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

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

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

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

fn(x)f(x)<ϵ3\mid f_n(x) - f(x) \mid < \frac{\epsilon}{3}

for all xEx\in E.

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

fn(x)fn(p)<ϵ3\mid f_n(x) - f_n(p)\mid<\frac{\epsilon}{3}

if xp<δ\mid x -p\mid < \delta for all pp in EE.

Then the following inequality f(x)f(p)  f(x)fn(x)+fn(x)fn(p)+fn(p)f(p)\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 nNn \geq N, f(x)f(p) < ϵ3+ϵ3+ϵ3 = ϵ\mid f(x)-f(p)\mid ~<~ \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3 ~}=~ \epsilon whenever xp<δ\mid x - p \mid < \delta.

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

Note by Tasha Kim
1 year, 5 months ago

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

Calvin Lin Staff - 1 year, 5 months ago

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Thanks for your comment. Following your suggestion, I have shown that ff is continuous for a fixed xx and pp of EE.

Tasha Kim - 1 year, 4 months ago

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The follow up question is: Is ff uniformly continuous?

Calvin Lin Staff - 1 year, 4 months ago

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

Tasha Kim - 1 year, 4 months ago

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@Tasha Kim Yes indeed. A similar approach works.

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

Calvin Lin Staff - 1 year, 4 months ago

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

Here’s the proof. Let (fi)iCu(X,Y)(f_{i})_{i}\subseteq C_{u}(X,Y) be a dd_{\infty}-Cauchy net. Then it is clear that the net is also pointwise convergent, since by definition d(fi(x),fj(x))d(fi,fj)i,j0d(f_{i}(x),f_{j}(x))\leq d_{\infty}(f_{i},f_{j})\longrightarrow_{i,j}0 for all xXx\in X. Since YY is complete under its metrisation, f(x):=limifi(x)Yf(x):=\lim_{i}f_{i}(x)\in Y exists for all xXx\in X. We will now show that ff is uniformly continuous, and furthermore that fiff_{i}\longrightarrow f within Cu(X,Y)C_{u}(X,Y).

Uniform continuity. Fix ε>0\varepsilon>0. Since the net is Cauchy, there exists a sufficiently large index i0i_{0}, such that (\star) d(fi,fj)<ε4d_{\infty}(f_{i},f_{j})<\frac{\varepsilon}{4} for all i,ji0i,j\geq i_{0}. Since fi0Cu(X,Y)f_{i_{0}}\in C_{u}(X,Y) there exists δ>0\delta>0 such that (\star\star) x,xX: d(x,x)<δd(fi0(x),fi0(x))<ε4\forall{x,x'\in X:~}d(x,x')<\delta\Rightarrow d(f_{i_{0}}(x),f_{i_{0}}(x'))<\frac{\varepsilon}{4}. Let x,xXx,x'\in X with d(x,x)<δd(x,x')<\delta. By (\star) and (\star\star) one has d(fi(x),fi(x))<3ε4d(f_{i}(x),f_{i}(x'))<\frac{3\varepsilon}{4} for all ii0i\geq i_{0}. Taking limits one has d(f(x),f(x))3ε4<εd(f(x),f(x'))\leq\frac{3\varepsilon}{4}<\varepsilon. Since ε>0\varepsilon>0 was arbitrarily chosen, this shows that ff is uniformly continuous.

Convergence. Let ε>0\varepsilon>0. Since the net is Cauchy, there exists a sufficiently large index i0i_{0}, such that d(fi,fj)<ε2d_{\infty}(f_{i},f_{j})<\frac{\varepsilon}{2} for all i,ji0i,j\geq i_{0}. For any xXx\in X and ii0i\geq i_{0} it thus follows that d(f(x),fi(x))=limjd(fj(x),fi(x))lim supjd(fj,fi)ε2d(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 ii0i\geq i_{0}, taking supremums yields d(f,fi)ε2<εd_{\infty}(f,f_{i})\leq\frac{\varepsilon}{2}<\varepsilon. Since ε>0\varepsilon>0 was arbitrarily chosen, this shows that fiff_{i}\longrightarrow f wrt the topology of uniform convergence.

R Mathe - 1 year, 4 months ago

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