Convolution on the line

Remarks

If $f$ and $g$ lie in $L^1(\R)$, then $f * g$ is defined at almost every $x$, it lies in $L^1(\R)$, and $\norm{f * g}_1 \le \norm{f}_1 \norm{g}_1$. If $f \in L^p(\R)$ and $g \in L^q(\R)$ with $1/p + 1/q = 1$, then $f * g$ is defined everywhere and is bounded and uniformly continuous.

Example: $f(x) = g(x) = e^{-x^2}$. Completing the square gives $y^2 + (x - y)^2 = 2(y - x/2)^2 + x^2/2$, so $$(f * g)(x) = e^{-x^2/2} \int_\R e^{-2(y - x/2)^2} \, dy = \sqrt{\pi/2} \, e^{-x^2/2}.$$ The convolution of two Gaussians is again a Gaussian, wider than both.

Non-example: $f = g = 1$ on $\R$. Then $\int_\R \abs{f(y) g(x - y)} \, dy = \int_\R 1 \, dy = \infty$ for every $x$, so $f * g$ is defined nowhere.

The integral can also be written $\int_\R f(x - y) g(y) \, dy$, by the substitution $y \mapsto x - y$.