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.)

  • $\begingroup$ Your problem is covered under this theorem "Sturm–Liouville theory". See en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory. See the different types of boundary conditions. $\endgroup$ Commented Jul 21, 2012 at 11:46
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    $\begingroup$ $\mathcal{L}$ and $\mathcal{L}^*$ are unbounded operators on $L^2$, and so you have to consider their domains. If they have been defined correctly, you should find that when $f$ is in the domain of $\mathcal{L}$ and $g$ is in the domain of $\mathcal{L}^*$, then the boundary term vanishes. $\endgroup$ Commented Jul 21, 2012 at 15:25
  • $\begingroup$ @Mhenni: I converted your post to a comment as we discourage posting just a link as an answer (except in the case of reference requests). If you have the time, please consider posting a more detailed answer. Thanks. $\endgroup$ Commented Jul 26, 2012 at 21:51


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