Let $G$ be a compact group and $\phi \in C(G)$. Given $f \in L^1(G)$, show that $$T_\phi f: G \to \mathbb{C}: g \mapsto \int_G \phi(gh^{-1})f(h) dh$$ is continuous. Here $dh$ is normalized Haar measure on $G$.
Attempt:
Let $g,g'\in G$. Then $$|\int_G \phi(gh^{-1})f(h) dh- \int_G \phi(g'h^{-1}) f(h) dh| \leq \int_G |\phi(gh^{-1}) - \phi(g'h^{-1})| |f(h)| dh$$ Somehow, I want to be able to estimate the factor $|\phi(gh^{-1})-\phi(g'h^{-1})|$ using continuity of $\phi$ but I don't quite see how this is possible.
If I was allowed to work with sequences, I could just apply DCT but here we need net continuity so that does not work as well.
Any help is appreciated!
First prove the following lemma (which basically says that $\phi$ is uniformly continuous).
For every $\epsilon > 0$ there exists an open neighborhood $V_{\epsilon}$ of $e$ in $G$ such that $|\phi(x_1) - \phi(x_2)| < \epsilon$ for every $x_1,x_2\in G$ satisfying $x_1x_2^{-1}\in V_{\epsilon}$.
Suppose now that $g' \in V_{\epsilon}g$ (note that $V_{\epsilon}g$ is an open neighborhood of $g$). Then $g'g^{-1}\in V_{\epsilon}$. Hence $$g'h^{-1}(gh^{-1})^{-1} = g'h^{-1}hg^{-1} = g'g^{-1} \in V_{\epsilon}$$ Thus $$|\phi(g'h^{-1}) - \phi(gh^{-1})| < \epsilon$$ Hence $$\int_G |\phi(gh^{-1}) - \phi(g'h^{-1})| |f(h)| dh < \epsilon \cdot \int_G |f(h)|dh = \epsilon \cdot ||f||_1$$