Let $Y$ be a complete metric space. Then $C^0 (X,Y)$ is complete under the uniform convergence metric.
$begingroup$
I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.
Just to be clear, the metric is:
$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$
And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.
So far what I got is:
$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?
Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.
Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:
$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$
$blacksquare$
We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.
And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.
real-analysis metric-spaces uniform-convergence
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
$endgroup$
This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.
The current answers do not contain enough detail.
add a comment |
$begingroup$
I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.
Just to be clear, the metric is:
$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$
And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.
So far what I got is:
$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?
Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.
Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:
$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$
$blacksquare$
We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.
And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.
real-analysis metric-spaces uniform-convergence
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
$endgroup$
This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.
The current answers do not contain enough detail.
$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday
add a comment |
$begingroup$
I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.
Just to be clear, the metric is:
$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$
And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.
So far what I got is:
$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?
Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.
Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:
$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$
$blacksquare$
We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.
And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.
real-analysis metric-spaces uniform-convergence
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
$endgroup$
I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.
Just to be clear, the metric is:
$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$
And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.
So far what I got is:
$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?
Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.
Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:
$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$
$blacksquare$
We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.
And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.
real-analysis metric-spaces uniform-convergence
real-analysis metric-spaces uniform-convergence
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
edited Dec 5 '18 at 4:44
Pedro Cavalcante Oliveira
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
asked Dec 4 '18 at 19:39
Pedro Cavalcante OliveiraPedro Cavalcante Oliveira
876
876
This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.
The current answers do not contain enough detail.
This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.
The current answers do not contain enough detail.
$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday
add a comment |
$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday
$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday
$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.
You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
$endgroup$
add a comment |
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$begingroup$
You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.
You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.
You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.
You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
$endgroup$
You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.
You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
answered Dec 4 '18 at 19:46
qbertqbert
22.1k32460
22.1k32460
add a comment |
add a comment |
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$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday