Second distributional derivative of $\ln x_+$

Question

Consider the function $g(x)=\ln x_+$ and we can form the distribution $$(\ln x_+,\varphi)=\int_{0}^{\infty}\ln x \,\varphi(x)\,\mathrm dx$$

My solution is \begin{align} \left<g'', \varphi \right> &= \left<g, \varphi''\right> \\&= \int^{\infty}_{0} \ln x \cdot \varphi''(x)\,\mathrm dx \\ &= \Big| \begin{array}{@{}cc@{}} u=\ln x & u'= \frac{1}{x} \\ v = \varphi'(x) & v'= \varphi''(x) \end{array} \Big|\\ &=\left[ \ln x \cdot \varphi'(x) \right]^\infty_{0} - \int^\infty_{0} \ln x \cdot \varphi'(x) \,\mathrm dx\\ \end{align}

But I guess my solution is not correct since when I partially integrate second time it gives me nothing. Any help would be great!

Answer 1

Let $$ \ln x_\epsilon = (\ln x) \, \mathbf{1}_{(\epsilon,\infty)}, $$ i.e. $$ \langle \ln x_\epsilon, \varphi \rangle = \int_\epsilon^\infty (\ln x) \, \varphi(x) \, dx . $$

Then $$ (\ln x_\epsilon)' = \frac{1}{x} \, \mathbf{1}_{(\epsilon,\infty)} + (\ln x) \, \delta_\epsilon = \frac{1}{x} \, \mathbf{1}_{(\epsilon,\infty)} + (\ln \epsilon) \, \delta_\epsilon $$ and $$ (\ln x_\epsilon)'' = \left( \frac{1}{x} \, \mathbf{1}_{(\epsilon,\infty)} + (\ln \epsilon) \, \delta_\epsilon \right)' = -\frac{1}{x^2} \, \mathbf{1}_{(\epsilon,\infty)} + \frac{1}{x} \, \delta_\epsilon + (\ln\epsilon) \, \delta_\epsilon' \\ = -\frac{1}{x^2} \, \mathbf{1}_{(\epsilon,\infty)} + \frac{1}{\epsilon} \, \delta_\epsilon + (\ln\epsilon) \, \delta_\epsilon' . $$

Now $\ln x_+ = \lim_{\epsilon\to 0} \ln x_\epsilon$ so $$ (\ln x_+)'' = (\lim_{\epsilon\to 0} \ln x_\epsilon)'' = \lim_{\epsilon\to 0} (\ln x_\epsilon)'' = \lim_{\epsilon\to 0} \left( -\frac{1}{x^2} \, \mathbf{1}_{(\epsilon,\infty)} + \frac{1}{\epsilon} \, \delta_\epsilon + (\ln\epsilon) \, \delta_\epsilon' \right) . $$

Answer 2

Here is one of the many possible ways to express the second derivative: let $ {\theta} \in {\mathcal{C}}_{0}^{\infty } \left(\mathbb{R}\right)$ such that $ {\theta} \left(0\right) = 1$ and $ {{\theta}'} \left(0\right) = 0$.

For every $ {\varphi} \in {\mathcal{C}}_{0}^{\infty } \left(\mathbb{R}\right)$ it holds

\begin{equation}\renewcommand{\arraystretch}{2.5} \begin{array}{rcl}\left\langle {g''} , {\varphi}\right\rangle &=&\left\langle {g''} \ , \ {\varphi}-{\varphi} \left(0\right) {\theta}-{{\varphi}'} \left(0\right) x {\theta}\right\rangle +{\varphi} \left(0\right) \left\langle {g''} , {\theta}\right\rangle +{{\varphi}'} \left(0\right) \left\langle {g''} , x {\theta}\right\rangle \\ &=&\displaystyle -\int_{0}^{\infty }\frac{{\varphi} \left(x\right)-{\varphi} \left(0\right) {\theta} \left(x\right)-{{\varphi}'} \left(0\right) x {\theta} \left(x\right)}{{x}^{2}} d x\\ &&\qquad \displaystyle +{\varphi} \left(0\right) \int_{0}^{\infty }\ln \left(x\right) {\theta}'' \left(x\right) d x+{{\varphi}'} \left(0\right) \int_{0}^{\infty }\ln \left(x\right) (x {\theta}(x))'' d x \end{array} \end{equation}

because $ {\psi}(x)= {\varphi}(x)-{\varphi} \left(0\right) {\theta}(x)-{{\varphi}'} \left(0\right) x {\theta}(x)\in {\mathcal{C}}_{0}^{\infty } \left(\mathbb{R}\right)$ satisfies

\begin{equation}{\psi} \left(0\right) = {{\psi}'} \left(0\right) = 0\end{equation}