# prove that $f_n$ conerges uniformly in $X$"

Please, help me to find a solution to this question:

"If a sequence of continuous functions $f_n:X \rightarrow \mathbb R$ converges uniformly in a dense set $D \subset X$, prove that $f_n$ conerges uniformly in $X$".

Obs:

Once $f_n$ converges uniformly in $D$, then $f_n$ is a Cauchy sequence, i.e., given $\varepsilon >0$, there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(d)-f_n(d)|<\varepsilon$, for every $d \in D$ Besides that, $f_n$ is continuous and every point of $X$ is the limit of a sequence of points pertaining to $D$. Then, $d_n \rightarrow x \Rightarrow |f(d_n) \rightarrow f(x)|$. How to conclude that there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(x)-f_n(x)|<\varepsilon$?

Hint: For $x\in X$, pick $d\in D$, and pick $n,m > N_{\epsilon/3}$, so large that $$|f_m(x) - f_m(d)| < \frac{\epsilon}{3}$$ $$|f_n(x) - f_n(d)| < \frac{\epsilon}{3}$$ And also so that $$|f_m(d) - f_n(d)| < \frac{\epsilon}{3}$$ (Why can we choose $d\in D$ so that this happens?)

Now, with these choices of $n$, $m$, and $x$, can you conclude that $|f_n(x) - f_m(x)| < \epsilon$?

• Hi, Nicholas. $|f_m(d)-f_n(d)|<\frac{\varepsilon}{3}$ is true because $f_n$ converges uniformly on $D$. But why is true that exists $n_0 \in \mathbb N$ such that $m>n_0 \Rightarrow |f_m(x)-f_m(d)|<\frac{\varepsilon}{3}$? Apr 24, 2014 at 1:20
• Actually, we don't even need $m>n_0$ for this to be true. What do you know about $f_m$? Apr 24, 2014 at 1:48
• In other words, the functions in the sequence $\{f_n\}$ have a certain property that tells us that, given $x\in X = \overline{D}$, there exists $d\in D$ so that $|f_m(x) - f_m(d)|< \frac{\epsilon}{3}$. We will need to use this. Apr 24, 2014 at 1:53
• would you be speaking about continuity? Apr 24, 2014 at 11:42
• Exactly correct. Apr 24, 2014 at 13:03

Fix $m$ and $n$, both greater than $n_0$, and let $x \in X$. You can pick $d \in D$ such that $|f_n(d) - f_n(x)| < \varepsilon$ and $|f_m(d) - f_m(x)| < \varepsilon$. Then you get $|f_m(x) - f_n(x)| \leq |f_m(x) - f_m(d)| + |f_m(d) - f_n(d)| + |f_n(d) - f_n(x)| < 3\varepsilon$.

For $$x \in X$$, pick a sequence $$d_k \in D$$ satisfying $$d_k \to x$$. Then $$|f_n(x) - f(x)| = \left|f_n\left(\lim_{k \to \infty} d_k\right) - f\left(\lim_{k \to \infty} d_k\right)\right| \\ = \left|\lim_{k \to \infty} f_n(d_k) - \lim_{k \to \infty}f(d_k)\right| \\ = \lim_{k \to \infty} |f_n(d_k) - f(d_k)| \\ \le \sup_{d \in D} |f_n(d) - f(d)| \\ = \|(f_n|_D) - (f|_D) \|_{\infty} \underset{n \to \infty}{\longrightarrow} 0.$$ The uniform convergence of $$f_n|_D$$ to $$f|_D$$ implies the last statement, and since $$\|(f_n|_D)-(f|_D)\|_{\infty}$$ is a quantity that depends only on $$n$$ and not on $$x$$, this proves uniform convergence of $$f_n$$ to $$f$$.

In fact, you can even show a bit more. Since $$h(x) \overset{def}= |f_n(x) - f(x)|$$ is a continuous function on $$X$$ and $$D$$ is dense in $$X$$, we have $$\sup_{d \in D} h(d) = \sup_{x \in X} h(x)$$ or in other words, $$\|(f_n|_D)-(f|_D)\|_{\infty} = \|f_n-f\|_{\infty}$$. Since uniform convergence is the same as convergence under the supremum norm, the fact that the statement you are asking about is true is no surprise anymore, since now the proof simplifies to $$\|f_n-f\|_{\infty} = \|(f_n|_D)-(f|_D)\|_{\infty} \to 0.$$

Hope that helps,