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Infrared
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wrobel said:It is clear there is an interval ##[a,b]\subseteq[0,1]## such that ##f(a)=f(b)=0,\quad |f|\mid_{(a,b)}>0##.
Consider a function ##F(x)=\ln|f(x)|-x##. We have ##F(x)\to-\infty## as ##x\to b-## or ##x\to a+##.
Thus for some ##\tilde x\in(a,b)## we have ##F'(\tilde x)=0##
This solves the problem
This is right, but I do think your first claim deserves a proof. If ##f## is the constant zero function, then the problem is solved. Otherwise, suppose ##f(p)\neq 0## for some ##p##. Then the set ##\{x\in [0,p]: f(x)=0\}## is nonempty, closed and so has a maximal element ##a## (that is not ##p##), and similarly, there is a minimal zero ##b## larger than ##p##. Then ##[a,b]## is the interval you are looking for.
Also, there still some outstanding problems in the March II thread (https://www.physicsforums.com/threads/math-challenge-march-2020-part-ii.985311/) that you might be interested in :)
This is not true, consider ##f(x)=\begin{cases}Not anonymous said:Is it valid to assume that for any differentiable function ##f:[0,1] \rightarrow \mathbb{R}## that is not the constant-value function but has infinite number of zeros, its zeros will form a countably infinite set?
0 & x\leq 1/2 \\
(x-1/2)^2 & x\geq 1/2\\
\end{cases}.##
Not anonymous said:Or that for any such function, there will be some interval ##I ⊂[0,1]## containing 2 or more zeros of ##f## with all zeros falling in that interval forming a finite or a countably infinite set (i.e. set of zeros of ##f## contains a finite or countably infinite subset, with all remaining zeros, if any, lying outside the range defined by this finite or countably infinite subset)?
The issue isn't having countable/uncountably many zeros, but see my above response to @wrobel to see how to find an interval that you're looking for. You should be wary of taking "consecutive elements" of a countable set. For example, there are no consecutive elements in ##\mathbb{Q}##.
Also, this problem has a cute one-line solution: apply Rolle's theorem to the function ##g(x)=e^{-x}f(x)## on the interval ##[0,1].## This is just the exponential of @wrobel's function, and works for the same reason, but has the advantage of being defined everywhere, so there isn't any need to find a sub-interval to work with.
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