Let $f(x) = log(x) \: \text{for } x \in (0,1) \text{ and } 0 \text{ for } x \in (-1,0) $. Compute the second weak derivative of $f$. In the lecture we computed the first weak derivative using a principal-value integral as: \begin{equation} \langle \partial T_{f}, \phi \rangle = \int_{0}^{1} \frac{\phi(x) - \phi(0)}{x} dx \: \forall \phi \in \mathcal{D}(\Omega) \end{equation}
As an exercise (not to be handed in) we should compute the second weak derivative of $f$. The solution should be: \begin{equation} \langle \partial^{2} T_{f}, \phi \rangle = \phi'(0) + \phi(0)+\int_{0}^{1} \frac{\phi(x)-\phi(0) - x\phi'(0)}{x^2} dx \end{equation}
It is also not clear to me why the last integral should be finite, since $\phi$ is not supposed to be real analytic, so we cannot use Taylor.
My idea would be the following: \begin{equation} \langle \partial^2 T_{f}, \phi \rangle = -\int_{0}^{1} \frac{\phi'(x)- \phi'(0)}{x} dx \end{equation} So first shifting one derivative to $\phi$ and then applying all the previous arguments. I don't see why this doesn't work but it doesn't give the desired result.
Thanks in advance for any help !
Thank you @md2perpe for the help. I think I have the answer now.
integrating by parts, we get: \begin{equation} -(\frac{\phi(x) - x\phi'(o)}{x}) \bigg {|}_{0}^{1} - \int_{0}^{1} \frac{\phi(x) - x \cdot \phi'(0)}{x^2} dx \end{equation}
Now, by adding $\lim_{\epsilon \to 0} \frac{\phi(0)}{x}$ and subtracting it again, we can make make both terms well-defined: \begin{equation} \phi(1) + \underbrace{\lim_{\epsilon \to 0}(\frac{\phi(\epsilon) - \phi(0)}{\epsilon})}_{\phi'(0)} - - \int_{0}^{1} \frac{\phi(x) - \phi(0) - x \cdot \phi'(0)}{x^2} dx \end{equation}
which is the desired expression. The last integral is well-defined since $\phi$ is assumed to be inifinitely, and thus also twice differentiable.