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 $K$: \begin{equation}\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 \end{equation}

$f_x(x)$ and $f_n$ are both continuous on the compact set $K$ and hence both are uniformly continuous on the compact set $K$: \begin{equation}\tag{2} \forall\epsilon > 0, ~~ \exists \delta>0,~~ s.t.~~|x-y|<\delta \rightarrow |f(x)-f(y)| < \epsilon \end{equation} and \begin{equation}\tag{3} \forall\epsilon > 0, ~~ \exists \bar{\delta}>0,~~ s.t.~~|x-y|<\bar{\delta} \rightarrow |f_n(x)-f_n(y)| < \epsilon \end{equation} 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: \begin{equation}\tag{5} K \subset \cup_{x \in K}V_{\delta_{min}(x)} \end{equation}

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

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 \begin{equation} \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, \end{equation} 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.

  • 1
    $\begingroup$ I believe the error is that $\overline{\delta}$ depends on $n$, right? $\endgroup$ Jun 13, 2022 at 1:09
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    $\begingroup$ Can you visualize what is $\delta_{min}$ in a counter-example? $\endgroup$ Jun 13, 2022 at 1:10
  • $\begingroup$ @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. $\endgroup$
    – abk
    Jun 13, 2022 at 1:33
  • 1
    $\begingroup$ Great question, and thanks for all the effort you put into it. $\endgroup$
    – K.defaoite
    Jun 13, 2022 at 1:36
  • $\begingroup$ 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 $f_n(2^{-1/n})=1.$ $\endgroup$ Jun 13, 2022 at 4:16


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