# Convergence in distribution (weak convergence)

1. Mar 8, 2013

### christoff

1. The problem statement, all variables and given/known data
Consider the function $$f_n(x)=n\cdot I\left[|x|<\frac{1}{2n}\right]$$ considered as a distribution in $D'(\mathbb{R})$, where $I$ denotes the indicator function. Recall that $f_n$ converges to $\delta_0$, the delta distribution, in $D'(\mathbb{R})$. Show that $f_n^2-n\delta_0$ converges in $D'(\mathbb{R})$.

3. The attempt at a solution

By a few calculations, we can show that for any test function $\phi$ and natural $n>0$, that
$$(f_n^2,\phi)=n(f_n-\delta_0,\phi).$$
At this point, I would be tempted to say that this converges to zero (as a sequence of real numbers) because $(f_n-\delta_0,\phi)$ "must" dominate the convergence of $n$ to infinity. However, this isn't rigorous, and is more of a hunch than anything. However, any attempts I've made to prove it have not gone anywhere. I feel as though it must have something to do with the smoothness of the test functions, but I can't see where to incorporate this.

At this point, I would appreciate a nudge in the right direction.

2. Mar 8, 2013

### Ray Vickson

Try using the second-order Taylor expansion of $\phi(x)$ in the evaluation of $(f_n^2,\phi)$.

3. Mar 11, 2013

### christoff

Yep, that worked. Thanks! It's actually very easy once you apply the Taylor expansion :)