Let $h:E \to E$ be a continous function (where $E:=[0,1]$) and $\epsilon >0$.
Is it possible to find a continous but nowhere differentiable function $f:E \to E$ such that $|f-h| < \epsilon$ on $E$?
Note: For the purposes of this question, it only makes sense to talk about differentiability on interior points of the domain. I know some people define one-sided "derivatives" at endpoints, but I do not.
WLOG, we can take $\epsilon< \frac 1 {100}$.
I already know that $v:E \to E$ by $v(x) = \sum_{i=1} ^ \infty 4^{-i} \phi(4^ix)$ is a continous but nowhere differentiable function. (Where $\phi$ is the zig-zag function with period $4$ which agrees with $|x|$ on $[-2,2]$.)
One of my first thoughts was to take $f(x)= \frac \epsilon 2 v(x)+h(x)$. This will work if $h$ is differentiable on $(a,b)$ and if the range stays small enough, but I don't think it works in general. [Look what happens when $h(x)=1- \frac \epsilon 2v(x).$]
Can we construct a continous but nowhere differentiable function which stays close enough to $h$? (Ideally, constructed from $v$?)
Best Answer
Let $\epsilon>0.$ Then by Weierstrass, there is a polynomial $p$ such that $|p-h|<\epsilon/2$ on $[0,1].$ Also by Weierstrass, there is a nowhere differentiable continuous function $v$ on $[0,1].$ Let $$M=\max_{[0,1]} |v|.$$ Then $p+\epsilon v/(2M)$ is continuous and nowhere differentiable, and for $x\in [0,1],$
$$\left |\left (p(x)+\epsilon v(x)/(2M)\right)-h(x)\right|$$ $$ \le |p(x)-h(x)| + |\epsilon v(x)/(2M)|$$ $$ < \epsilon/2 + (\epsilon/2)|v(x)/M| \le \epsilon.$$
Added later: So we have shown that there is a nowhere differentiable continuous $f_\epsilon$ on $[0,1]$ such that $|f_\epsilon-h|<\epsilon$ on $[0,1].$ This implies that on $[0,1],$
$$-\epsilon < f_\epsilon < 1+\epsilon\implies 0 < f_\epsilon+\epsilon < 1+2\epsilon $$ $$\implies 0<\frac{f_\epsilon+\epsilon}{1+2\epsilon}<1.$$
Verify that the last function is continuous, nowhere differentiable, maps $[0,1]$ to $[0,1],$ and on $[0,1],$
$$\left |\frac{f_\epsilon+\epsilon}{1+2\epsilon}-h\right| < 4\epsilon.$$