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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!

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$$\begin{align} \int_{-\infty}^{\infty}(f_n(x)-f(x))\phi(x)dx&=\int_{0}^{\infty} ((f_n(x)-f(x))\phi(x)+(f_n(-x)-f(-x))\phi(-x))dx\\\\ &=\int_{0}^{\infty} (f_n(x)-f(x))(\phi(x)-\phi(-x))dx \end{align}$$

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.

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  • $\begingroup$ 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? $\endgroup$ – 010514 May 4 '15 at 21:34
  • $\begingroup$ 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 :/ $\endgroup$ – 010514 May 4 '15 at 21:50
  • $\begingroup$ You are welcome. My pleasure. The minus sign is correct. Upon applying the absolute value, the minus sign is absorbed. $\endgroup$ – Mark Viola May 4 '15 at 21:51
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    $\begingroup$ 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) $1/\cosh(nx)<1$. $\endgroup$ – Mark Viola May 4 '15 at 21:54
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    $\begingroup$ $f_n-f = -\frac{e^{-nx}}{\cosh nx}$ is correct. There is indeed a minus sign. $\endgroup$ – Mark Viola May 4 '15 at 21:55
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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 $(-\delta,\delta)$, $$\left|\langle T_{f_n}, \phi \rangle - \langle T_f, \phi \rangle \right| \le \epsilon \int_{\mathbb R} |\phi(x)| \; dx + \int_{-\delta}^\delta |\phi(x)|\; dx \le (1 + \|\phi\|_{L^1}) \epsilon$$

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  • $\begingroup$ 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. $\endgroup$ – 010514 May 4 '15 at 19:59

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