# Alternative proof for Dini's theorem

I am attempting to prove Dini's theorem without consulting the exiting proof for it. However, in my proof I do not use the fact that $$f_n(x)$$ is an increasing sequence for a given x. So I am certain my proof is wrong but I need help to spot the flaw. Basically I prove the following hypothesis:

Hypothesis: If $$f_n \rightarrow f$$ pointwise on a compact set $$K$$, and both $$f_n$$ and $$f$$ are continuous on $$K$$ then $$f_n$$ uniformly converges to $$f$$.

Proof: $$f_n \rightarrow f$$ pointwise on compact set $$K$$: $$$$\tag{1} \forall\epsilon > 0, ~~ \forall x \in K, ~~ \exists N_x \in \mathbb{N},~~ s.t.~~ n \geq N_x \rightarrow |f_n(x)-f(x)| < \epsilon$$$$

$$f_x(x)$$ and $$f_n$$ are both continuous on the compact set $$K$$ and hence both are uniformly continuous on the compact set $$K$$: $$$$\tag{2} \forall\epsilon > 0, ~~ \exists \delta>0,~~ s.t.~~|x-y|<\delta \rightarrow |f(x)-f(y)| < \epsilon$$$$ and $$$$\tag{3} \forall\epsilon > 0, ~~ \exists \bar{\delta}>0,~~ s.t.~~|x-y|<\bar{\delta} \rightarrow |f_n(x)-f_n(y)| < \epsilon$$$$ I define $$\delta_{min}=\{\delta, \bar{\delta}\}$$, then using $$(1)-(3), ~~ \forall y \in V_{\delta_{min}}(x)$$ and $$n \geq N_x$$ we have: \begin{align}\tag{4} |f_n(y)-f(y)| &= |f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &< |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|\\&< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}= \epsilon \end{align}

Basically what $$(4)$$ says is that for all $$y$$ in the $$\delta_{min}$$ neighborhood of x (ball of radius $$\delta_{min}$$), there exists a single $$N_x$$ for which $$|f_n(y)-f(y)|$$ could be arbitrarily small.

Now we can construct an open cover for $$K$$ using $$V_{\delta_{min}(x)}$$ as follow: $$$$\tag{5} K \subset \cup_{x \in K}V_{\delta_{min}(x)}$$$$

Because $$K$$ is compact the above open cover has a finite subcover (Heine-Borel theorem), consequently $$$$\tag{6} K \subset \cup_{i=1}^{N}V_{\delta_{min}(x_i)}$$$$

As we argued in $$(4)$$ for all $$y \in V_{\delta_{min}(x_i)}$$ there exists a single $$N_{x_i}$$ such that $$n \geq N_{x_i}$$ then $$|f_n(y)-f(y)| < \epsilon$$. Hence, for $$N=max(N_{x_1}, \cdots, N_{x_N})$$ when $$n \geq N$$ then $$|f_n(y)-f(y)| < \epsilon,$$ $$\forall y \in \cup_{i=1}^{N}V_{\delta_{min}(x_i)}$$.

Finally, from the above and $$(6)$$ we can conclude that $$$$\forall\epsilon > 0, ~~ \exists N=max(N_{x_1}, \cdots, N_{x_N}),~~ when~~ n \geq N \rightarrow |f_n(x)-f(x)| < \epsilon, \forall x \in K,$$$$ hence uniform convergence.

I need help to the logical flaw in my proof above as I know the hypothesis is wrong. I have also found this counter-example for the proof but still unable to spot the problem.

• I believe the error is that $\overline{\delta}$ depends on $n$, right? Jun 13, 2022 at 1:09
• Can you visualize what is $\delta_{min}$ in a counter-example? Jun 13, 2022 at 1:10
• @Dabouliplop great point! I can see $\bar{\delta}$ could depend on n but need to try to visualize it to see where exactly the problem is.
– abk
Jun 13, 2022 at 1:33
• Great question, and thanks for all the effort you put into it. Jun 13, 2022 at 1:36
• A bunch of counterexamples: take a continuous function $f(x)$ on $[0,1]$ such that $f(0)=f(1)=0$ and $f(1/2)=1.$ Then $f_n(x)= f(x^n)\to 0$ pointwise but $f_n(2^{-1/n})=1.$ Jun 13, 2022 at 4:16