# computation of the norm

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 $$2$$.
• thank you @md2prepe yes this is very nice but what we can do if we have a closed interval $[-1,1]$. – Bernstein Apr 8 '19 at 15:45
• @Bernstein : $\int_{[-1,1]}f=\int_{(-1,1)}f$ always. – user657324 Apr 8 '19 at 16:25
• @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... – Bernstein Apr 8 '19 at 17:11
• @user657324 $\int_{[-1,1]}\delta(x-1)=1$ but $\int_{(-1,1)}\delta(x-1)=0$. – user647486 Apr 8 '19 at 17:36
• @user647486: $\delta$ is not a function, whereas $f'(x)=sgn(x)$ is a function. – user657324 Apr 8 '19 at 18:05