Smoothing by convolution

Remarks

For $k = 0$, continuity of $f * g$ follows from the uniform continuity of $g$ and the integrability of $f$ on compact sets. The case $k \ge 1$ follows by induction from the derivative of a convolution, applied to the $C^{k-1}$ functions $\partial_j g$. The case $k = \infty$ is the conjunction of all finite cases.

Example: $f = \mathbf{1}_{[0, 1]}$ is discontinuous on $\R$. If $0 < \varepsilon < 1/2$ and $\varphi$ is a smooth function with support in $[-\varepsilon, \varepsilon]$ and integral $1$, then $f * \varphi$ is smooth, equals $1$ on $[\varepsilon, 1 - \varepsilon]$ and $0$ outside $[-\varepsilon, 1 + \varepsilon]$.

The smoothness comes only from $g$: if $g$ is merely continuous, $f * g$ need not be differentiable. For $f = g = \mathbf{1}_{[0, 1]}$, neither factor is continuous, yet $f * g$ is the continuous tent function, which is not differentiable at $0$, $1$, $2$.

Going deeper

For a student

With a mollifier $\varphi_\varepsilon$, the functions $f * \varphi_\varepsilon$ are smooth and tend to $f$ in $L^p$ as $\varepsilon \to 0$ when $f \in L^p$, $1 \le p < \infty$. This is the standard way to approximate integrable or continuous functions by smooth ones. It is used to prove density theorems, to define weak derivatives through smooth approximations, and to regularize distributions.