# Example of a continuous but no where differentiable R to R function that is not expressed as a series of functions?

A famous example is the Weierstrass function, which can be found here https://en.wikipedia.org/wiki/Weierstrass_function. I'm curious as to if there exists such functions that are not a series of functions. Especially, are there elementary functions that are continuous but no where differentiable?https://en.wikipedia.org/wiki/Elementary_function

• Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on this problem are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Commented May 18, 2018 at 23:57
• Well, let $f$ be such a function. Then $f=f+0 + 0 + 0 + \cdots$
– zhw.
Commented May 19, 2018 at 0:02
• The notion "expressed as a series of functions" is not clearly defined, is it? Anyway take a look here: pdfs.semanticscholar.org/8cfb/… Wen function (3.18) is expressed as infinite product, not a series :) Commented May 19, 2018 at 0:12
• @Virtuoz Very interesting thesis! Commented May 19, 2018 at 1:29

Let $f:[0,1]\to [0,1]^2$ be a continuous surjection of the type described by Peano or Hilbert (see "Space-filling curve" in Wikipedia) with $f(0)=(0,0)$ and $f(1)=(1,1).$ Let $g:[0,1]^2\to [0,1]$ be the projection onto the first co-ordinate. That is, $g(x,y)=x.$ Then $gf:[0,1]\to [0,1]$ is a continuous surjection and is nowhere differentiable.

We can extend $gf$ to the domain $\Bbb R$ by letting $gf(t+n)=n+gf(t)$ for $t\in [0,1]$ and $n\in \Bbb Z$ to get a continous nowhere-differentiable surjection from $\Bbb R$ to $\Bbb R.$

The idea is that $f$ is the uniform limit of a sequence $(f_n)_{n\in \Bbb N}$ of continuous functions (so that $f$ is continuous), such that for every $n\in \Bbb N$ and any non-negative integers $j,k<2^n$ there exists a unique non-negative integer $l<4^n$ such that $\{f(t): t\in [l4^{-n},(l+1)4^{-n}]\}=$ $= [j2^{-n},(j+1)2^{-n}]\times [k2^{-n},(k+1)2^{-n}].$

Let $t\in [0,1].$ For any $n\in \Bbb N$ there exists a non-negative integer $l$ such that $t\in T=[l4^{-n},(l+1)4^{-n}]\subset [0,1]$ and unique non-negative integers $j,k$ such that $f(T)=[j2^{-n}, (j+1)2^{-n}]\times [k2^{-n},(k+1)2^{-n}].$

Let $f(t)=(x,y).$ We have $gf(t)=x\in [j2^{-n}, (j+1)2^{-n}].$ Now take $m\in \{0,1\}$ such that $|x-(j+m)2^{-n}|\geq 2^{(-n-1)}.$ There exists $t'\in [l4^{-n},(l+1)4^{-n}]$ such that $f(t)=((j+m)2^{-n},y)$ so $gf(t')=(j+m)2^{-n}.$

So $0<|t'-t|\leq 4^{-n}$ while $|gf(t')-gf(t)|\geq 2^{(-n-1)}.$ Therefore $$\left | \frac {gf(t')-gf(t)}{t'-t}\right |\geq \frac {2^{(-n-1)}}{4^{-n}}=2^{n-1}.$$ Letting $n\to \infty$ we have $\lim \sup_{t'\to t}\left |\frac {gf(t')-gf(t)}{t'-t}\right |=\infty.$

Another interesting property of $gf$ is that, although it is continuous but not constant on any interval of positive length, if $r>0$ and $t\in [0,1]$ then the cardinal of the set $A=\{u\in [0,1]\cap [t-r,t+r]: gf(u)=gf(t)\}$ is the cardinal of $\Bbb R.$ In the notation of the previous paragraph, take $n\in \Bbb N$ with $4^{-n}\leq r .$ Then $A\supset f^{-1}(\{f(t)\}\times [k2^{-n},(k+1)^{-n}]).$