Let $G$ be a locally compact Hausdorff unimodular topological group, and $\phi: G \rightarrow \mathbb C$ a continuous function. For any continuous and compactly supported function $f: G \rightarrow \mathbb C$, define
$$\rho(f)\phi(x) = \int\limits_G \phi(xy)f(y)dy.$$
The integral converges absolutely, since $y \mapsto \phi(xy)f(y)$ is a continuous function vanishing off of a compact set. Is $\rho(f)\phi$ actually a continuous function?
When $G$ is first countable, I can think of a very simple proof using the dominated convergence theorem. I wondered if there is an easy way to prove this when $G$ is not necessarily first countable.
Proof when $G$ is first countable:
In this case, continuity can be checked on sequences, so it suffices to show that for any $x_0 \in G$ and sequence $x_n \to x_0$, that we have $\lim\limits_n \rho(f)\phi(x_n) = \rho(f)\phi(x)$.
Let $\Omega$ be a compact set containing the support of $f$, so that
$$\rho(f)\phi(x) = \int\limits_{\Omega} \phi(xy)f(y)dy.$$
If we define a function $h_n: \Omega \rightarrow \mathbb C$ by $h_n(y) = \phi(x_ny)f(y)$, then there is a uniform bound $M$ on all the functions $h_n$ (since the $x_n$ and $y$ all belong to a compact set). Then $h_n(y)$ converges pointwise to $\phi(x_0y)f(y)$, with $|h_n| \leq M 1_{\Omega}$. The dominated convergence theorem implies that $$\lim\limits_n \int\limits_{\Omega} h_n(y)dy = \lim\limits_n \rho(f)\phi(x_n) = \int\limits_{\Omega} \phi(x_0y)f(y)dy.$$