# Approximation of a continuous function with a particular sequence of smooth functions

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

$$\sup_n\|f-f_n\|_\infty<\varepsilon$$?

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

• You are using $f$ in two different ways. – zhw. Jun 14 at 15:47
• Do you know the Weierstrass approximation theorem? – 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 $$\|f-p_n \| < \epsilon.$$
• 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 $g\circ b_n.$ – zhw. Jun 18 at 18:51