Let the interval $[0,1]$ be divided into $n$ subintervals each of length $\frac{1}{n}$. Let $f\in C([0,1], \mathbb R)$ a continuous functions in $[0,1]$ and consider the set $\Omega_n=\big(f\in C^2([0,1], \mathbb R) : f'({\frac{k}{n}})=0, \forall k=0,\ldots, n\big)$. Is that true that $\forall \varepsilon>0$ there exists $f_n\in\Omega_n$ such that


Which kind of functions $f_n$ I can try to define?

  • $\begingroup$ You are using $f$ in two different ways. $\endgroup$ – zhw. Jun 14 at 15:47
  • $\begingroup$ Do you know the Weierstrass approximation theorem? $\endgroup$ – zhw. Jun 14 at 16:35

Consider $b_n(t) = 2\int_0^t \sin^2 (\pi nx)$, note that $b_n \in \Omega_n$, $b_n$ is strictly increasing, $b_n(0) = 0, b_n(1) = 1$.

If you plot $b_n$ for a few values of $n$ you will notice that it is a smooth staircase function whose steps are getting smaller and smaller with increasing $n$.

Note that $b_n(t) = t$ for $t ={k \over n}$ and combine this with monotonicity to show that $b_n(t) \to t$ uniformly.

Choose $\epsilon>0$ and find a polynomial $p$ such that $\|f-p\| < { 1\over 2 } \epsilon$. Now consider $p_n = p \circ b_n$ and use uniform continuity to conclude that $\|p-p_n\| \to 0$. Note that $p_n \in \Omega_n$. Choose $n$ large enough such that $\|p-p_n\| < { 1\over 2 } \epsilon$, then $\|f-p_n \| < \epsilon.$

  • 1
    $\begingroup$ But $f$ is merely continuous, so $f\circ b_n\notin C^2$ may happen. So I think you first want $g\in C^2$ with $\|f-g\|_\infty$ small, then look at $g\circ b_n.$ $\endgroup$ – zhw. Jun 18 at 18:51
  • $\begingroup$ @zhw.: You are correct. Thanks for catching that. I added a hack fix. $\endgroup$ – copper.hat Jun 18 at 18:55

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