Prove that the parity operator is Hermitian

Question

We know that an operator is Hermitian when:
$langle f|hat{O}grangle$ =  $langle hat{O} f|grangle$
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?

mike stone · Accepted Answer

Set $x=-xi$ in
$$
int_{-infty}^{infty} f(x)g(-x),dx
$$
to get
$$ int_{-infty}^{infty} f(x)g(-x),dx=int_{+infty}^{-infty}f(-xi)g(xi)d(-xi) = - int_{+infty}^{-infty}f(-xi)g(xi)dxi= int_{-infty}^{+infty}f(-xi)g(xi)dxi
$$
so $langle Pf,grangle = langle f,Pgrangle$

daydreamer · Answer

Regarding eigenvalues, notice that the parity operator is an involution, in the present context means it is it's own inverse. Next, use that every function can be expressed as the sum of its symmetric and antisymmetric part. Think that it does the job.
Please note that we assume the operator is hermitian with respect to some integration interval. For an accessible discussion on this, check Shankar's Principles of Quantum Mechanics chapter 2 or 3 I think

Prove that the parity operator is Hermitian

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