Proof for Uniform Convergence for $\{f_n\}$

Question

Suppose $\{f_n\}$ is an equicontinuous sequence of functions defined on $[0,1]$ and $\{f_n(r)\}$ converges $∀r ∈ \mathbb{Q} \cap [0, 1]$. Prove that $\{f_n\}$ converges uniformly on $[0, 1]$.

Since I know that $\mathbb{Q} \cap [0, 1]$ is not compact, I am a bit stuck on my proof.

So far I have:

Let $f_n \to f$ pointwise on $\mathbb{Q} \cap [0, 1]$

Since $\{f_n\}_n$ is equicontinuous and point-wise bounded (it’s pointwise convergent, so in particular), there exists a subsequence $\{f_{n_k}\}_k$ such that $f_{n_k} \to f$ uniformly.

Since each $f_n$ is continuous, $f$ is then continuous.

Now take $\varepsilon > 0$. Using equicontinuity of $\{f_n\}_n$, we find $\delta_1 > 0$ such that if $d(x, y) < δ_1$, $x, y \in K$, then $|f_n(x) − f_n(y)| < \varepsilon/3$ for all $n ∈ \mathbb{Z}^+$.

Using continuity of $f$, for each $x \in K$, let $\delta_2 = \delta_2(x) > 0$ be such that if $|x − y| < \delta_2(x)$, $y \in \mathbb{Q} \cap [0, 1]$, then $|f(x) − f(y)| < \varepsilon/3$. For $x \in \mathbb{Q} \cap [0, 1]$, let $\delta(x) = \min(\delta_1, \delta_2(x)) > 0$

I am not sure how to continue nor am I too sure I am on the right path.

Answer 1

Step 1: For every $x\in [0,1]$, $\{f_n(x)\}_n$ converges.

Fix $\varepsilon >0$. Pick $\delta>0$ witnessing the definition of equicontinuity for $\varepsilon/3$. Pick a rational number $r$ with $|x-r|<\delta$. Fix $N$ such that $|f_n(r)-f_m(r)|<\varepsilon/3$ for every $n,m\ge N$.

If $n,m\ge N$, then $$ |f_n(x)-f_m(x)|\le |f_n(x)-f_n(r)|+|f_n(r)-f_m(r)|+|f_m(r)-f_m(x)|\le \varepsilon $$ Thus $\{f_n(x)\}_n$ is a Cauchy sequence, and we're done by equicontinuity.

Let $f(x):=\lim_n f_n(x)$.

Step 2: The convergence is uniform. See this answer

Answer 2

Firstly, we show that $\lim_{n\rightarrow\infty}f_{n}(x)$ exists for all $x\in[0,1]$. Let $x\in[0,1]$. Let $\varepsilon>0$ be arbitrary. By equicontinuity, there exists $\delta>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$ whenenver $n\in\mathbb{N}$ and $y\in[0,1]\cap(x-\delta,x+\delta)$. By density of $\mathbb{Q}$, there exists $r\in[0,1]\cap(x-\delta,x+\delta)$. Choose $N$ such that $|f_{n+p}(r)-f_{n}(r)|<\varepsilon$ whenever $n\geq N$ and $p\in\mathbb{N}$ (this is possible because $\{f_{n}(r)\}_{n}$ is convergent). For any $n\geq N$ and $p\in\mathbb{N}$, we have \begin{eqnarray*} & & |f_{n+p}(x)-f_{n}(x)|\\ & \leq & |f_{n+p}(x)-f_{n+p}(r)|+|f_{n+p}(r)-f_{n}(r)|+|f_{n}(r)-f_{n}(x)|\\ & < & 3\varepsilon. \end{eqnarray*} This shows that $\{f_{n}(x)\}$ is a Cauchy sequence and hence it converges.

Next, we show that $\{f_{n}(x)\}$ converges uniformly in $x$. Let $\varepsilon>0$ be arbitrary. By equicontinuity, for each $x\in[0,1]$, there exists $\delta_{x}>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$ whenever $y\in[0,1]\cap(x-\delta_{x},x+\delta_{x})$ and $n\in\mathbb{N}$. Note that $\{(x-\delta_{x},x+\delta_{x})\mid x\in[0,1]\}$ is an open convering for the compact set $[0,1]$, so it has a finite subcover $\{(x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})\mid i=1,\ldots,K\}$. Choose $N$ such that $|f_{n+p}(x_{i})-f_{n}(x_{i})|<\varepsilon$ whenever $n\geq N$, $p\in\mathbb{N}$, and $i=1,2,\ldots,K$. Now, let $x\in[0,1]$, $n\geq N$, and $p\in\mathbb{N}$ be arbitrary. Choose $i$ such that $x\in (x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})$. We have \begin{eqnarray*} & & |f_{n+p}(x)-f_{n}(x)|\\ & \leq & |f_{n+p}(x)-f_{n+p}(x_{i})|+|f_{n+p}(x_{i})-f_{n}(x_{i})|+|f_{n}(x_{i})-f_{n}(x)|\\ & < & 3\varepsilon. \end{eqnarray*} This shows that $\{f_{n}(x)\}$ is uniformly Cauchy in $x$ and hence $\{f_n\}$ converges uniformly.

Answer 3

This is actually a simple case of applying the Arzela-Ascoli Propagation Theorem which states:

Point-wise convergence of an equicontinuous sequence of functions on a dense subset of the domain propagates to uniform convergence on the whole domain.

The rational numbers, $\mathbb{Q}$, are dense in the interval $[0,1]\subset \mathbb{R}$. So if <${f_n(r)}$> is converging to some function $f$ for every $r\in\mathbb{Q}$ $\cap$ $[0,1]$, the convergence will propagate to uniform convergence on all of $[0,1]$.

A proof of this theorem can be found on page 227 of Real Mathematical Analysis by Charles Pugh.