$\lim_{|x|\rightarrow\infty}(f\cdot g)(x)=0$

Question

suppose $f,g\in L^1(\mathbb{R}^d)$ and $g$ bounded. then $\lim_\limits{{|x|\to\infty}}(f\cdot g)(x)=0$?

since $|(f\cdot g)(x)|=|\int_{\mathbb{R}^d}f(x-y)\,g(y)\,dy|\leq ||g||_\infty\int_{\mathbb{R}^d}|f(x-y)|\,dy=||g||_\infty||f||_1<\infty$,

at least the limit is bounded. I guess I need to know if $\int_{\mathbb{R}^d}|f(x-y)|\,dy$ goes to $0$ as $|x|\rightarrow \infty$, but I got nothing until now.

if $x$ moves finite distance, then the integral won't change. would it be different near $\infty$?

Answer 1

The convolution is both uniformally continuous and $L^1$.

To see this, note the other answer and, for every $x,h\in \mathbb{R}$ we have

$$|(f*g)(x)-(f*g)(x-h)| = |(f*g)(x)- (f*g)_h(x)| = |(f*g)(x)-(f_h*g)(x)|$$ Next, by lineary we have $(f*g)(x)-(f_h*g)(x) = ((f-f_h)*g )(x)$ and finally

$$|(f*g)(x)-(f*g)(x-h)| \leq \|f-f_h\|_{L^1} \|g\|_{L^\infty}$$ which tends to $0$ as $h\rightarrow 0$. since $x$ was arbitrary, we have uniform convergence and so the tail must decay.

Answer 2

here is my idea: note that for summable $f,g$ the following equations holds: $$\int_{\mathbb R^d}|f*g| \space dx = \bigg(\int_{\mathbb R^d}|f| dx\bigg)\bigg(\int_{\mathbb R^d}|g| dx\bigg) < \infty $$ this follows easily with Fubini. Now assume $lim_{|x| \rightarrow \infty}f*g \space(x) \neq 0$ and your left hand side would be an unbounded integral.

Answer 3

We have \begin{align} (f\ast g)(x) &= (g\ast f)(x) \\ &= \int_{\mathbb{R}^n}g(x-y)f(y)dy \\ &= \int_{B_R(x)}g(x-y)f(y)dy + \int_{\mathbb{R}^n \setminus B_R(x)}g(x-y)f(y)dy \\ &= \int_{B_R(x)}g(x-y)f(y)dy + \int_{\mathbb{R}^n \setminus B_R(x)}g(x-y)f(y)1_{\{ f\leq k\}}dy +\int_{\mathbb{R}^n \setminus B_R(x)}g(x-y)f(y)1_{\{ f > k\}}dy, \end{align} where $B_R(x)$ denotes the closed ball with radius $R>0$ and center $x\in\mathbb{R}^n$ and $k>0$ is also arbitrary.

We can estimate the third term as $$\lvert \int_{\mathbb{R}^n \setminus B_R(x)}g(x-y)f(y)1_{\{ f > k\}}dy\rvert\leq \lVert g \rVert_\infty \int_{\mathbb{R}^n}|f(y)|1_{\{ f > k\}}dy $$ $f\in L^1$ yields $$1_{\{ f > k\}}\rightarrow 0,\; k\rightarrow \infty ,$$ and dominated convergence thus implies $$ \int_{\mathbb{R}^n}|f(y)|1_{\{ f > k\}}dy \rightarrow 0,\; k\rightarrow \infty.$$ Note that this convergence is independent of $x$ and $R$.

The second term is much easier to estimate: We write \begin{align} \lvert \int_{\mathbb{R}^n \setminus B_R(x)}g(x-y)f(y)1_{\{ f\leq k\}}dy \rvert &\leq k \int_{\mathbb{R}^n \setminus B_R(x)}|g(x-y)|dy \\ &= k \int_{\mathbb{R}^n \setminus B_R(0)}|g(y)|dy \end{align} Letting $R\rightarrow \infty$ yields convergence towards $0$ since $g\in L^1$. Note that this convergence is independent of $x$.

For the first term we have, similarily, \begin{align} \lvert \int_{B_R(x)}g(x-y)f(y)dy \rvert &\leq \lVert g \rVert_\infty \int_{B_R(x)}|f(y)|dy. \end{align} If $R$ is fixed, the RHS converges to $0$ for $|x|\rightarrow \infty$ since $f\in L^1$.

Now let $\varepsilon >0$. We first chose $k>0$ such that the third term is smaller than $\frac \varepsilon 3$. Remember that this choice is independent from $R$ and $x$, so we can now chose $R$ sufficiently big for the second term to be smaller than $\frac \varepsilon 3$. Since this choice was also independent of $x$, we can finally chose $x_\varepsilon\in\mathbb{R}$ such that the first term is smaller than $\frac \varepsilon 3$ for all $x\in\mathbb{R}^n$ with $\Vert x \Vert \geq x_\varepsilon$.