An important example in Dirac's delta, which is defined as a linear functional acting on continuous test functions, for $\Omega\subset\mathbb{R}^d$, $$\delta(\phi)=\phi(0), \quad \forall \phi\in C_0^0(\Omega)$$ Let now $d=1,\Omega=(-1,1)$ and $$f(x)=\begin{cases} x,&x\ge 0\\ 0,&x\le 0 \end{cases}\qquad g(x)=\begin{cases} 1,&x>0\\ 0,&x<0. \end{cases}$$ Show that $f'=g,g'=\delta$ in the sense of generalized derivative, i.e., $$f'(\phi)=-\int_\Omega\,f\phi'\,dx=\int_\Omega\,g\phi\,dx \qquad \forall\phi\in C_0^1(\Omega)$$ $$g'(\phi)=-\int_\Omega\,g\phi'\,dx=\phi(0) \qquad\forall\phi\in C_0^1(\Omega)$$

Conclude that the generalized derivative $f'=g$ belongs to $L_2$, but that $g'=\delta$ does not. For the latter statement, you must show that $\delta$ is not bounded with respect to the $L_2$-norm,i.e., you need to find a sequence of test functions such that $\lVert\phi_i\rVert_{L_2}\to 0$, but $\phi(0)=1$ as $i\to\infty$. Thus $f\in H^1(\Omega)$ and $g\not\in H^1(\Omega)$.

I did everything except the part where we need to find a sequence such that $\lVert\phi_i\rVert_{L_2}\to 0$, but $\phi(0)=1$ as $i\to\infty$.

Any help would be greatly appreciated. Thanks in advance.

  • $\begingroup$ Do want to prove it exactly like that? In this link math.stackexchange.com/questions/922708/… it is shown that $\delta$ is not locally integrable and therefore $\delta$ can't be in $L^2$ $\endgroup$ – Fritz Apr 21 '17 at 16:34

The idea is to make $\phi_n$ a triangle with base length $\frac{2}{n}$ and height one. This way $\phi_n(0) = 1$ for each $n$ and the area of the triangle (which is related to the $L^2$ norm) will go to zero.

Define \begin{equation} \phi_n(x) = \begin{cases} 1 - n|x| &\text{ if } |x| \in {[0, \frac{1}{n})}\\ 0 &\text{ if } |x| \in {[\frac{1}{n},1)} \end{cases} \end{equation} which is exactly this triangle described. I recommend drawing a picture of this function to see that it is continuous. You can also sketch what happens as $n$ increases to see that it gives the result you want.

It should be clear that $\phi_n(0) = 1$ for each $n$. The condition on the $L^2$ norm can be seen by explicitly computing the integral and taking a limit. Or, if you are lazy, you can do it as follows: \begin{equation} \int_{-1}^{1} |\phi_n(x)|^2\mathrm{d}x = \int_{-\frac{1}{n}}^{\frac{1}{n}} |\phi_n(x)|^2\mathrm{d}x\leq \frac{2}{n}\sup_{x\in{(-1,1)}} |\phi_n(x)|^2 = \frac{2}{n} \xrightarrow{n \to \infty} 0. \end{equation}


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