- #1
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Does there exist a continuous function [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex] such that f is nowhere constant and [tex]\{x:f(x)=0\}[/tex] is uncountable?
Take the function f(x) = 0 on [0,1]. Now replace f on the interval [1/3,2/3] with a triangle wave. Now replace f on the intervals [1/9,2/9] and [7/9,8/9] with a similar triangle wave. Repeat this process for every interval on which f is zero, and we have a continuous function that vanishes on the Cantor set (which is uncountable), and is nowhere constant. Replacing the triangle wave with a suitable [itex]C^{\infty}[/itex] function yields an infinitely differentiable function that vanishes uncountably many times but is not constant. If you want it to be continuous but nowhere differentiable, replace each triangle wave with such a function instead.
"if we infinitely repeat..." simply refers to the limit function of the sequence of functions I described.
Define
[tex]
f_1[a,b](x) = \left\{
\begin{array}{cc}
0 & x \in \left[a,\frac{2a+b}{3}\right]\cup\left[\frac{a+2b}{3},b\right]\\
x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\
\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]
\end{array}
[/tex]
and
[tex]
f_{n+1}[a,b](x) = \left\{
\begin{array}{cc}
f_n\left[a,\frac{2a+b}{3}\right](x) & x \in \left[a,\frac{2a+b}{3}\right]\\
x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\
\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]\\
f_n\left[\frac{a+2b}{3},b\right](x) & x \in \left[\frac{a+2b}{3},b\right]
\end{array}
[/tex]
Then [tex]f(x) = \lim_{n\to\infty}f_n[0,1](x)[/tex] is the function I just described.