[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.

Note by Tasha Kim
4 months, 1 week ago

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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 - 4 months, 1 week ago

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

Tasha Kim - 4 months, 1 week ago

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The follow up question is: Is \(f\) uniformly continuous?

Calvin Lin Staff - 4 months, 1 week ago

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@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 - 4 months, 1 week ago

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@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?

Calvin Lin Staff - 4 months ago

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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 - 3 months, 3 weeks ago

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