Conjugates in integrals Is it always true that $\int f(x)^*M^\dagger Mf(x) dx=\int (Mf(x))^*(Mf(x)) dx$? where $M$ is an operator. If so, is there a simple proof that this is so? THanks.
 A: The operator Hermitian adjoint $X^\dagger$ is defined to be the operator such that $\langle X \circ,\bullet\rangle=\langle\circ,X^\dagger\bullet\rangle$ for any two appropriate inputs $\circ,\bullet$. The conjugate symmetry of the inner product yields $$\langle X \circ,\bullet\rangle=\langle\circ,X^\dagger\bullet\rangle=\langle X^\dagger\bullet,\circ\rangle^*=\langle\bullet,(X^\dagger)^\dagger\circ\rangle^*=\langle (X^\dagger)^\dagger\circ,\bullet\rangle\quad\forall\circ,\bullet$$
$$\implies X=(X^\dagger)^\dagger.$$
Therefore $\langle M^\dagger (M f),f\rangle=\langle Mf,(M^\dagger)^\dagger f\rangle=\langle Mf,Mf\rangle$, which proves the identity after specializing the inner product to $\langle f,g\rangle=\int fg^*d\mu$. This fact goes beyond just integration as it applies to arbitrary inner products and can only be derived by using the definition of the adjoint ${}^\dagger$, after which nothing is really necessary. For specific types of operators and function spaces the identity may be check-able using by-parts integration. (For example, $M=d/dx$ for vanishing boundary conditions.)
