How to show that such a function is continuous Let $G$ be a compact group with a fixed Haar measure $dy$. Let $F_1$ be a complex-valued continuous function on $G$. Set $F_2(x):=\int_G F_1(yxy^{-1})dy$ for $x\in G$. Show that $F_2$ is continuous. 

Could anyone give a hint? Thanks!
 A: Define the group $H = G \times G$ and the map $L : H \to \mathbb R$ by $L(x,y)= F_1(yxy^{-1})$
Cover $\mathbb R$ by balls of radius $\epsilon$. Each ball has an open preimage under $L$. Let $\mathcal U$ be the cover of $H$ by there preimages. By the Lebesgue number lemma there is an open neighborhood $U \subset H$ of the identity whose every translate is contained entirely in an element of $\mathcal U$. That means:
$(x,y)(a,b)^{-1} \in U \implies |F_1(yxy^{-1}) - F_1(bab^{-1})| < 2\epsilon$
That means there are neighborhoods $V$ and $W$ of the identity in $G$ such that. . . 
$(xa^{-1},yb^{-1}) \in V \times W \implies |F_1(yxy^{-1}) - F_1(bab^{-1})| < 2\epsilon$
In particular when $y=b$ we have. . . 
$xa^{-1} \in V \implies |F_1(yxy^{-1}) - F_1(yay^{-1})| < 2\epsilon$
Observe the LHS is independent of $y$.
Now integrate the RHS over $G$ to recover the definition of $F_2$ being continuous at $a \in G$.
A: Suppose $x_n \to x$. Then $F_2(x_n) = \int_G F_1(yx_ny^{-1})dy$. Since $F_1$ is continuous and $G$ is compact, say $|F_1(z)| \le C$ for all $z \in G$. The point is that since $C \in L^1(G)$ (since Haar measure on compact group is finite), we can use dominated convergence, noting $F_1(yx_ny^{-1}) \to F_1(yxy^{-1})$ for each $y$, by the continuity of $F_1$ and the group operations. So, $F_2(x_n) \to \int_G F_1(yxy^{-1})dy = F_2(x)$. 
Disclaimer: I might have cheated a bit. It might not be true that "if $x_n \to x$, then $F_2(x_n) \to F_2(x)$" implies $F_2$ is continuous. We might need to use nets...
