[proofplan] Starting from any Hermitian inner product $(\cdot, \cdot)$ on $V$, we average it against the action of $G$ using the normalised Haar measure: $\langle v, w \rangle := \int_G (\rho(g)v, \rho(g)w)\, dg$. The averaged form is well-defined because the integrand is continuous on the compact group $G$, sesquilinear in $(v, w)$ because integration is linear in the integrand, Hermitian because the original form is, positive definite because the integrand is non-negative and strictly positive at $v = w \neq \mathbf{0}$, and $G$-invariant because the Haar measure is left-invariant. The compactness of $G$ is essential: without it the integral could diverge. [/proofplan] [step:Fix data and define the averaged form $\langle \cdot, \cdot \rangle$] Let $G$ be a compact group with normalised Haar measure $\mu$ (so $\mu(G) = 1$) and let $(\rho, V)$ be a continuous finite-dimensional representation of $G$ over $\mathbb{C}$. Fix any Hermitian inner product \begin{align*} (\cdot, \cdot): V \times V \to \mathbb{C} \end{align*} on $V$ (e.g. the standard inner product after choosing a basis). Define \begin{align*} \langle \cdot, \cdot \rangle: V \times V &\to \mathbb{C}, \\ (v, w) &\mapsto \int_G (\rho(g)v,\, \rho(g)w)\, d\mu(g). \end{align*} We must check that this definition makes sense: that the integral converges. For fixed $v, w \in V$, the map \begin{align*} F_{v, w}: G &\to \mathbb{C}, \\ g &\mapsto (\rho(g)v,\, \rho(g)w) \end{align*} is continuous: $g \mapsto \rho(g)$ is continuous from $G$ into $\operatorname{End}(V)$ by the assumption that $\rho$ is a continuous representation, the maps $A \mapsto Av$ and $A \mapsto Aw$ are continuous from $\operatorname{End}(V)$ to $V$ (linear maps on finite-dimensional spaces are continuous), and the inner product $(\cdot, \cdot): V \times V \to \mathbb{C}$ is continuous (sesquilinear forms on finite-dimensional spaces are continuous). Since $G$ is compact, $F_{v, w}$ is bounded, hence integrable against the finite measure $\mu$. So $\langle v, w \rangle \in \mathbb{C}$ is well-defined. [/step] [step:Verify sesquilinearity and Hermitian symmetry of $\langle \cdot, \cdot \rangle$] *Linearity in the first slot.* For $v_1, v_2, w \in V$ and $\alpha_1, \alpha_2 \in \mathbb{C}$, sesquilinearity of $(\cdot, \cdot)$ in its first slot together with linearity of $\rho(g)$ gives \begin{align*} (\rho(g)(\alpha_1 v_1 + \alpha_2 v_2),\, \rho(g)w) = \alpha_1 (\rho(g)v_1, \rho(g)w) + \alpha_2 (\rho(g)v_2, \rho(g)w). \end{align*} Integrating against $\mu$ and using linearity of the integral, \begin{align*} \langle \alpha_1 v_1 + \alpha_2 v_2,\, w \rangle = \alpha_1 \langle v_1, w \rangle + \alpha_2 \langle v_2, w \rangle. \end{align*} *Conjugate-linearity in the second slot.* The analogous argument applies, using that $(\cdot, \cdot)$ is conjugate-linear in its second slot. *Hermitian symmetry.* For $v, w \in V$, Hermitian symmetry of $(\cdot, \cdot)$ gives $(\rho(g)v, \rho(g)w) = \overline{(\rho(g)w, \rho(g)v)}$. Integrating and using that integration commutes with complex conjugation (the integral of a real-valued function is real, the integral of an imaginary-valued function is imaginary, and Haar measure is a positive real measure), \begin{align*} \langle v, w \rangle = \int_G (\rho(g)v, \rho(g)w)\, d\mu(g) = \overline{\int_G (\rho(g)w, \rho(g)v)\, d\mu(g)} = \overline{\langle w, v \rangle}. \end{align*} [/step] [step:Verify positive definiteness using continuity of $g \mapsto (\rho(g)v, \rho(g)v)$] Fix $v \in V$. The map $f_v: G \to \mathbb{R}_{\geq 0}$, $g \mapsto (\rho(g)v, \rho(g)v)$, is continuous (as in Step 1) and non-negative (positive-definiteness of $(\cdot, \cdot)$). Therefore \begin{align*} \langle v, v \rangle = \int_G f_v(g)\, d\mu(g) \geq 0. \end{align*} Suppose $\langle v, v \rangle = 0$ but $v \neq \mathbf{0}$. Evaluating $f_v$ at $g = e$ (the identity of $G$), \begin{align*} f_v(e) = (\rho(e)v, \rho(e)v) = (v, v) > 0 \end{align*} since $\rho(e) = \mathrm{id}_V$ and $(\cdot, \cdot)$ is positive definite. By continuity of $f_v$, there exists an open neighbourhood $U \subseteq G$ of $e$ with $f_v(g) \geq f_v(e)/2 > 0$ for $g \in U$. Then \begin{align*} \int_G f_v\, d\mu \geq \int_U f_v\, d\mu \geq \frac{f_v(e)}{2} \cdot \mu(U). \end{align*} Haar measure on a compact group is strictly positive on every nonempty open set: this is a standard property (the support of Haar measure on $G$ is all of $G$). Hence $\mu(U) > 0$, and the right-hand side is strictly positive, contradicting $\langle v, v \rangle = 0$. Therefore $\langle v, v \rangle > 0$ for every $v \neq \mathbf{0}$, completing positive definiteness. [guided] We need to rule out the pathology $\langle v, v \rangle = 0$ with $v \neq \mathbf{0}$. The non-negative integrand $f_v(g) = (\rho(g)v, \rho(g)v)$ has $\int_G f_v\, d\mu = 0$ only if $f_v = 0$ $\mu$-almost everywhere on $G$. Why does this fail when $v \neq \mathbf{0}$? At the identity, $f_v(e) = (v, v) > 0$, and by continuity of $f_v$ this strict positivity persists in a small neighbourhood $U \ni e$. The set $U$ is open and nonempty, so it has *positive Haar measure* — this is a key property of Haar measure on a compact (in fact, locally compact) group: every nonempty open set has positive measure. Hence the integral is bounded below by $\frac{f_v(e)}{2} \mu(U) > 0$, contradicting $\langle v, v \rangle = 0$. This is the only place continuity of $\rho$ is consumed: without continuity, $f_v$ could vanish on a set of full measure but be nonzero on a $\mu$-null set, breaking positive definiteness. [/guided] [/step] [step:Verify $G$-invariance using left-invariance of Haar measure] Let $h \in G$ and $v, w \in V$. Substituting $g' := hg$ (so $g = h^{-1} g'$) and using *left-invariance* of $\mu$, i.e., $d\mu(hg) = d\mu(g)$ (equivalently, $\int_G f(hg)\, d\mu(g) = \int_G f(g)\, d\mu(g)$ for all integrable $f$), \begin{align*} \langle \rho(h)v, \rho(h)w \rangle &= \int_G (\rho(g)\rho(h)v,\, \rho(g)\rho(h)w)\, d\mu(g) \\ &= \int_G (\rho(gh)v,\, \rho(gh)w)\, d\mu(g), \end{align*} using the homomorphism property $\rho(g)\rho(h) = \rho(gh)$. Performing the change of variable $g \mapsto gh^{-1}$ (right translation) and invoking *right-invariance* of Haar measure on the compact group $G$ (compact groups are unimodular: every left Haar measure is also right-invariant), \begin{align*} \int_G (\rho(gh)v,\, \rho(gh)w)\, d\mu(g) = \int_G (\rho(g)v,\, \rho(g)w)\, d\mu(g) = \langle v, w \rangle. \end{align*} Hence $\langle \rho(h)v, \rho(h)w \rangle = \langle v, w \rangle$ for all $h \in G$ and all $v, w \in V$, which is the required $G$-invariance. [guided] The substitution $g \mapsto gh^{-1}$ uses *right-invariance* of Haar measure, while the symbol $d\mu(g)$ is the left Haar measure. Why is right-invariance available? A topological group is *unimodular* if its left Haar measure is also right-invariant; equivalently, the modular function $\Delta: G \to \mathbb{R}_{>0}$ is identically $1$. Compact groups are unimodular: the modular function is a continuous homomorphism $G \to \mathbb{R}_{>0}$ from a compact group to a non-compact group, so its image is a compact subgroup of $\mathbb{R}_{>0}$, which is necessarily $\{1\}$. Hence $\Delta \equiv 1$ and right- and left-Haar measures coincide. This is where compactness of $G$ enters in a load-bearing way. For a non-compact non-unimodular group (such as the affine group $\{x \mapsto ax + b\}$), the averaging trick must be modified or fails altogether. [/guided] [/step] [step:Conclude: $\langle \cdot, \cdot \rangle$ is a $G$-invariant Hermitian inner product on $V$] Combining Steps 2, 3, and 4: the form $\langle \cdot, \cdot \rangle$ is sesquilinear (Step 2), Hermitian symmetric (Step 2), positive definite (Step 3), and $G$-invariant (Step 4). It is therefore a $G$-invariant Hermitian inner product on $V$. [/step]