Pointwise convergence and the convergence of a distribution sequence. The sequence of functions $f_n(x)=\{tanh(nx)\}_{n=0}^{\infty}$ does not converge uniformly to $f(x)=sgn(x)$ but only pointwise.
Is it then, still possible that the sequence of distributions
\begin{equation}
\langle T_{f_n},\phi(x)\rangle=\int_{-\infty}^{\infty}f_n(x)\phi(x)dx
\end{equation}
 to converge to the correspoding distribution 
\begin{equation}
\langle T_f,\phi(x)\rangle=-\int_{-\infty}^{0}\phi(x)dx+\int_{0}^{\infty}\phi(x)dx
\end{equation}
of $f(x)=sgn(x)$? 
The question rises from a Theorem which suggests that since we do know that $\{f_n\}_{n=0}^{\infty}$ consists of continuous functions which converge uniformly to a funtion $f$, then the correspoding sequence of distributions converges to the distribution $T_f$.
Thank you for your time!
 A: Yes, it does.  Note that for any $\delta > 0$, convergence is uniform outside $(-\delta, \delta)$.  
For any test function $\phi$ and $\epsilon > 0$, take 
$\delta$ so $\int_{-\delta}^\delta |\phi(x)|\; dx < \epsilon$.  Then
if $n$ is large enough that $|f_n - f| < \epsilon$ outside $(-\delta,\delta)$,
$$\left|\langle T_{f_n}, \phi \rangle - \langle T_f, \phi \rangle \right| 
\le \epsilon \int_{\mathbb R} |\phi(x)| \; dx + \int_{-\delta}^\delta |\phi(x)|\; dx \le (1 + \|\phi\|_{L^1}) \epsilon$$
A: $$\begin{align}
\int_{-\infty}^{\infty}(f_n(x)-f(x))\phi(x)dx&=\int_{0}^{\infty} ((f_n(x)-f(x))\phi(x)+(f_n(-x)-f(-x))\phi(-x))dx\\\\
&=\int_{0}^{\infty} (f_n(x)-f(x))(\phi(x)-\phi(-x))dx
\end{align}$$
where we used the fact that $f_n$ and $f$ are odd to arrive at the last equality.
Now, $f_n(x)-f(x)=-\frac{e^{-nx}}{\cosh(nx)}$.  Let's look at the following
$$\begin{align}
\left|\int_{-\infty}^{\infty}(f_n(x)-f(x))\phi(x)dx\right|&=\left|\int_{0}^{\infty}(f_n(x)-f(x))\left(\phi(x)-\phi(-x)\right)dx\right|\\\\
&\le \int_{0}^{\infty} \frac{e^{-nx}}{\cosh(nx)} \left|\phi(x)-\phi(-x)\right|\,dx\\\\
&\le 2M \int_{0}^{\infty} e^{-nx}\,dx\\\\
&=2M/n
\end{align}$$
where $M$ is a finite, upper bound of $\phi$.  Recall that $\phi$ is $L^1$ and continuous on $(-\infty,\infty)$, and therefore is bounded.  
Now, given $\epsilon>0$, choose an $N\ge 2M/\epsilon$ so that $\int_{0}^{\infty} \frac{e^{-nx}}{\cosh(nx)} \left|\phi(x)-\phi(-x)\right|\,dx<\epsilon$ whenever n>N.  This completes the proof.
