Convergence of a sequence of distributions to $\delta'$

Question

I'm trying to show that the sequence of functions $$f_n(x)=\begin{cases}n^2, & -1/n< x< 0\\ -n^2, & 0<x< 1/n\\ 0, & \text{otherwise}\end{cases}$$ converges to $\delta'$ in $\mathcal S^*$. For any $\varphi\in\mathcal S$, Using integration by parts, I can show that $$\left<f_n, \varphi\right>=n\left[\varphi(-1/n)-\varphi(1/n)\right]+n^2\int_0^{1/n}x\left[\varphi'(x)-\varphi'(-x)\right]\,dx$$ So the first term converges to $-2\varphi'(0)$ as $n\to\infty$. I think the second term should converge to $\varphi'(0)$, but I can't seem to find a way to show that. Any hint would be appreciated.

Answer 1

$$\begin{align} \langle f_n, \varphi\rangle &= -n^2 \left( \int_{0}^{1/n} \varphi(x) \, dx - \int_{-1/n}^{0} \varphi(x) \, dx \right) \\ &= \{ t = nx \} \\ &= -n \left( \int_{0}^{1} \varphi(t/n) \, dt -\int_{-1}^{0} \varphi(t/n) \, dt \right) \\ &= \{ t \to -t \text{ in second integral} \} \\ &= -n \left( \int_{0}^{1} \varphi(t/n) \, dt -\int_{0}^{1} \varphi(-t/n) \, dt \right) \\ &= - \int_{0}^{1} \frac{\varphi(t/n)-\varphi(-t/n)}{1/n} dt \\ &\to -\int_{0}^{1} \varphi'(0) \, dt \\ &= - \varphi'(0) \\ &= \langle \delta', \varphi \rangle \end{align}$$

Answer 2

The second term converges to $0$. Let $\epsilon >0$ and choose $n$ so large that $|\phi'(x)-\phi'(-x)| <\epsilon$ for $0<x <\frac 1 n$. This is possible by continuity of $\phi'$. Hence the second term is bounded in absolute value by $\epsilon n^{2}\int_0^{1/n} x dx=\frac 1 2 \epsilon$.