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

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...

• The $g$ in question is $-\frac{1}{2} x^{-3/2}$ anyway, the issue is whether it is as regular as you want it to be. For $C^\infty_c$ test functions it's a valid distribution.
– Ian
Commented Feb 15, 2022 at 21:56
• I fixed the interval and changed to $f(x)=\sqrt{|x|}$. Everything should make sense now. The choice above does not work because of lack of integrability (that's the point of the question). Commented Feb 16, 2022 at 21:51
• Sorry, $f(x) = 1/\sqrt{|x|}$, of course. Commented Feb 16, 2022 at 23:24

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.

• This is not really a rigorous proof and your conclusion is false (the limit as $\delta \to 0$ of the right hand side will diverge). But the idea may actually work, if argued correctly. I will try to construct a rigorous argument later today. Commented Feb 18, 2022 at 19:49
• This is a rigorous proof. I didn’t say that you let $\delta$ go to zero in the integral. I just said that if the weak derivative $g$ exists, then it coincides with the classical derivative $f’$ in $(-1,1)\setminus (-\delta,\delta)$. So if you let $\delta$ go to zero or use the arbitrariness of $\delta$, you find that $g$ has to be $f’$ almost everywhere. Commented Feb 18, 2022 at 21:22
• Thank you for the edit, that looks a lot better. Using the fundamental lemma of the calculus of variations was another good idea. Please allow me to clarify the argument at the end: For all $1>\delta >0$ we have $f'=g$ on $(-1, \delta) \cup (\delta,1)$. As $g$ is integrable on $(-1,1)$, $$\lim_{\delta \to 0^+} \int_{(-1, \delta) \cup (\delta,1)} g(x) \, \phi(x) \, d x = \int_{-1}^{1} g(x) \, \phi(x) \, d x < \infty \, .$$ As $f'$ is not integrable on $(-1,1)$, the other integral does not converge. Contradiction. Your argument answers the question completely, thank you! Commented Feb 19, 2022 at 3:15
• Sorry to open this up again, but the closing argument in my comment above does not work, because the integral will only diverge if we also increase the support of $\phi$ with $\delta$ decreasing. Otherwise, all we have obtained is that the limit of the two integrals as $\delta \to 0^+$ is equal for all $\phi \in C^1_c((-1,1)\setminus (- \delta_0,\delta_0))$---a correct statement. The last sentence you wrote down seems to be a pseudo-proof, for it is not clear what taking the limit is supposed to mean mathematically. Commented Mar 3, 2022 at 22:28
• I tried to fix your proof by taking two $C^\infty$-bump functions that approach each other as $\delta \to 0$ and then look at the asymptotic behavior of the product with $f'(x)$ at zero. Unfortunately, the `zero of the bump function at the edges' seems to be too strong, so that one cannot get any asymptotic behavior of $x^{-\alpha}$ with $\alpha >0$, let alone $\alpha \geq 1$. The question explicitly asked for the test function space to be $C^\infty_c(-1,1)$. As of now, the proof is incomplete. Commented Mar 3, 2022 at 23:13

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.

• A wonderful modification.
– xxxg
Commented Jun 26 at 13:38