Let $(G,K)$ be a Gelfand pair. Why, for a function ( f ) ( K )-binvariant with respect to a compact subgroup ( K ) of a group ( G ), do we have the following equality: $$ f(xy) = \int_K f(xky) \, dk$$

A function $ f : G \to \mathbb{C} $ is $ K $-binvariant if for all $ k_1, k_2 \in K $ and $ x \in G $ $$ f(k_1 x k_2) = f(x). $$

Let $(G,K)$ be a Gelfand pair. Why, for a function $f$ $K$-binvariant with respect to a compact subgroup $K$ of a group $G$, do we have the following equality: $$ f(xy) = \int_K f(xky) \, dk$$

A function $ f : G \to \mathbb{C} $ is $ K $-binvariant if for all $ k_1, k_2 \in K $ and $ x \in G $ $$ f(k_1 x k_2) = f(x). $$

Why, for a function ( f ) ( K )-binvariant with respect to a compact subgroup ( K ) of a group ( G ), do we have the following equality: $$ f(xy) = \int_K f(xky) \, dk$$

A function $ f : G \to \mathbb{C} $ is $ K $-binvariant if for all $ k_1, k_2 \in K $ and $ x \in G $ $$ f(k_1 x k_2) = f(x). $$

Let $(G,K)$ be a Gelfand pair. Why, for a function ( f ) ( K )-binvariant with respect to a compact subgroup ( K ) of a group ( G ), do we have the following equality: $$ f(xy) = \int_K f(xky) \, dk$$

A function $ f : G \to \mathbb{C} $ is $ K $-binvariant if for all $ k_1, k_2 \in K $ and $ x \in G $ $$ f(k_1 x k_2) = f(x). $$