[proofplan] We compute the right-hand side $\langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G$ and show it equals the left-hand side $\langle \operatorname{Res}^G_H \varphi, \psi \rangle_H$. Expanding the definition of induction unfolds the inner product into a double sum over $g, x \in G$ of $\overline{\varphi(g)}\,\mathring{\psi}(x^{-1}gx)$. The substitution $y = x^{-1} g x$ converts the conjugate $\mathring{\psi}(x^{-1}gx)$ into $\mathring{\psi}(y)$ and replaces $\overline{\varphi(g)}$ with $\overline{\varphi(y)}$ — using that $\varphi$ is a $G$-class function. The sum over $x$ then contributes a constant factor of $|G|$. Finally, $\mathring{\psi}$ is supported on $H$, so only terms $y \in H$ survive, and we recognise the $H$-inner product of $\operatorname{Res}^G_H \varphi$ with $\psi$. [/proofplan] [step:Expand $\langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G$ using the definitions of the $G$-inner product and induction] The $G$-inner product on $\mathcal{C}(G)$ is sesquilinear (linear in the first argument, conjugate-linear in the second). For class functions $\alpha, \beta \in \mathcal{C}(G)$, \begin{align*} \langle \alpha, \beta \rangle_G = \frac{1}{|G|} \sum_{g \in G} \alpha(g)\, \overline{\beta(g)}. \end{align*} Applied to $\alpha = \varphi$ and $\beta = \operatorname{Ind}^G_H \psi$, but with $\varphi$ in the second slot to match the conventional Frobenius statement, we instead expand \begin{align*} \langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G = \frac{1}{|G|} \sum_{g \in G} \varphi(g)\, \overline{\operatorname{Ind}^G_H \psi(g)}. \end{align*} Because the values of $\psi$ are complex and the result must symmetrise correctly, we work with the convention \begin{align*} \langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G = \frac{1}{|G|} \sum_{g \in G} \overline{\varphi(g)}\, \operatorname{Ind}^G_H \psi(g), \end{align*} which corresponds to the conjugate-linearity-in-first-argument convention used throughout these notes for character pairings (and is the convention in which the Frobenius identity is symmetric in its two sides). Substituting the definition of [Induced Class Function](/theorems/2447) — namely $\operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|}\sum_{x \in G} \mathring{\psi}(x^{-1} g x)$ — yields \begin{align*} \langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G = \frac{1}{|G|\,|H|} \sum_{g \in G} \sum_{x \in G} \overline{\varphi(g)}\, \mathring{\psi}(x^{-1} g x). \end{align*} The double sum is finite, so reordering is unrestricted. [/step] [step:Substitute $y = x^{-1} g x$ and use $G$-class-functionality of $\varphi$ to replace $\varphi(g)$ by $\varphi(y)$] Fix $x \in G$. The map \begin{align*} G &\to G \\ g &\mapsto x^{-1} g x \end{align*} is a bijection (its inverse is $y \mapsto x y x^{-1}$). So as $g$ ranges over $G$, the element $y = x^{-1} g x$ also ranges over $G$ exactly once. Furthermore, $g$ and $y = x^{-1} g x$ are $G$-conjugate, and $\varphi \in \mathcal{C}(G)$ is constant on $G$-conjugacy classes, so \begin{align*} \varphi(g) = \varphi(x^{-1} g x) = \varphi(y) \qquad \implies \qquad \overline{\varphi(g)} = \overline{\varphi(y)}. \end{align*} Substituting in the inner sum (with $x$ fixed), \begin{align*} \sum_{g \in G} \overline{\varphi(g)}\, \mathring{\psi}(x^{-1} g x) = \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y). \end{align*} The right-hand side is independent of $x$. Therefore \begin{align*} \sum_{x \in G} \sum_{g \in G} \overline{\varphi(g)}\, \mathring{\psi}(x^{-1} g x) = \sum_{x \in G} \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y) = |G| \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y). \end{align*} Plugging into Step 1 and cancelling $|G|$: \begin{align*} \langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G = \frac{1}{|H|} \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y). \end{align*} [guided] This is the heart of the proof. The double sum $\sum_{g, x} \overline{\varphi(g)}\, \mathring{\psi}(x^{-1} g x)$ has a built-in symmetry: $\mathring{\psi}$ depends only on the conjugate $x^{-1}gx$, and $\varphi$ depends only on the conjugacy class of $g$. The substitution $y = x^{-1}gx$ exposes this symmetry by making $y$ the natural summation variable. **Why $\varphi(g) = \varphi(y)$ matters.** Without class-functionality of $\varphi$, the substitution would change the integrand. With it, the dependence on $x$ disappears entirely from the inner sum, and we get $|G|$ identical inner sums when we sum over $x$. **Why we substitute and don't just relabel.** The substitution $y = x^{-1}gx$ for a fixed $x$ is a bijection of $G$, so $\sum_g f(x^{-1} g x) = \sum_y f(y)$ for any function $f$. Here $f(y) = \overline{\varphi(y)}\, \mathring{\psi}(y)$ — the post-substitution integrand, valid because $\overline{\varphi(g)} = \overline{\varphi(x^{-1} g x)}$ by class-functionality. **Why the inner sum is independent of $x$ — the crucial cancellation.** After substitution, the integrand is $\overline{\varphi(y)} \mathring{\psi}(y)$, which depends only on $y$, not on $x$. So summing over $x \in G$ multiplies by $|G|$, killing the $\frac{1}{|G|}$ from the $G$-inner product and leaving $\frac{1}{|H|} \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y)$. [/guided] [/step] [step:Restrict the sum from $G$ to $H$ using $\operatorname{supp}(\mathring{\psi}) \subseteq H$] By the definition of the zero extension, \begin{align*} \mathring{\psi}: G &\to \mathbb{C} \\ y &\mapsto \begin{cases} \psi(y) & y \in H, \\ 0 & y \in G \setminus H. \end{cases} \end{align*} Hence terms with $y \in G \setminus H$ contribute zero to the sum, and we may restrict the summation to $y \in H$: \begin{align*} \sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y) = \sum_{y \in H} \overline{\varphi(y)}\, \psi(y). \end{align*} On $H$, the value $\varphi(y)$ is by definition $(\operatorname{Res}^G_H \varphi)(y)$, the restriction of $\varphi$ to $H$. So \begin{align*} \frac{1}{|H|}\sum_{y \in G} \overline{\varphi(y)}\, \mathring{\psi}(y) = \frac{1}{|H|} \sum_{y \in H} \overline{(\operatorname{Res}^G_H \varphi)(y)}\, \psi(y) = \langle \operatorname{Res}^G_H \varphi, \psi \rangle_H, \end{align*} where the last equality is the definition of the $H$-inner product. Combining with Step 2, \begin{align*} \langle \varphi, \operatorname{Ind}^G_H \psi \rangle_G = \langle \operatorname{Res}^G_H \varphi, \psi \rangle_H, \end{align*} as claimed. This completes the proof of Frobenius reciprocity. [/step]