Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to construct a sequence $f_n$ of continuous functions in $[0,1]$ such that $||f_n||=1$ (so a bounded sequence) and $||f_n-f_m||=1$ (it doesn't have any convergent subsequences). The norm is the maximum norm defined as:$$ ||f||=max_{x \in [0,1]} |f(t)|$$

share|cite|improve this question
up vote 1 down vote accepted

Let $f_n(x)=|\sin(2^n\pi x)|$. For $m>n$, you have $f_n(2^{-(n+1)})=|\sin(\pi/2)|=1$ but $f_m(2^{-(n+1)})=|\sin(2^{m-n-1}\pi)|=0$ since $2^{m-n-1}\pi$ is an integer multiple of $\pi$.

share|cite|improve this answer
Thanks. This is the kind of example I was looking for. – user73793 May 23 '14 at 15:08

Given a closed interval $X$ and an open interval $Y$ containing $X$, you can construct a function $\phi$ such that $\phi(X)=1$ and $\phi([0,1]\setminus Y)=0$. (This is called a "bump function"; there is a standard construction involving the function defined as $0$ for $x≤0$ and $e^{-1/x}$ for $x>0$.) Now if you make disjoint open intervals in $[0,1]$, you can use them as $Y$ in the above construction which will give you the desired construction.

share|cite|improve this answer
Ok. I think I got what you mean. Do you know if there are other simpler examples expressible in function of $x$ and $n$? – user73793 May 23 '14 at 14:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.