So in my textbook i get this expression that i just quite dont understand. Let $G$ be a compact group, and $\mu$ the normalised Haar measure on G. Let $(\pi, \mathcal H)$ Be a unitary representation of $G$. For $v \in \mathcal H$ cosider the operator $K_v$ of $\mathcal H$ $$ K_vw = \int_G (w | \pi(g)v)\ \pi(g)v \ \mu(dg) $$ Here $(\cdot| \cdot)$ is an innerproduct. So the book says this also can be written as $$ (K_vw | w' ) = \int_G (w|\pi(g)v)\overline{(w'|\pi(g)v)} \mu(dg)$$
I just dont see how these two are the same? The first $K_vw$ isn't it an operator from $\mathcal H \to \mathbb{C}?$ and the second one needs to take to inputs, a $w$ and a $w'$??
No, $K_w$ is not an operator from $\mathcal H$ into $\Bbb C$. It's from $\mathcal H$ into $\mathcal H$. Note that each $\bigl(w\mid \pi(g)v\bigr)\pi(g)v$ is an element of $\mathcal H$.
Knowing a vector $v$ is the same thing as knowing the map $(v\mid\cdot)$. So, knowing the vector $K_vw$ is the same thing as knowing the value of $\left(K_vw\mid w'\right)$ for every vector $w'$. And it turns out that$$\left(K_vw\mid w'\right)=\int_G\bigl(w\mid\pi(g)v\bigr)\overline{\bigl(w'\mid\pi(g)v\bigr)}\,\mu(dg).$$