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
 A: 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 $A\supset f^{-1}(\{f(t)\}\times [k2^{-n},(k+1)^{-n}]).$
