Show that there is sequence of homeomorphism polynomials on [0,1] that converge uniformly to homeomorphism

Solution 1:

We can show the result along the lines you sketched. First, we note that it suffices to consider monotonically increasing $f$, for the transformation $g \mapsto 1-g$ is an isometry that preserves polynomials.

Now we need to approximate the homeomorphism $f$ by continuously differentiable homeomorphisms. For that, extend $f$ to $\mathbb{R}$ by setting

$$g(x) = \begin{cases} f(x) &, x \in [0,1] \\ x &, x \notin [0,1].\end{cases}$$

Further, let $\varphi$ be a non-negative even smooth function with compact support and $\int_\mathbb{R} \varphi(x)\,dx = 1$. Convoloution yields a family

$$g_\eta(x) = \int_\mathbb{R} g(x-\eta y)\varphi(y)\,dy$$

of strictly increasing smooth functions converging uniformly to $g$ on $\mathbb{R}$ for $\eta \searrow 0$, with $g_\eta(x) = x$ for $x \leqslant -\eta K$ or $x \geqslant 1+\eta K$ if the support of $\varphi$ is contained in $[-K,K]$.

Given $0 < \varepsilon < \frac{1}{2}$, choose $\eta > 0$ so small that $\lvert g_\eta(x)-g(x)\rvert < \frac{\varepsilon}{10}$ for all $x\in\mathbb{R}$. Then

$$h(x) = \frac{g_\eta(x) - g_\eta(0)}{g_\eta(1) - g_\eta(0)}$$

defines a smooth homeomorphism of $[0,1]$. We have

$$\begin{align} \lvert g_\eta(x) - h(x)\rvert &= \left\lvert \frac{g_\eta(x)(g_\eta(1)-g_\eta(0)) - (g_\eta(x)-g_\eta(0))}{g_\eta(1)-g_\eta(0)}\right\rvert\\ &\leqslant \frac{\lvert g_\eta(x)\rvert\cdot\lvert g_\eta(1)-1\rvert + \lvert g_\eta(0)\rvert\cdot \lvert 1-g_\eta(x)\rvert}{g_\eta(1)-g_\eta(0)}\\ &\leqslant 2\frac{\varepsilon}{10}\frac{1+\frac{\varepsilon}{10}}{1-2\frac{\varepsilon}{10}}\\ &< \frac{\varepsilon}{4}, \end{align}$$

so $\lvert g(x) - h(x)\rvert < \frac{\varepsilon}{2}$ for all $x\in [0,1]$.

Now, since $h$ is smooth and strictly increasing, $h'$ is continuous and strictly positive, hence you can uniformly approximate $h'$ by positive polynomials. If $Q$ is a positive polynomial such that $\lvert Q(x)-h'(x)\rvert < \frac{\varepsilon}{5}$ for $x\in [0,1]$, and $P(x) = \int_0^x Q(t)\,dt$, then $P$ is a strictly increasing (on $[0,1]$) polynomial with $\lvert P(x) - h(x)\rvert < \frac{\varepsilon}{5}$ for all $x\in [0,1]$ and $R(x) = \frac{P(x)}{P(1)}$ is a polynomial homeomorphism of $[0,1]$ with

$$\lvert P(x) - R(x)\rvert = \left\lvert \frac{P(x)(P(1)-1)}{P(1)}\right\rvert\leqslant \frac{\varepsilon}{5}\cdot\frac{1+\frac{\varepsilon}{5}}{1-\frac{\varepsilon}{5}} < \frac{\varepsilon}{4},$$

so

$$\lvert f(x) - R(x)\rvert \leqslant \lvert f(x) - h(x)\rvert + \lvert h(x) - P(x)\rvert + \lvert P(x) - R(x)\rvert < \varepsilon.$$