that $f_k\to f$ in $L^2(\mathbb{R})$ and $f_k$'s can be chosen to be in $L^2(\mathbb{R})$.

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Since simple function with bounded support is dense in $L^2(\mathbb{R})$, there exists $f_k\to f$ in $L^2$, where $f_k$'s are simple functions with bounded support in $L^2$. This shows $$|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|$$ Similarly, we can find a sequence of simple function $g_k$ with bounded support and $g_n\to g$ in $L^2$. Then, $$\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|\leq \lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$ Note that $f_k(x-y)g_n(y)=0$ for large enough $|x|$. This is because if the radius of the support of $f_k$ is $r_k$ and the radius of support of $g_n$ is $R_n$, then if $|x|>r_k+R_n$, either $|y|>R_n$ or $|x-y|>r_k$, so either $f_k(x-y)=0$ or $g_n(y)=0$. This implies that $$|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$ First take $|x|\to\infty$ on both sides, we have $$\lim_{|x|\to\infty}|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2$$ Then take $n\to\infty$ on both sides, since LHS is independent of $n$, we have $$\lim_{|x|\to\infty}|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2$$ Finally, take $k\to\infty$ on both sides, since LHS is independent of $k$, we obtain $\lim_{|x|\to\infty}|(f*g)(x)|\leq 0$, i.e., $(f*g)(x)\to 0$ as $|x|\to \infty$.

Answer 2

If we have: $$I(x)=\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$ this denotes our convolution. Now we can say that we want $|x|\to\infty$ so we can write this as: $$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$ we can make the substitution $u=x-y,du=-dy$ and so: $$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x+n}^{x-n}f(x-u)g(u)du=-\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(x-u)g(u)du$$ also: $$I_{\infty}=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(-y)g(x+y)dy$$ $$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(u-x)g(u)du=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(-(x-u))g(u)du$$ this suggests that $f(-(x-u))=-f(x-u)$ and since we are talking about $|x|\to\infty$ we can say that $f(-)=-f(+)$. If $g$ is an even function then we have an odd integrand over the real domain which will equal zero.