Functions which are continuous but non differentiable at a point except modulus function

Question

I have noticed one thing during solving problems : That is, wherever I find a function which is continuous but non differentiable at a point, there has always been some |.|(modulus) function or [.](greatest integer function) in it.

I feel that's the only way one can have a sharp point in the graph of a function. Every other function tend to be smooth at all points.I am unable to manipulate them into a non differentiable continuous function by adding, multiplying, squaring or any other operations.

Is my assumption true ? If not, can you give an example of a function which is continuous but non differentiable at a point except modulus function or GIF function ?

√x is actually √|x|. So, you are using |.| @MichaelHoppe — Shubham, Oct 03 '15 at 10:54
I guess, in your graph you have used more than one function: from infinity to zero you have a function and after zero you have a different function, I am talking about one single function @ThePortakal — Shubham, Oct 03 '15 at 10:59
https://www.google.co.in/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=x%5E2%2F3 - its differentiable at 0 @DavidMitra — Shubham, Oct 03 '15 at 11:02
i guess its still incorrect....@DavidMitra check https://www.google.co.in/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=x%5E1%2F3 — Shubham, Oct 03 '15 at 11:06
Those aren't the correct graphs. (Input "y= x^(2/3)" for $f(x)=x^{2/3}$). — David Mitra, Oct 03 '15 at 11:08
Google interpreted "x^2/3" as "(x^2)/3". You have to enter "x^(2/3)" to get 2/3 as the exponent. — David Mitra, Oct 03 '15 at 11:12
Indeed, most continuous functions are far worse than you could possibly imagine. Almost all of them are nowhere differentiable. We deal only with the nice ones because that's all our feeble minds can handle. — Matt Samuel, Oct 03 '15 at 11:21

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

Consider the Thomae function; $f:[0,1]\to\mathbb R$ where $$f(x)=\begin{cases} \frac1q,& x\in\mathbb Q,\ x=\frac pq, \ p,q\in\mathbb Z, \ p\not\lvert q, \ q>0\\ 0,& x\notin\mathbb Q. \end{cases}$$ Then $f$ is continuous on $[0,1]\setminus\mathbb Q$ but not continuous on $[0,1]\cap\mathbb Q$. See this MSE question for a proof: Proving Thomae's function is nowhere differentiable.

For an even more pathological example, we have the Weierstrass function $w:\mathbb R\to\mathbb R$, defined by $$w(x) = \sum_{n=0}^\infty a^n\cos(b^n\pi x), $$ where $0<a<1$, $b$ is a positive odd integer, and $$ab > 1 + \frac32\pi. $$ This function is continuous everywhere but differentiable nowhere. The proof of this is not trivial; see this paper: https://pure.ltu.se/ws/files/30923977/LTU-EX-03320-SE.pdf

score 0 · Answer 2 · answered Oct 03 '15 at 18:49

0

Consider $f(x)=max(x, x^2)$.It is continuous but not differentiable at x=1.

answered Oct 03 '15 at 18:49

Aditya Anand

101

Functions which are continuous but non differentiable at a point except modulus function

2 Answers2