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 $x<y$) and remove continuous of $f_n$.
My question is ,
- 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.
- 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. $$ a=x_0<x_1<\cdots<x_k=b,\quad\Delta x_i=x_i-x_{i-1}=\frac{b-a}{k},\quad i=1,2,\cdots,k. $$ 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 $$ \mid f_{n}(x)-f(x)\mid\leqslant\max\{\mid f_{n}(x_{i+1})-f(x)\mid,\mid f_{n}(x_{i})-f(x)\mid\}<\varepsilon. $$
