# Uniform convergence of a sequence of monotonic functions on closed intervals?

The problem is,

$$f_n$$ is monotonic on $$\left [ a,b \right ]$$ for all $$n$$, and $$\left \{ f_n(x) \right \}$$ converges to a continuous function $$f$$ on this closed interval, then the convergence is uniform.

Compared to Dini's Theorem, the problem above change the condition of monotonicity on $$n$$ i.e. ($$f_n \le f_{n+1},\; \forall x$$) to that on $$x$$ i.e. ($$f_n(x) < f_n(y)$$ if $$x) and remove continuous of $$f_n$$.

My question is ,

1. Can this question be proved by Heine–Borel theorem(as in Dini's Theorem), or when(or what conditions does $$f_n$$ satisfy) can we use Heine–Borel theorem to prove uniform convergence on a closed intervel.
2. Is this proof below related to Heine–Borel theorem in some way?

This problem can be proved by this method,

Since $$f \in C\left [ a,b \right ]$$ then $$f$$ is uniformly continuous on $$\left [ a,b \right ]$$ , thus $$\forall \varepsilon >0,\; \exists\delta >0$$, for $$|x^{\prime}-x^{\prime\prime}|<\delta$$, we have $$\mid f(x^{\prime})-f(x^{\prime\prime})\mid<\varepsilon.\quad (x^{\prime}, x^{\prime\prime} \in \left [ a,b \right ])$$ Take the natural number $$k$$, s.t. $$\frac{b-a}k<\delta$$ then divide the interval into $$k$$ equal parts. $$a=x_0 Then, \begin{aligned} (1)& \left|f\left(x_{i}\right)-f\left(x_{i-1}\right)\right|<\varepsilon ;\\ (2)& \forall x \in\left[x_{i-1}, x_{i}\right], \left|f(x)-f\left(x_{i}\right)\right|<\frac{\varepsilon}{2},\left|f(x)-f\left(x_{i-1}\right)\right|<\frac{\varepsilon}{2} ;\\ (3)& \exists N_{i} , n>N_{i} , \left|f_{n}\left(x_{i}\right)-f\left(x_{i}\right)\right|<\frac{\varepsilon}{2}(i=0,1, \cdots, k) .\\ \end{aligned}

Let $$N=\max\left\{N_{0},N_{1},\cdots,N_{k}\right\}$$, $$\forall x{\in}[x_{i-1},x_{i}]$$ when $$n>N$$ we have \begin{aligned} &\mid f_n(x_{i-1})-f(x)\mid\leqslant\mid f_n(x_{i-1})-f(x_{i-1})\mid+ \mid f(x_{i-1})-f(x)\mid<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon, \\ &\mid f_n(x_i)-f(x)\mid\leqslant\mid f_n(x_i)-f(x_i)\mid+ \mid f(x_i)-f(x)\mid<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}

Then $$\forall x \in[a, b] ,\exists i (1 \leqslant i \leqslant k)$$, s.t. $$x \in\left[x_{i-1}, x_{i}\right]$$, By the monotonicity of $$f_n$$, we have $$\mid f_{n}(x)-f(x)\mid\leqslant\max\{\mid f_{n}(x_{i+1})-f(x)\mid,\mid f_{n}(x_{i})-f(x)\mid\}<\varepsilon.$$

• (Side comment, of no help for your question) Though seemingly due to Pólya, in France this is called Dini's second theorem. Commented Nov 28, 2023 at 15:41