# generalized derivative of $\log |x|$ (sobolev derivative), where $x\in (-1,1)$

Let $$u\in L_{loc}(a,b)$$ and $$\phi \in C_0^{\infty}$$. Function $$v$$ is generalized derivative of $$u$$, if $$1)v\in L_{loc}(a,b)$$ $$2)\int_{a}^bu(x)\phi'(x)dx=-\int_{a}^bv(x)\phi(x)dx$$ for $$\forall \phi \in C_0^{\infty}$$

I am trying to find the generalized derivative of $$ln|x|$$ when $$x\in (-1,1)$$. There is one problem is point $$0$$. I tried to cut special point by using limit.

By definition: \begin{align*} \int_{-1}^1 \log|x|\phi'(x)dx &=\int_{-1}^0 \log(-x)\phi'(x)dx+\int_0^1 \log(x)\phi'(x)dx \\ &=\lim_{\epsilon\to0}\int_{-1}^{-\epsilon}\log(-x)\phi'(x)dx+\lim_{\delta\to0}\int_{\delta}^1\log(x)\phi'(x)dx \\ &=\lim_{\epsilon\to0}\left[\log(-x)\phi(x)|_{-1}^{-\epsilon}-\int_{-1}^{-\epsilon}\frac{\phi(x)}{x}dx\right]+\lim_{\delta\to0}\left[\log(x)\phi(x)|_{\delta}^{1}-\int_{\delta}^1\frac{\phi(x)}{x}dx\right] \\ &=\lim_{\epsilon\to 0}\left[\log(\epsilon)\phi(-\epsilon)-\int_{-1}^{-\epsilon}\frac{\phi(x)}{x}dx\right]+\lim_{\delta\to0}\left[-\log(\delta)\phi(\delta)-\int_{\delta}^1\frac{\phi(x)}{x}dx\right] \\ &=\lim_{\epsilon\to 0, \delta\to 0}[\log(\epsilon)\phi(-\epsilon)-\log(\delta)\phi(\delta)]-\lim_{\epsilon\to 0, \delta\to 0}\left[\int_{-1}^{-\epsilon}\frac{\phi(x)}{x}dx+\int_{\delta}^1\frac{\phi(x)}{x}dx\right] \end{align*} For existence the generalized derivative must be $$\log(\epsilon)\phi(-\epsilon)-\log(\delta)\phi(\delta) = 0$$ and integrals must converge. But $$\frac{1}{x}\notin L_{loc}(-1,1)$$ and the equality with logarithms is not right for all $$\phi$$. Then I conclude that the derivative does not exist. Is it right?

Your conclusion is correct: $$\ln|x|$$ does not possess a generalized derivative in $$L^1_{loc}(-1,1)$$ as per you own definition.

Although $$\ln|x|$$ does not possess a generalized derivative, it is a distribution and as such it has a distributional derivative (see Derivative of ln|x| in the distributional sense).