I need help seeing that $$ \mathcal{L}^* g = -\frac{\partial (bg)}{\partial x} + \frac{1}{2}\frac{\partial^2(\sigma^2g)}{\partial x^2} $$ is the adjoint operator of $$ \mathcal{L} = b\frac{\partial f}{\partial x} + \frac{1}{2}\sigma\frac{\partial ^2 f}{\partial x^2} $$ in the $L^2$ sense $\langle \mathcal{L}f,g\rangle = \langle f,\mathcal{L}^*g\rangle$, where $b(x)$ and $\sigma(x)$ are some suitable functions. Doing the computations I arrive to $$ \eqalign{ \langle \mathcal{L}f,g\rangle &= \int_{\mathbb{R}} \left(b\frac{\partial f}{\partial x} + \frac{1}{2}\sigma\frac{\partial ^2 f}{\partial x^2}\right)g dx = \cdots\text{by parts x2} \cr &= \int_{\mathbb{R}}f\left(-\frac{\partial (bg)}{\partial x} + \frac{1}{2}\frac{\partial^2(\sigma^2g)}{\partial x^2}\right)dx + \left[ bfg + \frac{1}{2}\sigma^2g\frac{\partial f}{\partial x} - \frac{1}{2}\frac{\partial(\sigma^2g)}{\partial x^2}f\right]_{-\infty}^{+\infty}\ . } $$ So basically I need help understanding in what circumstances $$ \left[ bfg + \frac{1}{2}\sigma^2g\frac{\partial f}{\partial x} - \frac{1}{2}\frac{\partial(\sigma^2g)}{\partial x^2}f\right]_{-\infty}^{+\infty} = 0 $$ so that $\mathcal{L}^*$ is the adjoint of $\mathcal{L}$.
(Reference Robert V.Kohn ch1 pg 14.)
