# 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

$$\langle T_{f_n},\phi(x)\rangle=\int_{-\infty}^{\infty}f_n(x)\phi(x)dx$$ to converge to the correspoding distribution

$$\langle T_f,\phi(x)\rangle=-\int_{-\infty}^{0}\phi(x)dx+\int_{0}^{\infty}\phi(x)dx$$

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!

## 2 Answers

\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.

• Thank you so much for your detailed proof! I am going throught it atm and I would like to ask: Is it $f_n(x)-f(x)=-\frac{e^{-nx}}{cosh(nx)}$ or should it be without the minus sign? – 010514 May 4 '15 at 21:34
• Also, I would like to ask why the inequality: 1.78€M\int_{0}^{\infty}\frac{e^{-nx}}{cosh(nx)}dx\leq 2M\int_{0}^{\infty}e^{-nx}dx$holds. I am not able to see it :/ – 010514 May 4 '15 at 21:50 • You are welcome. My pleasure. The minus sign is correct. Upon applying the absolute value, the minus sign is absorbed. – Mark Viola May 4 '15 at 21:51 • The first inequality comes from applying$|\int g dx|\le\int |g| dx$. The second inequality comes from applying ($1$)$|\phi_{+}-\phi_{-}|< 2M$and (2)$1/\cosh(nx)<1$. – Mark Viola May 4 '15 at 21:54 •$f_n-f = -\frac{e^{-nx}}{\cosh nx}$is correct. There is indeed a minus sign. – Mark Viola May 4 '15 at 21:55 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)\$, $$\left|\langle T_{f_n}, \phi \rangle - \langle T_f, \phi \rangle \right| \le \epsilon \int_{\mathbb R} |\phi(x)| \; dx + \int_{-\delta}^\delta |\phi(x)|\; dx \le (1 + \|\phi\|_{L^1}) \epsilon$$

• Can you be a little more detailed? I am new to the whole concept of distributions and I am not yet able to understand the part with the inequalities. – 010514 May 4 '15 at 19:59