Uniform convergence on closure of a set

Question

Suppose $(f_n)$ is a sequence of continuous functions on a subset X of ℝ which converges uniformly on Y ⊆ X. Let $\overline{Y}$ denote the closure of Y. Prove that there exists a function f such that $(f_n)$ converges uniformly to f on $\overline{Y}$ ∩ X.

Answer 1

I do not think that the proposition is true as you did not require continuity of the $f_n$.

Take $X:=[0,1]$, $Y:=(0,1]$, then for $n\in \mathbb{N}$ set $f_n(0): =n$ and $f_n(x)=1$ for $x\in(0,1]$. Then $f_n$ converges uniformly on $Y$ to the constant function $1$, but not on $[0,1] =\bar{Y}\cap X$.

Answer 2

Let us suppose $f_n$ all map to $\mathbb{R}$ (or another complete metric space) As $(f_n)$ uniformly converges on $Y$, then there is a limit function $f:Y\to \mathbb{R}$ with $f(y)=\lim_{n\to \infty}f_n(y)$ for every $y\in Y$.

Due to the uniform convergence of the continuous $(f_n)$, then $f$ is continuous on $Y$ as well.

Now we aim to extend $f$ to the boundary of $Y$: Let $x \in X$ and $(y_n)$ be a sequence with members from $Y$ such that $x=\lim_{n\to \infty}y_n$. let $\epsilon >0$. Due to the uniform convergence there is an $N>0$ such that for $n\geq N,m\geq N$ it holds $$|f_m(y)-f_n(y)| < \epsilon/3 \qquad y\in Y$$ If we let go $m\to \infty$ and set $n:=N$ then the inequality above becomes $$|f(y)-f_N(y)| \leq \epsilon/3 \qquad y\in Y$$ As $f_N$ is continuous on the compact set $K:=\{x, y_1, y_2, y_3, ...\}$ it is uniformly continuous on $K$. Thus there is a $\delta>0$ such that for $a,b\in K$ with $|a-b| <\delta$ implies $|f_N(a)-f_N(b)| < \epsilon/3$. As $(y_n)$ converges to $x$ it is a Cauchy sequence as well, so there is some $M>0$ such that for $n\geq M,m\geq M$ it holds $|y_n-y_m|<\delta$

With this preparation we can show that $(f(y_n))$ is a Cauchy sequence. Then if $n\geq N,m\geq M$ it holds $$|f(y_n)-f(y_m)| \leq |f(y_n) - f_N(y_k)| + |f_N(y_k)-f_N(y_m)|+|f_N(y_m)-f(y_m)| \\ \le \epsilon/3 +\epsilon/3+\epsilon/3$$ So $(f(y_n))$ is convergent, and let us denote $f(x):=\lim f(y_n)$. From the inequality above it is easy to deduct that $f(x)$ does not depend on the special choice of $(y_n)$. Furthermore we have proven (by construction) that we extended $f$ to a continuous function on $\overline{Y}$.

Now the uniform convergence on $\overline{Y}$ follows from the uniform convergence on $Y$ and the continuous extension of $f$ to $\overline{Y}$: Let $\epsilon > 0$ then there is a $L>0$ such that for all $n>L$ and $y\in Y$ it holds: $|f_n(y)-f(y)| \leq\epsilon$. Again let $(y_n)$ be a sequence with members from $Y$ such that $x=\lim_{n\to \infty}y_n$. Thus we get $$|f(x) -f_n(x)|= |\lim f(y_m) -f_n(\lim y_m)| =\lim |f(y_m) - f_n(y_m)| \leq\epsilon$$ This renders uniform convergence on $\overline{Y}$.