The operator $F$ is defined by $F\psi(x)=\psi(x+a)+\psi(x-a) $, where $a$ is a nonzero constant. Determine whether or not $F$ is a Hermitian operator.
If the condition for $F$ to be Hermitian is $(\psi|F\psi)=(F\psi|\psi)$,
then $(\psi|F\psi)=\int\psi^*(\psi(x+a)+\psi(x-a))dx$
and $(F\psi|\psi)=\int(\psi^*(x+a)+\psi^*(x-a))\psi dx$
how can I show equality?
By using linearity of the inner product and change of variables in the integral. $$ (F\psi \mid \phi) = (\psi_a + \psi_{-a}\mid\phi) = (\psi_a\mid\phi) +(\psi_{-a}\mid\phi) $$ Where the subscript in $\psi_a$ means translation by $a$. Change of variables ($x'=x+a$) in the integral implies $(\psi_a\mid\phi) = (\psi\mid\phi_{-a})$ and vice versa. $$ (\psi_a\mid\phi) +(\psi_{-a}\mid\phi) = (\psi\mid\phi_{-a})+(\psi\mid\phi_{a}) = (\psi\mid F\phi) $$