Must any $\phi \in \operatorname{Hom}_G(V, L^2(G))$ have continuous values? Let $G$ be a compact group and $V$ a finite-dimensional vector space with a continuous $G$-action. Consider a linear map $\phi: V \to L^2(G)$ satisfying that for any $v \in V, h \in G$:
$$
    \phi(v)(g h) = \phi(h \cdot v)(g) \quad \text{for almost all $g \in G$}
$$
Must $\phi(v)$ be continuous for any $v$?
This is used (implicitly) in Serganova's A Journey Through Representation Theory (Chapter 2, Lemma 2.3) to prove that matrix coefficients
$$
\begin{align}
    V^* \times V \times G &\to \mathbb{C} \\
    \alpha, v, g &\mapsto \alpha(g \cdot v)
\end{align}
$$
provide an isomorphism
$V^* \cong \operatorname{Hom}_G(V, L^2(G))$, and ultimately prove the Peter-Weyl theorem.
 A: The answer is yes, $\phi(v)$ is equal to a continuous function a.e. for all $v$.
Here is a proof.
The assumption about $\phi$ precisely says that it is a covariant map from $V$ to $L^2(G)$, where $L^2(G)$ is equipped with the right regular representation of $G$.
Therefore the range of $\phi$, henceforth denoted $R$, is an invariant subspace.
Splitting $R$ as a direct sum of irreducible subspaces we may assume WLOG that $R$ itself is irreducible.
By the Peter Weyl Theorem
(https://en.m.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem)
we have that $R$ is generated by
the matrix coefficients in $R$, which are known to be continuous.  Therefore $R$ is formed by continuous functions.

EDIT: Let me expand a bit on the assertion above.
The conceptual reason is the uniqueness of the decomposition of a unitary representation into irreducible ones.  To better
explain this let $\{\pi _i:i\in I\}$ be a set of representatives for the equivalence classes of irreducible
representations of $G$.
Decomposing $R^\perp$ into irreducibles, say $\bigoplus_{i\in I} n_iH_{\pi _i}$,  we have that
$$
  L^2(G) = R\oplus \bigoplus_{i\in I} n_iH_{\pi _i},
  $$
at the same time that we have the standard decomposition of $L^2(G)$ given by the Peter Weyl Theorem
$$
  L^2(G) = \bigoplus_{i\in I} d(\pi _i)H_{\pi _i}.
  $$
Thus, if $\pi _{i_0}$ is the class of the representation of $G$ on $R$, we have  that
$$
  R\oplus n_{i_0}H_{\pi _{i_0}} = d(\pi _{i_0})H_{\pi _{i_0}},
  $$
by uniqueness, and in particular
$
  R\subseteq  d(\pi _{i_0})H_{\pi _{i_0}}.
  $
Since $d(\pi _{i_0})H_{\pi _{i_0}}$ is spanned by the (continuous) matrix coefficients associated to $\pi _{i_0}$,  we conclude
that  $R$ is formed by continuous functions.

A more pedestrian approach  is as follows: denote by $\rho $ the representation of $G$ on $R$, and let
$\{e_i\}_{1\leq i\leq n}$ be an orthonormal basis for $R$.
We will show that each $e_i$ is orthogonal to every matrix
coefficient associated to any irreducible representation $\pi $ not equivalent to $\rho $.
By Peter-Weyl we will then deduce that each $e_i$ is a finite linear combination of  matrix coefficients associated to $\rho $,
which are continuous functions, thus
proving the $e_i$ to be continuous.
Observe that since the regular representation restricts  to $\rho $ on $R$,
for every $g$ and $h$ in $G$,  we have that
$$
  e_j(hg) =
  \rho_g e_j(h) =\sum_{i=1}^n u_{ij}(g)e_i(h),
  $$
where the $u_{ij}$ are the matrix coefficients of $\rho $ in the given basis.
Let $\pi $ be another irreducible representation of $G$ which is
inequivalent to $\rho $,  and let $x$ and $y$ be vectors in the space of $\pi $,
so that
$$
  c(g) := \langle x, \pi _{g}(y)\rangle
  $$
defines a matrix coefficient for $\pi $.  As we already said,  we will next prove  that $c$ is orthogonal
to each $e_j$.
By invariance of the Haar measure we have for every $g$ that
$$
  \langle c, e_j\rangle  =
  \int_G \overline{c(h)} e_j(h)\, dh =
  \int_G \overline{c(hg)} e_j(hg)\, dh = $$$$ =
  \sum_{i=1}^n   \int_G \overline{ c(hg)   } u_{ij}(g)e_i(h)\, dh =  \cdots
  $$
Observing that this  does not depend on $g$, we may integrate it against $g$.  After doing so and exchanging the order
of integration we see that the above equals
$$
  \cdots  =
  \sum_{i=1}^n   \int_G e_i(h)\left(\int_G \overline{  c(hg)  } u_{ij}(g)\, dg\right) \, dh.
  $$
Notice that the term within parenthesis  is the inner-product in $L^2(G)$ of the matrix coefficient
$$
  g\mapsto c(hg) = \langle x, \pi _{hg}(y)\rangle  = \langle \pi _{h^{-1}}(x), \pi _{g}(y)\rangle
  $$
by the matrix coefficient $u_{ij}$, so it vanishes by the Peter-Weyl orthogonality relations since $\pi $ and $\rho $ are
inequivalent.
