Show that $f_{n}^2(x)$ does not converge in $D^1({\Omega})$ Let 
$$ f_n(x) = \left\{ \begin{array}{ll}
         n & \mbox{if $0 \lt x \lt \frac{1}{n}$};\\
        0 & \mbox{otherwise}.\end{array} \right. \ $$
I have to show that 


*

*$\lim_{n \to \infty}f_n(x)$ exists in terms of distributions 

*$\lim_{n \to \infty}f_n^2(x)$ doesn't exist in terms of distributions. 

*$\lim_{n \to \infty}(f_n^2(x)-n\delta)$  exists in terms of distributions
For $(1)$
$$n\int_{0}^{\frac{1}{n}}\phi(x) dx \to \phi(0) , \text{for},  n \to \infty$$ 
For $(2)$
$$n^2\int_{0}^{\frac{1}{n}}\phi(x) dx $$ Now there exists a $C^{\infty}$ cut off function such that $\phi(x) =1$ for $x \in (0,\frac{1}{n+1})$ and $\phi(x)=0, x \gt \frac{1}{n}$. For this particular choice of $\phi$ this integral goes to $\infty$. 
For $(3)$ This will be 
$$\int_{-n}^{\frac{1}{n}} -(n\delta)\phi(x) dx +(n^2-n\delta)\int_{0}^{\frac{1}{n}}\phi(x) dx+\int_{\frac{1}{n}}^{n} -(n\delta)\phi(x) dx $$
I am unable to see where this would converge. 
Thanks for the help!!

A quick remark on your solution of (2): You need to say that $\phi = 1$ on $[0;1]$ or something like that, as $\phi$ has to be independent of $n$. I know it's pedantic, but this is how you can loose marks on exams. 
For (3): you actually want to show that the map $$T(\phi) = \lim_{n \to \infty} \int_{0}^{\frac{1}{n}} n^2f(x)dx - n \delta(\phi)$$ is a distribution. 
Because $\delta(\phi) = \phi(0) = \int_{0}^{\frac{1}{n}}n\phi(0)dx$, the above map is actually 
$T(\phi) = n^2\int_{0}^{\frac{1}{n}}\phi(x) - \phi(0) dx$
There are many ways to find this limit. I will use Taylor's theorem: $\phi(x)  \phi(0) + x \phi'(0) + O(n^{-2})$ where I used that in the integral $|x| \leq \frac{1}{n}$. The remainder term is bounded by $\frac{C}{n^2}$ so when you multiply by $n^2$ and integrate from $0$ to $\frac{1}{n}$, this will converge to $0$. 
So, $$ T(\phi) = \lim n^2\int_{0}^{\frac{1}{n}} x \phi'(0) dx = \frac{\phi'(0)}{2}$$
Indeed, this is a distribution: it is a linear functional and it is continuous: if  $\phi_k \to \phi$ in the topology of $D$ then $\sup_{\mathbb{R}} |\phi_k' - \phi'|\to 0$, i.e $T(\phi_k) \to T(\phi)$. If the last explanation looks meaningless you might use a different definition of the topology of $\mathbb{D}$: you can look at Distributions for more information.
I hope this helps.
