[proofplan] Choose an arbitrary Hermitian [inner product](/page/Inner%20Product) on $V$ and average its pullbacks by the representation over the compact group $G$ using normalized Haar measure. The averaged form is Hermitian by linearity of the integral and positive definite because the integrand is continuous, non-negative, and strictly positive near the identity whenever $v\ne 0$. Finally, bi-invariance of normalized Haar measure on compact groups shows that the averaged form is preserved by every $\rho(h)$. [/proofplan] [step:Average an arbitrary Hermitian inner product over the group] Let $(\cdot,\cdot)_0$ be a Hermitian inner product on $V$, linear in the first variable and conjugate-linear in the second. By the existence of normalized Haar measure on compact groups (citing a result not yet in the wiki: Existence and uniqueness of normalized Haar measure on compact groups), let $\mu$ denote the normalized Haar probability measure on $G$. For $v,w\in V$, define \begin{align*}(v,w)_V := \int_G (\rho(g)v,\rho(g)w)_0\,d\mu(g).\end{align*} For fixed $v,w\in V$, the map \begin{align*}F_{v,w}: G \to \mathbb C, \qquad g \mapsto (\rho(g)v,\rho(g)w)_0\end{align*} is continuous because $\rho:G\to GL(V)$ is continuous and $(\cdot,\cdot)_0$ is continuous on the finite-dimensional [vector space](/page/Vector%20Space) $V\times V$. Since $G$ is compact, $F_{v,w}$ is bounded and Borel measurable, so the integral defining $(v,w)_V$ is finite. [/step] [step:Verify that the averaged form is Hermitian] Let $u,v,w\in V$ and let $a,b\in\mathbb C$. Linearity of $(\cdot,\cdot)_0$ in the first variable and linearity of the integral give \begin{align*}(au+bv,w)_V = a(u,w)_V+b(v,w)_V.\end{align*} Conjugate-linearity in the second variable gives \begin{align*}(u,av+bw)_V = \overline{a}(u,v)_V+\overline{b}(u,w)_V.\end{align*} For conjugate symmetry, use conjugate symmetry of $(\cdot,\cdot)_0$ pointwise: \begin{align*}(w,v)_V = \int_G (\rho(g)w,\rho(g)v)_0\,d\mu(g) = \int_G \overline{(\rho(g)v,\rho(g)w)_0}\,d\mu(g) = \overline{(v,w)_V}.\end{align*} Thus $(\cdot,\cdot)_V$ is a Hermitian sesquilinear form. [/step] [step:Prove positive definiteness using continuity near the identity] Let $v\in V$ with $v\ne 0$. Since $\rho(e)=\operatorname{id}_V$, where $e\in G$ is the identity element, we have \begin{align*}(\rho(e)v,\rho(e)v)_0=(v,v)_0>0.\end{align*} Define \begin{align*}F_v:G \to [0,\infty), \qquad g \mapsto (\rho(g)v,\rho(g)v)_0.\end{align*} The function $F_v$ is continuous, and $F_v(e)>0$. Hence there exists an open neighbourhood $U\subset G$ of $e$ such that \begin{align*}F_v(g)>\frac{1}{2}F_v(e)\end{align*} for every $g\in U$. Haar measure on a compact group is positive on non-empty open sets, so $\mu(U)>0$. Therefore \begin{align*}(v,v)_V = \int_G F_v(g)\,d\mu(g) \ge \int_U F_v(g)\,d\mu(g) \ge \frac{1}{2}F_v(e)\mu(U) > 0.\end{align*} Thus $(\cdot,\cdot)_V$ is positive definite. [guided] We must check more than non-negativity. For each $g\in G$, the vector $\rho(g)v$ is non-zero whenever $v\ne 0$, because $\rho(g)\in GL(V)$ is invertible. This gives pointwise positivity, but the integral is strictly positive most cleanly by using continuity near the identity. Define \begin{align*}F_v:G \to [0,\infty), \qquad g \mapsto (\rho(g)v,\rho(g)v)_0.\end{align*} This map is continuous because $g\mapsto \rho(g)v$ is continuous and the inner product $(\cdot,\cdot)_0$ is continuous. At the identity element $e\in G$, the representation satisfies $\rho(e)=\operatorname{id}_V$, so \begin{align*}F_v(e)=(v,v)_0>0.\end{align*} By continuity, there is an open neighbourhood $U\subset G$ of $e$ such that \begin{align*}F_v(g)>\frac{1}{2}F_v(e)\end{align*} for every $g\in U$. The relevant Haar-measure fact is that normalized Haar measure on a compact group has full support, so every non-empty open subset has positive measure. Thus $\mu(U)>0$. Now integrate the non-negative function $F_v$. Restricting the integral from $G$ to $U$ lowers the value because $F_v\ge 0$ everywhere: \begin{align*}(v,v)_V = \int_G F_v(g)\,d\mu(g) \ge \int_U F_v(g)\,d\mu(g).\end{align*} On $U$, the lower bound for $F_v$ gives \begin{align*}\int_U F_v(g)\,d\mu(g) \ge \int_U \frac{1}{2}F_v(e)\,d\mu(g) = \frac{1}{2}F_v(e)\mu(U).\end{align*} Since both $F_v(e)>0$ and $\mu(U)>0$, the final quantity is strictly positive. Hence $(v,v)_V>0$ for every non-zero $v\in V$, which is exactly positive definiteness. [/guided] [/step] [step:Use Haar invariance to prove that every representation operator preserves the averaged form] Fix $h\in G$ and $v,w\in V$. Since $\rho$ is a [group homomorphism](/page/Group%20Homomorphism), \begin{align*}(\rho(h)v,\rho(h)w)_V = \int_G (\rho(g)\rho(h)v,\rho(g)\rho(h)w)_0\,d\mu(g) = \int_G (\rho(gh)v,\rho(gh)w)_0\,d\mu(g).\end{align*} Normalized Haar measure on a compact group is bi-invariant, so it is invariant under the right translation \begin{align*}R_h:G \to G, \qquad g \mapsto gh.\end{align*} Changing variables by $k=gh$, with $d\mu(k)=d\mu(g)$, gives \begin{align*}\int_G (\rho(gh)v,\rho(gh)w)_0\,d\mu(g) = \int_G (\rho(k)v,\rho(k)w)_0\,d\mu(k) = (v,w)_V.\end{align*} Therefore \begin{align*}(\rho(h)v,\rho(h)w)_V=(v,w)_V\end{align*} for every $h\in G$ and all $v,w\in V$. Thus each $\rho(h)$ is unitary with respect to $(\cdot,\cdot)_V$, completing the proof. [/step]