Show that $|x|^{-1/2}$ is not weakly differentiable

Question

My general question is how one shows the non-existence of weak derivatives for $L^1$-functions where integration by parts fails because the strong derivative, though it exists, is not $L^1$-integrable.

As an example, consider the real interval $(-1,1)$ and the function $f$ thereon with $f(x) = |x|^{-1/2}$. Then $f$ is an $L^1$ function, but its strong derivative (away from $0$) fails to be locally $L^1$-integrable. How do I show that it is not weakly differentiable, i.e. that there does not exist any locally $L^1$-integrable function $g$ such that for all $\phi \in C_c^\infty(-1,1)$: $$\int_{-1}^1 |x|^{-1/2} \, \phi'(x) \, \operatorname{d} \negthinspace \, x = - \int_{-1}^1 g(x) \phi(x) \, \operatorname{d} \negthinspace \, x \, .$$

It is clear to me that one needs to argue via a contraction, assuming the existence of $g$. I know in similar problems one chooses a suitable sequence of functions $\phi_k \in C_c^\infty(-1,1)$ with $k \in \mathbb N$ to obtain the contradiction, but I have not gotten it to work here...

Answer 1

Start by taking functions $\phi$ that are zero in $(-\delta,\delta)$. Then you can integrate by parts in the classical sense and the function $g$ has to be the standard derivative of $f$ outside of $(-\delta,\delta)$. Then you let $\delta$ go to zero and find that $g$ is the standard derivative of $f$ except at one point, which is irrelevant when it comes to integrability. EDIT: I added more details, in view of the comments

Assume that $f$ has a weak derivative $g$ in $L^{1}((-1,1))$. Take $0<\delta<1$ and a function $\phi\in C_{c}^{1}((-1,1))$, which is zero in $(-\delta,\delta)$. Then $$ \int_{-1}^{1}f(x)\phi^{\prime}(x)\,dx=-\int_{-1}^{1}g(x)\phi(x)\,dx $$ by the definition of weak derivative, and by standard integration by parts $$ \int_{-1}^{1}f(x)\phi^{\prime}(x)\,dx=-\int_{-1}^{1}f^{\prime}(x)\phi(x)\,dx $$ because $f$ is $C^{1}$ in $[-1,1]\setminus(-\delta,\delta)$. Subtracting these two identities, you get $$ 0=\int_{-1}^{1}(g(x)-f^{\prime}(x))\phi(x)\,dx $$ for all $\phi\in C_{c}^{1}((-1,1))$ that are zero in $(-\delta,\delta)$. Since these functions are dense in $L^{1}((-1,1)\setminus(-\delta,\delta))$, you get that $g(x)-f^{\prime}(x)=0$ a.e. in $(-1,1)\setminus(-\delta,\delta)$. Hence, if you let $\delta\rightarrow0^{+}$, you get that $g=f^{\prime}$ a.e., which is a contradiction, since $f^{\prime}$ is not integrable.

Answer 2

This is just a cleaned-up version of Gio67's argument above, as I did not want to edit his post and comment there further:

Assume such an integrable $g$ exists. Then for all $\delta >0$ and all $\phi \in C^\infty((-1,1)\setminus(-\delta, \delta))$ we have $$\int_{-\infty}^\infty f(x) \phi'(x) \, d x = - \int_{-\infty}^\infty f'(x) \phi(x) \, d x = -\int_{-\infty}^\infty g(x) \phi(x) \, d x \, .$$ Taking the difference and applying the fundamental lemma of the calculus of variations, we find that for all $\delta >0$ and for almost all $x \in (-1,1) \setminus (-\delta,\delta)$ we have $f'(x)=g(x)$. But this implies that $f'(x)=g(x)$ for almost all $x \in (-1,1) \setminus \lbrace 0 \rbrace$. So $f'=g$ almost everywhere on $(-1,1)$. But $g$ was assumed to be integrable, which is a contradiction.