If we have a sequence of functions {fn} that converges uniformly on E⊂X, and each fn is continuous on E, is the limit function
f(x)=n→∞limfn(x)
also continuous?
Let's fix ϵ>0. Since we know that the sequence of functions {fn} is uniformly convergent on E, we can find an integer N so that n≥N implies
∣fn(x)−f(x)∣<3ϵ
for all x∈E.
Additionally, we know that each of these functions fn is continuous, so for all ϵ>0 we can find a δ>0 so that
∣fn(x)−fn(p)∣<3ϵ
if ∣x−p∣<δ for all p in E.
Then the following inequality ∣f(x)−f(p)∣ ≤ ∣f(x)−fn(x)∣+∣fn(x)−fn(p)∣+∣fn(p)−f(p)∣
shows that for n≥N, ∣f(x)−f(p)∣ < 3ϵ+3ϵ+3 ϵ= ϵ
whenever ∣x−p∣<δ.
Therefore, we conclude that the limit function f(x)=limn→∞fn(x) is continuous on E under the following conditions.
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Top NewestWhat 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.
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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.
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The follow up question is: Is f uniformly continuous?
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f will only be uniformly continuous if the each fn is uniformly continuous.
That's a good question! I believe the limit functionLog in to reply
Can you find a counterexample, namely a sequence of continuous fn that converge uniformly, but whose limit is not uniformly continuous?
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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 Cu(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)):x∈X}. So, yes, it is well known that if (fn)n is Cauchy, then it converges to limnfn (pointwise limit), and that this object itself is (uniformly!) continuous.
Here’s the proof. Let (fi)i⊆Cu(X,Y) be a d∞-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,j0 for all x∈X. Since Y is complete under its metrisation, f(x):=limifi(x)∈Y exists for all x∈X. We will now show that f is uniformly continuous, and furthermore that fi⟶f within Cu(X,Y).
Uniform continuity. Fix ε>0. Since the net is Cauchy, there exists a sufficiently large index i0, such that (⋆) d∞(fi,fj)<4ε for all i,j≥i0. Since fi0∈Cu(X,Y) there exists δ>0 such that (⋆⋆) ∀x,x′∈X: d(x,x′)<δ⇒d(fi0(x),fi0(x′))<4ε. Let x,x′∈X with d(x,x′)<δ. By (⋆) and (⋆⋆) one has d(fi(x),fi(x′))<43ε for all i≥i0. Taking limits one has d(f(x),f(x′))≤43ε<ε. Since ε>0 was arbitrarily chosen, this shows that f is uniformly continuous.
Convergence. Let ε>0. Since the net is Cauchy, there exists a sufficiently large index i0, such that d∞(fi,fj)<2ε for all i,j≥i0. For any x∈X and i≥i0 it thus follows that d(f(x),fi(x))=limjd(fj(x),fi(x))≤jlimsupd∞(fj,fi)≤2ε. Hence for all i≥i0, taking supremums yields d∞(f,fi)≤2ε<ε. Since ε>0 was arbitrarily chosen, this shows that fi⟶f wrt the topology of uniform convergence.
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