Given the sequence of functions:
\begin{equation} f_n(x)=tanh(nx) \end{equation}
and knowing that:
\begin{equation} \lim_{x \to \pm \infty}f_n(x)=f(x)=\begin{cases} -1, & x<0 \\ 1, & x>0 \end{cases} \end{equation}
prove that the correspoding sequence of distributions: \begin{equation} \langle T_{f_n},\phi\rangle=\int_{-\infty}^{+\infty}f_n(x)\phi(x)dx \end{equation} converges to the distribution: \begin{equation} \langle T_f,\phi\rangle=\int_{-\infty}^{+\infty}f(x)\phi(x)dx \end{equation} In fact, all I want to prove is that: \begin{equation} |\langle T_{f_n},\phi\rangle-\langle T_f,\phi\rangle|\to 0 \end{equation} But when I reach a point where: \begin{equation} \langle T_{f_n},\phi\rangle-\langle T_f,\phi\rangle=\int_{-\infty}^{0}(tanh(nx)+1)\phi(x)dx+\int_{0}^{+\infty}(tanh(nx)-1)\phi(x)dx \end{equation} , then I have no idea how to proceed. Can someone help? Thanks!
***UPDATE: In order to prove that the integral: \begin{equation} \int_{0}^{\infty}(tanh(nx)-1)\phi(x)dx \to 0 \end{equation} I must show the following:
1) The sequence: $\{g_n\}_{n\in \mathbb{N}}=(tanh(nx)-1)\phi(x)$ converges uniformly to a function $f$, $\forall \phi \in D{\mathbb{(R)}}$.
2)The functions $\{g_n\}_{n\in \mathbb{N}}$ are Riemann integrable.
I would like your help! Thanks a lot!
Hint
We have an odd function here: $$\tanh -x = -\tanh x$$ This gives rise to $$\int_0^\infty (\tanh nx - 1)\phi(x) - (\tanh nx - 1)\phi(-x)\ \mathrm dx = \int_0^\infty (\tanh nx - 1)(\phi(x) - \phi(-x))\ \mathrm dx$$ The second factor is another testing function, so all you need to do is show that $$\int_0^\infty \tanh nx - 1 \ \mathrm dx \to 0$$