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?
Best Answer
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).