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I am reading a book on Sobolev and having trouble understanding a notion of weak derivative. I consider a function $(x-1)^+=\max(x-1,0),x\in[0,2]$, I have a problem at $x=1$, so it is continuous and "almost everywhere differentiable" now, does it have a weak derivative everywhere? Can we "quantify" that? I know that the derivative in classical sense is either $0$ or $1$ and delta function at one point...can I say similar statements about weak?

Now I take a discontinuous function, say indicator function, would it have a weak derivative?

thanks!

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What does $+$ in the exponent mean? –  Rudy the Reindeer Apr 23 '12 at 15:59

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The weak derivative of a locally integrable $u$, denoted $u'$, is a function such that $$\forall \varphi\in C^{\infty}_0(\Omega),\quad\int_{\Omega}u'(x)\varphi(x)\lambda(dx)=-\int_{\Omega}u(x)\varphi'(x)\lambda(dx),$$ where $\Omega$ is the open on which we are working and $C^{\infty}_0(\Omega)$ is the vector space of real-valued functions defined on $\Omega$, which have a compact support and are infinitely differentiable.

When $u$ is a $C^1$ function, the weak derivative is equal to the usual derivative almost everywhere.

In this case, let us compute the weak derivative. Fix $\varphi\in C^{\infty}_0(\Omega)$ and define $u(x)=(x-1)^+$. We should have $$\int_0^2u'(x)\varphi(x)dx=\int_0^1(x-1)^+\varphi'(x)dx+\int_1^2(x-1)\varphi'(x)dx=-\int_1^2\varphi(x)dx,$$ so the weak derivative is almost-everywhere equal to the characteristic function of $(1,2)$.

Indicator functions have a weak derivative since these one are locally integrable functions. For example, with $u=\chi_{(0,1)}$, we have $$\int_0^2u(x)\varphi'(x)dx=\int_0^1\varphi'(x)dx=\varphi(1)=\delta_1(\varphi),$$ so the weak derivative of $u$ is $-\delta_1$ (which is not locally integrable, but it's still a distribution).

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Firts, in the first function, so what is the weak derivative at 0, or we just define them in a.s. sense and don't specify at that point? Because classical derivative is also a.s. a characteristic function of $(1,2)$. And second, in the last equality, you have integrated from $0$ to $2$, however, if the domain is [0,1], then RHS=$0$, since $\phi(1)$ is. What is the weak derivative then? Just $0$ on $[0,1]$? Thanks! –  Medan Jul 6 '12 at 13:45
    
Yes, it's defined almost everywhere. In the last equality, I considered the domain $(0,2)$, not $(0,1)$. –  Davide Giraudo Jul 6 '12 at 13:52
    
As a sum up, in the first case the weak derivative is defined a.e. and equal to the classical derivative, in the second case it is zero on (0,2) except the point $1$, where it is a delta function. Can we proceed with a first case and take more derivatives? Can we define a second weak derivative of it? What I require is local integrability then. Is this satisfied for a.e. defined weak derivative? –  Medan Jul 6 '12 at 19:53
    
Yes, you can take more derivatives (since the $d$-th derivative of a distribution $u$ is defined as $\langle u^{(d)},\varphi\rangle:=(-1)^d\langle u,\varphi^{(d)}\rangle$. –  Davide Giraudo Jul 6 '12 at 20:35
    
Davide, a related question: For the case of a step function I can see that the weak derivative is 0 a.e. except a point where it is a delta function. So, I want to know whether the step function belongs to the $H^1$ space. But since the weak derivative $f'$ is a distribution, what the integral $$\int_0^2 f^{\prime}^2dx$$ then means? In the first case it is easier since I can compute the integral of a square of a indicator function $1_{\{1,2\}}$, but I am confused how to deal with delta function. thanks! –  Medan Jul 9 '12 at 21:43

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