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 \begin{equation} g_\varepsilon(x) = |x|+\frac{2\varepsilon}{|x|+\sqrt{\varepsilon}}\hspace{2cm} h_\varepsilon(x) = |x|+\frac{\varepsilon}{|x|+2\sqrt{\varepsilon}} \end{equation} 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.

  • $\begingroup$ 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 $a\to 0$? $\endgroup$ – Ilya Dec 4 '14 at 12:33
  • $\begingroup$ 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. $\endgroup$ – slo Dec 4 '14 at 14:29
  • $\begingroup$ @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: $$f_n(x) = |x|+\frac{n^{-1}}{|x|+n^{-1/2}} $$ $\endgroup$ – 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.

  • $\begingroup$ 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 :-). $\endgroup$ – Eff Dec 4 '14 at 18:10
  • $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Dec 4 '14 at 19:41
  • $\begingroup$ I have not gone through the computations yet, but this is closer to what I'm looking for, +1. $\endgroup$ – 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 $$ \lim_{n\rightarrow\infty} h_n(x) = x\lim_{n\rightarrow\infty} x^\frac{1}{2n-1}= |x|,\qquad \forall x\in[-1,1]. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.