# Sequence of differentiable functions converging to non-differentiable function

Purely out of interest, I wanted to try and construct a sequence of differentiable functions converging to a non-differentiable function. I began with the first non-differentiable function that sprung to my mind, namely \begin{align} &f:\mathbb{R}\to\mathbb{R}\\ &f(x)=|x|. \end{align} After some testing I considered the function defined by $$f_\varepsilon(x) = |x|+\frac{\varepsilon}{|x|+\sqrt{\varepsilon}}$$ for some $\varepsilon>0$. Then $\lim\limits_{\varepsilon\to0^+}f_\varepsilon(x)=f(x)$, and $f_\varepsilon(x)$ looks smooth, i.e. differentiable for every $\varepsilon>0$ on the entire domain.

Question: How can I prove that $f_\varepsilon$ is differentiable for every $\varepsilon>0$ (or disprove) using the definition of the derivative?

If this assertion is true, then I construct the sequence simply by setting $\varepsilon = 1/n$ for $n\in\mathbb{N}$.

Attempt: I set up the definition for the derivative \begin{align} \frac{\mathrm{d}f_\varepsilon}{\mathrm{d}x} &= \lim_{h\to 0}\frac{1}{h}\left[\left(|x+h|+\frac{\varepsilon}{|x+h|+\sqrt{\varepsilon}}\right)-\left(|x|+\frac{\varepsilon}{|x|+\sqrt{\varepsilon}}\right)\right]\\ &=\lim_{h\to 0}\frac{1}{h}\left[|x+h|-|x|+\frac{\varepsilon}{|x+h|+\sqrt{\varepsilon}}-\frac{\varepsilon}{|x|+\sqrt{\varepsilon}}\right], \end{align} but I could not figure out how to proceed.

Sidenotes: An interesting thing I discovered when constructing $f_\varepsilon$, was that almost any small change removes its smoothness, for example $$g_\varepsilon(x) = |x|+\frac{2\varepsilon}{|x|+\sqrt{\varepsilon}}\hspace{2cm} h_\varepsilon(x) = |x|+\frac{\varepsilon}{|x|+2\sqrt{\varepsilon}}$$ do both not look smooth at all. Similarly for the other terms; changing the coefficients will remove the smoothness. I am also somewhat intrigued by this. So if anyone can shed some light on this, even better.

• Unrelated to your construction, what about taking $f(x) = |x|$ and fitting it with a polynomial on $[-a,a]$ with a smooth fit, and letting $a\to 0$? – Ilya Dec 4 '14 at 12:33
• Interestingly, in Cauchy's Cours d'Analyse de l'École Polytechnique there is a theorem stating that the limit of a sequence of continuous functions is continuous as well. Abel proved this wrong with the example $f_n(x)=\sum_{i=1}^n (-1)^{n-1}\frac{\sin nx}{n}$ which converges to a saw-tooth function. If you want to construct your own sequence, I suggest looking at Fourier series. – slo Dec 4 '14 at 14:29
• @slo Interesting. Right now I am looking mostly at differentiability, not just continuity. Perhaps I will try with Fourier series, but right now I'm interested in the sequence that I chose: $$f_n(x) = |x|+\frac{n^{-1}}{|x|+n^{-1/2}}$$ – Eff Dec 4 '14 at 14:41

Why do you think your function is differentiable? Did you calculate the derivatives? What does the picture look like for $ε=1$ or $ε=10$?

An easier differentiable approximation of the abs function is $$\sqrt{ε^2+x^2}$$ or $$\sqrt{ε^2+x^2}-ε.$$

Differentiability here is obvious by the chain rule.

Added: Close to zero, $|x|\lt \sqrt ε$, one can use the binomial formula to get \begin{align} |x|+\frac{ε}{|x|+\sqrt{ε}} &= |x|+ε\frac{\sqrt{ε}-|x|}{ε-x^2} =\frac{ε\sqrt{ε}-x^2|x|}{ε-x^2} \end{align} which tells that the function is twice continuously differentiable and symmetric at the origin (which we knew before), so that it has horizontal slope there.

• Good idea using $\sqrt{x^2}=|x|$ as a way to approximate the function, but I'm not looking for another approximating function. The function looks smooth (plot). Using WolframAlpha: $$f_\varepsilon'(x) = \frac{x(|x|+2\sqrt{\varepsilon})}{(|x|+\sqrt{\varepsilon})^2}$$ which is a continuous function when $\varepsilon>0$, supporting the idea that $f_\varepsilon\in\mathcal{C}^1$. If you have any tip for calculating the derivative with the definition let me know :-). – Eff Dec 4 '14 at 18:10
• Using the binomial formula, one can find an expression for the original formula with smooth denominator, so that the differentiability can be read off from the numerator. – Lutz Lehmann Dec 4 '14 at 19:41
• I have not gone through the computations yet, but this is closer to what I'm looking for, +1. – Eff Dec 4 '14 at 21:10

Consider the sequence of differentiable functions $h_n(x) = x^{1+\frac{1}{2n-1}}$ defined on $[-1,1]$ and note $$\lim_{n\rightarrow\infty} h_n(x) = x\lim_{n\rightarrow\infty} x^\frac{1}{2n-1}= |x|,\qquad \forall x\in[-1,1].$$