[Real Analysis] Does Uniform Convergence for a Sequence of Functions preserve Continuity?

If we have a sequence of functions $\{f_n\}$ that converges uniformly on $E\subset X$ , and each $f_n$ is continuous on E, is the limit function

$f(x)=\lim_{n\to\infty}f_n(x)$ also continuous?

Let's fix $\epsilon>0$ . Since we know that the sequence of functions $\{f_n\}$ is uniformly convergent on E, we can find an integer N so that $n\geq N$ implies

$\mid f_n(x) - f(x) \mid < \frac{\epsilon}{3}$

for all $x\in E$ .

Additionally, we know that each of these functions $f_n$ is continuous, so for all $\epsilon>0$ we can find a $\delta>0$ so that

$\mid f_n(x) - f_n(p)\mid<\frac{\epsilon}{3}$

if $\mid x -p\mid < \delta$ for all $p$ in $E$ .

Then the following inequality $\mid f(x)-f(p)\mid ~\leq ~\mid f(x) - f_n(x)\mid + \mid f_n(x)-f_n(p)\mid + \mid f_n(p) - f(p)\mid$

shows that for $n \geq N$ , $\mid f(x)-f(p)\mid ~<~ \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3 ~}=~ \epsilon$ whenever $\mid x - p \mid < \delta$ .

Therefore, we conclude that the limit function $f(x)=\lim_{n\to\infty}f_n(x)$ is continuous on $E$ under the following conditions.

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What have you tried to do? For a fixed $x$ and $y$ , how can we get from $f(x)$ to $f(y)$ with the least amount of error.

Calvin Lin Staff - 1 year, 5 months ago

Thanks for your comment. Following your suggestion, I have shown that $f$ is continuous for a fixed $x$ and $p$ of $E$ .

Tasha Kim - 1 year, 4 months ago

The follow up question is: Is $f$ uniformly continuous?

Calvin Lin Staff - 1 year, 4 months ago

@Calvin Lin – That's a good question! I believe the limit function $f$ will only be uniformly continuous if the each $f_n$ is uniformly continuous.

@Tasha Kim – Yes indeed. A similar approach works.

Can you find a counterexample, namely a sequence of continuous $f_n$ that converge uniformly, but whose limit is not uniformly continuous?

The framework of what you proved is very unclear. Here’s what it is: Let $X,Y$ be bounded metric spaces, and assume that $Y$ is complete under its metrisation. Let $C_{u}(X,Y)$ be the space of uniformly continuous functions. It can be shown that this space is complete under the metric $d_{\infty}(f,g)=\sup\{d(f(x),g(x)):x\in X\}$ . So, yes, it is well known that if $(f_{n})_{n}$ is Cauchy, then it converges to $\lim_{n}f_{n}$ (pointwise limit), and that this object itself is (uniformly!) continuous.

Here’s the proof. Let $(f_{i})_{i}\subseteq C_{u}(X,Y)$ be a $d_{\infty}$ -Cauchy net. Then it is clear that the net is also pointwise convergent, since by definition $d(f_{i}(x),f_{j}(x))\leq d_{\infty}(f_{i},f_{j})\longrightarrow_{i,j}0$ for all $x\in X$ . Since $Y$ is complete under its metrisation, $f(x):=\lim_{i}f_{i}(x)\in Y$ exists for all $x\in X$ . We will now show that $f$ is uniformly continuous, and furthermore that $f_{i}\longrightarrow f$ within $C_{u}(X,Y)$ .

Uniform continuity. Fix $\varepsilon>0$ . Since the net is Cauchy, there exists a sufficiently large index $i_{0}$ , such that ( $\star$ ) $d_{\infty}(f_{i},f_{j})<\frac{\varepsilon}{4}$ for all $i,j\geq i_{0}$ . Since $f_{i_{0}}\in C_{u}(X,Y)$ there exists $\delta>0$ such that ( $\star\star$ ) $\forall{x,x'\in X:~}d(x,x')<\delta\Rightarrow d(f_{i_{0}}(x),f_{i_{0}}(x'))<\frac{\varepsilon}{4}$ . Let $x,x'\in X$ with $d(x,x')<\delta$ . By ( $\star$ ) and ( $\star\star$ ) one has $d(f_{i}(x),f_{i}(x'))<\frac{3\varepsilon}{4}$ for all $i\geq i_{0}$ . Taking limits one has $d(f(x),f(x'))\leq\frac{3\varepsilon}{4}<\varepsilon$ . Since $\varepsilon>0$ was arbitrarily chosen, this shows that $f$ is uniformly continuous.

Convergence. Let $\varepsilon>0$ . Since the net is Cauchy, there exists a sufficiently large index $i_{0}$ , such that $d_{\infty}(f_{i},f_{j})<\frac{\varepsilon}{2}$ for all $i,j\geq i_{0}$ . For any $x\in X$ and $i\geq i_{0}$ it thus follows that $d(f(x),f_{i}(x))=\lim_{j}d(f_{j}(x),f_{i}(x))\leq\limsup_{j}d_{\infty}(f_{j},f_{i})\leq\frac{\varepsilon}{2}$ . Hence for all $i\geq i_{0}$ , taking supremums yields $d_{\infty}(f,f_{i})\leq\frac{\varepsilon}{2}<\varepsilon$ . Since $\varepsilon>0$ was arbitrarily chosen, this shows that $f_{i}\longrightarrow f$ wrt the topology of uniform convergence.

R Mathe - 1 year, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$