We know that an operator is Hermitian when:
$\langle f|\hat{O}g\rangle$ = $\langle \hat{O} f|g\rangle$
Parity operator in 1D is simply defined as:
$\hat{\Pi} f(x) = f(-x)$
I don't know anything about the eigenvalues of parity operator (that is asked in the next problem).
How can I show it is Hermitian?
$\langle f(x)|\hat{\Pi}|g(x)\rangle$ = $\langle f(x)|g(-x)\rangle$
$\langle \hat{\Pi} f(x)|g(x)\rangle$ = $\langle f(-x)|g(x)\rangle$
These last integrals are not equal, unless both functions are symmetric.
How can I prove it?