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.
Best Answer
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