Problem 1

Exercise 11.1 (Dini's Theorem)   Suppose that $\{f_n\}$ is a sequence of continuous functions from $[0,1]$, where each $f_n(x)$ is monotone increasing. And suppose that $f_n(x)$ converges to a continuous function $f(x)$ pointwisely on $[0,1]$.

1. Show that $f_n(x)$ in fact uniformly converges to $f(x)$ on $[0,1]$.
2. Give an example where the uniform convergence fails if the limit function $f(x)$ is not continuous.

Proof. For (a), let $f_n\to f$ as described. Define

$$E_n = \{x\in [0,1] \;\vert\; f(x) < f_n(x) + \epsilon\}.$$ (11.1)

Since $f_1(x) \leq f_2(x) \leq \cdots \leq f(x)$, we have

$$E_1\subseteq E_2 \subseteq \cdots \subseteq [0,1]$$ (11.2)

and by pointwise convergence it follows that

$$[0,1] = \bigcup_{n=1}^\infty E_n.$$ (11.3)

Extracting a finite subcover, we find

$$[0,1]=\bigcup_{n=1}^N E_n = E_N$$ (11.4)

Now show the convergence is uniform. Let $n>N$. Then $x\in E_N$ and

$$\Vert f_n-f\Vert = \sup_{x\in[0,1]}\vert f_n(x)-f(x)\vert$$ (11.5)

$$= \sup f(x)-f_n(x)$$ (11.6)

$$\leq \epsilon$$ (11.7)

For (b), consider $f_n(x)=1-x^n$, converging to $1-1_{\{1\}}$. A uniformly converging sequence of continuous functions converges to a continuous function, so we can see that the convergence must not be uniform by contradiction. $\qedsymbol$