Dirac's delta defined as a linear functional acting on a continuous test functions 
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.
 A: 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}
