Uniform convergence via Arzela-Ascoli

Question

I want to show that $u_\epsilon = -\epsilon \log\left(\frac{ e^{\frac{x}{\epsilon}} + e^{-\frac{x}{\epsilon}}}{e^{\frac{1}{\epsilon}} + e^{-\frac{1}{\epsilon}}} \right)$ converges uniformly to $1-|x|$ on $[-1,1]$ as $\epsilon \downarrow 0$. I have shown the pointwise convergence for all $x\in[-1,1]$ - now I want to deduce the uniform convergence.

We know that $(u_\epsilon(x))' = -\tanh(\frac{x}{\epsilon})$. Now we could use Arzela-Ascoli and get a subsequence (subfamily) from $(u_\epsilon)'$ that converges uniformly on $[-1,1]$. Then we could use this subsequence (subfamily) and deduce (I don't know the english name for this theorem - the theorem states basically: If $(u_\epsilon)'$ converges uniformly and $u_\epsilon$ converges pointwise for at least one point, then $u_\epsilon$ converges uniformly to a continuosly differentiable function) that $u_\epsilon$ converges uniformly (to $1-|x|$). But this limit is not continuosly differentiable in $x=0$. Did I make a mistake somewhere or can the uniform convergence be shown differently?

Thanks along!

Answer 1

You don't need the sledgehammer of the Arzelà-Ascoli theorem to get uniform convergence. A little algebraic manipulation and a few elementary inequalities will give it to you: \begin{align} \big|\, u_\epsilon(x)-(1-|x|)\,\big|&=\epsilon\left|\,\log\left(e^\frac{|x|-1}{\epsilon}\right)-\log\left(\frac{e^\frac{|x|}{\epsilon}+e^{\frac {-|x|}{\epsilon}}}{e^\frac{1}{\epsilon}+e^\frac{-1}{\epsilon}}\right)\,\right|\\ &=\epsilon\left|\,\log\left(\frac{e^\frac{|x|}{\epsilon}+e^{\frac {-|x|}{\epsilon}}}{e^\frac{1}{\epsilon}+e^\frac{-1}{\epsilon}}\right)-\log\left(e^\frac{|x|-1}{\epsilon}\right)\,\right|\\ &=\epsilon\left|\,\log\left(\frac{e^\frac{|x|-1}{\epsilon}+e^{\frac {-|x|-1}{\epsilon}}}{1+e^\frac{-2}{\epsilon}}\right)-\log\left(e^\frac{|x|-1}{\epsilon}\right)\,\right|\\ &=\epsilon\left|\,\log\left(\frac{1+e^{\frac {-2|x|}{\epsilon}}}{1+e^\frac{-2}{\epsilon}}\right)\,\right|\\ &\le\epsilon\left|\,\log\left(1+e^{\frac {-2|x|}{\epsilon}}\right)\,\right|\\ &\le\epsilon e^{\frac {-2|x|}{\epsilon}}\\ &\le\epsilon\ . \end{align}