Let $C\left( a,b \right)$ be the space of differentiable functions on the interval $\left[ a,b \right]$. Let $B(f,g)=\int_{a}^{b}f(t)g(t)w(t)dt$ be a scalar product over $C\left( a,b \right)$, with $w(t)>0$ for all $t\in\left[ a,b \right]$. For p and q fixed functions in $\left[ a,b \right]$ define $$T(f)=\frac{{(p{f}')}'+qf}{w}$$ Show that T is a adjoint operator.
My attemp: If T is a adjoint operator, T should satisfy $B(T(f),g)=B(f,T^{*}(g))=B(f,T(g)$), i.e. $$\int_{a}^{b}\left[{(p{f}')}'g+qfg \right]dt=\int_{a}^{b}\left[{(p{g}')}'f+qfg \right]dt$$ I used integration by parts, but it do not helped me. Can you give me a hint? Thanks.
By parts gives
$$ \int_a^b (pf')'g\,\mathrm{d}x \quad = \quad \big[(pf')g\big]_a^b-\int_a^b pf'g'\,\mathrm{d}x $$
and
$$ \int_a^b (pg')'f\,\mathrm{d}x \quad = \quad \big[(pg')f\big]_a^b -\int_a^b pg'f'\mathrm{d}x $$
These should be equal as long as the evaluated terms (in brackets) are zero. I assume this follows from a condition given in the problem (either $f,g$ or $p,q$ ought to vanish at $a,b$).