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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:$$ ||f||=max_{x \in [0,1]} |f(t)|$$

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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 $\pi$.

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Thanks. This is the kind of example I was looking for. –  user73793 May 23 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.

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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 at 14:44

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