Exercise 6.2.11

Question

[Dini's Theorem]
  Assume $f_{n} \rightarrow f$ pointwise on a compact set $K$ and assume that for each $x \in K$ the sequence $f_{n}(x)$ is increasing. Follow these steps to show that if $f_{n}$ and $f$ are continuous on $K$, then the convergence is uniform.
  \begin{enumerate}[label=(\alph*)] 
  \item Set $g_{n}=f-f_{n}$ and translate the preceding hypothesis into statements about the sequence $\left(g_{n}\right)$.
  \item Let $\epsilon>0$ be arbitrary, and define $K_{n}=\left\{x \in K: g_{n}(x) \geq \epsilon\right\} .$ Argue that $K_{1} \supseteq K_{2} \supseteq K_{3} \supseteq \cdots$, and use this observation to finish the argument.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We want $(g_n) 	o 0$ uniformly, where $g_n$ is continuous, $g_n(x)$ is decreasing and $g_n(x) 	o 0$.
  \item $x \in K_{n+1}$ by definition has $g_{n+1}(x) \ge \epsilon$, since $(g_n(x))$ is decreasing we must also have $g_{n}(x) \ge \epsilon$. Thus $K_{n+1} \subseteq K_n$. Now, if each $K_n 
e \emptyset$ the nested compact set property (Theorem 3.3.5) would imply there exists an $x_0 \in \bigcap_{n=1}^\infty K_n$. But this is impossible because $g_n(x_0) 	o 0$ implies eventually $g_n(x_0)<\epsilon$.
    Therefore since the compact set property doesn't apply, there must exist an $N$ with $K_N = \emptyset$, implying every $n \ge N$ has (by subsets) $K_n = \emptyset$ and thus $|g_n| < \epsilon$.
   \end{enumerate}
  % TODO: Understand intuitively, I proved it but I have no idea why it's true

Exercise 6.2.11

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Exercise 6.2.11

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