Uniform convergence of the inverse (solution verification)

Question

Consider a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$. Let $f_n$ be a sequence of non-decreasing functions converging uniformly to $f$. Show that $f_n^{-1}$ converges uniformly to $f^{-1}$ over the open interval $(0, 1)$ where we define $g^{-1}(\alpha) := \inf \{y | g(y) \geq g(\alpha)\}$.

What I've tried

We know that $f^{-1}$ is continuous (Inverse of a continuous increasing function is continuous (solution verification)).

We need to show that: $\sup_{\alpha \in (0,1)}|f_n^{-1}(\alpha) - f^{-1}(\alpha)| \rightarrow 0$ as $n\rightarrow \infty$.

Notice that: $$|f_n^{-1}(\alpha) - f^{-1}(\alpha)| = |f_n^{-1}(f(x)) - f^{-1}(f(x))|$$

for some $x$ such that $f(x) = \alpha$ (since $f$ is onto $(0, 1)$). Now we have:

\begin{equation}\tag{1} = |f_n^{-1}(f(x)) - x| \end{equation}

Applying the definition of $g^{-1}$:

$$f_n^{-1}(f(x)) = \inf\{ y | f_n(y) \geq f(x)\} = \inf\{ y | f(y) + o(1) \geq f(x)\}$$

where the second equality comes from the uniform convergence of $f_n$ to $f$. Finally, since $f^{-1}$ is continuous: $y \geq f^{-1}(f(x) + o(1)) \rightarrow x$ so the RHS above converges to $x$. So we have that expression (1) goes to $0$ showing the desired convergence.

I could used some feedback on my reasoning.

Answer 1

Having proved the pointwise convergence you are in fact done, because of the following

Claim: Let $(h_n)$ be a sequence of non-decreasing functions, each mapping $(0,1)$ into $\Bbb R$ and converging pointwise to a continuous strictly increasing function $h$ mapping $(0,1)$ onto $\Bbb R$. Then the convergence is uniform on compact subsets of $(0,1)$.

Fix $0<a<b<1$ and consider $h$ on the compact interval $[a,b]$. For a large integer $m$ partition $[h(a),h(b)]$ into $m$ equal-length subintervals, with endpoints $h(a)=y_0<y_1<\cdots<y_m=h(b)$. Define $x_k:=h^{-1}(y_k)$. Suppose $u\in[x_k,x_{k+1}]$ for some $k\in\{0,1,2,\ldots,m-1\}$. Then $$ h_n(x_k)-h(x_{k+1})\le h_n(u)-h(u)\le h_n(x_{k+1})-h(x_k), $$ and so $$ \sup_{a\le u\le b}|h_n(u)-h(u)|\le\left[\max_{0\le k\le m-1}|h(x_k)-h_n(x_{k+1})|\vee \max_{0\le k\le m-1}|h_n(x_k)-h(x_{k+1})|\right]. $$ Sending $n\to\infty$ we obtain $$ \limsup_n\sup_{a\le u\le b}|h_n(u)-h(u)|\le\max_{1\le k\le m}|h(x_k)-h(x_{k+1})|=1/m. $$ As $m$ can be made as large as we please, $\lim_n\sup_{u\in[a,b]}|h_n(u)-h(u)|=0$.