I try to understand the notions of weak derivative and Sobolev space

I take this example:

$f(x)= |x|\quad $ for $ \quad x\in [-1, 1]$

The derivative in the sense of distributions is $(T_f)^{'}= T_{f^{'}}+\delta_{-1}f(-1)+\delta_{1}f(1),$ where $T_f$ is the distributions associated to $f$ and $\delta_{a}$ is the Dirac distributions.

Now I want to compute $$\int_{-1}^{1}(f^{'}(x))^{2} dx$$ where$ f^{'}$ is the weak derivative of $f$ . but I don't know from where to start because this for me doesn't make sense.

Thank you in advance!


On $(-1, 1)$ the weak derivative of $f(x) = |x|$ is $f'(x) = \operatorname{sign}(x),$ which equals $1$ if $x>0$ and equals $-1$ if $x<0.$

The square of this equals $1$ on all of $(-1, 1)$ so the integral has value $2$.

  • $\begingroup$ thank you @md2prepe yes this is very nice but what we can do if we have a closed interval $[-1,1]$. $\endgroup$ – Bernstein Apr 8 '19 at 15:45
  • $\begingroup$ @Bernstein : $\int_{[-1,1]}f=\int_{(-1,1)}f$ always. $\endgroup$ – user657324 Apr 8 '19 at 16:25
  • $\begingroup$ @md2prepe yes , but in my opinion to define the weak derivative for some functions in [-1,1] you can not use the space of smooth function with compact support in [-1,1] , in my opinion there exist some things that we need to do like the extention of this fuction in $\mathbb{R}$ otherwise we will have some problems... $\endgroup$ – Bernstein Apr 8 '19 at 17:11
  • 1
    $\begingroup$ @user657324 $\int_{[-1,1]}\delta(x-1)=1$ but $\int_{(-1,1)}\delta(x-1)=0$. $\endgroup$ – user647486 Apr 8 '19 at 17:36
  • $\begingroup$ @user647486: $\delta $ is not a function, whereas $f'(x)=sgn(x)$ is a function. $\endgroup$ – user657324 Apr 8 '19 at 18:05

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