# Is it possible for function $f : \mathbb{R} \to \mathbb{R}$ have a maxima at every point in a countable dense subset of its domain?

Is it possible for function $f : \mathbb{R} \to \mathbb{R}$ have a maxima at every point in a countable dense subset of its domain ? The motivation for this question is I have a sequence of functions $\{f_n\}$ where the number of maxima increase with $n$ and I am interested to know what happens to the sequence of functions.

PS : every function of the sequence has a finite number of maxima.

EDIT : $f$ should not be constant function.

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What if the function $f$ was constant?... ;) Or do you want to have the set of maximas to be countable? If you can allow ''more'' maximas, the constant function would be an example of one such function. –  Patrick Da Silva Jun 3 '11 at 5:04
@Patrick: Or, if you insist on countable, the characteristic function of the rationals. –  Ross Millikan Jun 3 '11 at 5:06
@Patrick : maxima means second derivative should be negative. –  Rajesh D Jun 3 '11 at 5:07
@Rajesh: So you want your function to be twice differentiable. Please include that in the statement of your question. –  Yuval Filmus Jun 3 '11 at 5:08
@Yuval : Ok, i do not need it to be differentiable but i do not want constant function either. –  Rajesh D Jun 3 '11 at 5:10

Thomae's function has a strict local maximum at each rational number.

I believe the Weierstrass function is another example.

Another question on this site posed the problem of showing that if $f$ is continuous and not monotone on any interval, then $f$ has a local maximum at each point in a dense subset of $\mathbb{R}$.

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You're quick, aren't you. XD –  Patrick Da Silva Jun 3 '11 at 5:13
I'd like to know if it's possible for the sequence of functions $\{f_n\}$ where each one is smooth to converge to such a function ? –  Rajesh D Jun 3 '11 at 5:20
@Rajesh: The Weierstrass function is a uniform limit of smooth functions. –  Jonas Meyer Jun 3 '11 at 5:22
Sample paths of Brownian motion have this property (with probability $1$), see here.