Suppose we have continuous $f_{n} : D rightarrow mathbb{R}$ converging to $f : D rightarrow mathbb{R}$...
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$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
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Suppose we have continuous $f_{n} : D rightarrow mathbb{R}$
converging to $f : D rightarrow mathbb{R}$ uniformly on $D$. Then,
is $f$ continuous?
I know that if we have $f_{n}$ converging to $f$ and $f_{n}$ are all continuous then it is not necessary for $f$ to be continuous. This is shown by
$$x^{n} hspace{1cm} 0 < x leq 1 $$
for $n geq 0$. But, it is the uniformly part that gets me here. Is this statement true?
real-analysis uniform-convergence
asked Dec 14 '18 at 1:02
josephjoseph
496111
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Dec 14 '18 at 3:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
2 answers
Suppose we have continuous $f_{n} : D rightarrow mathbb{R}$
converging to $f : D rightarrow mathbb{R}$ uniformly on $D$. Then,
is $f$ continuous?
I know that if we have $f_{n}$ converging to $f$ and $f_{n}$ are all continuous then it is not necessary for $f$ to be continuous. This is shown by
$$x^{n} hspace{1cm} 0 < x leq 1 $$
for $n geq 0$. But, it is the uniformly part that gets me here. Is this statement true?
real-analysis uniform-convergence
asked Dec 14 '18 at 1:02
josephjoseph
496111
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marked as duplicate by RRL real-analysis
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Dec 14 '18 at 3:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
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Is $D$ a metric space?
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– Guido A.
Dec 14 '18 at 1:08
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I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
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– joseph
Dec 14 '18 at 1:09
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Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
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– Guido A.
Dec 14 '18 at 1:10
add a comment |
$begingroup$
This question already has an answer here:
$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
2 answers
Suppose we have continuous $f_{n} : D rightarrow mathbb{R}$
converging to $f : D rightarrow mathbb{R}$ uniformly on $D$. Then,
is $f$ continuous?
I know that if we have $f_{n}$ converging to $f$ and $f_{n}$ are all continuous then it is not necessary for $f$ to be continuous. This is shown by
$$x^{n} hspace{1cm} 0 < x leq 1 $$
for $n geq 0$. But, it is the uniformly part that gets me here. Is this statement true?
real-analysis uniform-convergence
asked Dec 14 '18 at 1:02
josephjoseph
496111
$endgroup$
This question already has an answer here:
$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
2 answers
Suppose we have continuous $f_{n} : D rightarrow mathbb{R}$
converging to $f : D rightarrow mathbb{R}$ uniformly on $D$. Then,
is $f$ continuous?
I know that if we have $f_{n}$ converging to $f$ and $f_{n}$ are all continuous then it is not necessary for $f$ to be continuous. This is shown by
$$x^{n} hspace{1cm} 0 < x leq 1 $$
for $n geq 0$. But, it is the uniformly part that gets me here. Is this statement true?
This question already has an answer here:
$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
2 answers
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Dec 14 '18 at 1:02
josephjoseph
496111
asked Dec 14 '18 at 1:02
josephjoseph
496111
asked Dec 14 '18 at 1:02
josephjoseph
496111
asked Dec 14 '18 at 1:02
josephjoseph
496111
asked Dec 14 '18 at 1:02
josephjoseph
496111
496111
marked as duplicate by RRL real-analysis
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Dec 14 '18 at 3:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by RRL real-analysis
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Dec 14 '18 at 3:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
$begingroup$
Is $D$ a metric space?
$endgroup$
– Guido A.
Dec 14 '18 at 1:08
$begingroup$
I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
$endgroup$
– joseph
Dec 14 '18 at 1:09
$begingroup$
Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
$endgroup$
– Guido A.
Dec 14 '18 at 1:10
add a comment |
$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
$begingroup$
Is $D$ a metric space?
$endgroup$
– Guido A.
Dec 14 '18 at 1:08
$begingroup$
I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
$endgroup$
– joseph
Dec 14 '18 at 1:09
$begingroup$
Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
$endgroup$
– Guido A.
Dec 14 '18 at 1:10
$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
$begingroup$
Is $D$ a metric space?
$endgroup$
– Guido A.
Dec 14 '18 at 1:08
$begingroup$
Is $D$ a metric space?
$endgroup$
– Guido A.
Dec 14 '18 at 1:08
$begingroup$
I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
$endgroup$
– joseph
Dec 14 '18 at 1:09
$begingroup$
I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
$endgroup$
– joseph
Dec 14 '18 at 1:09
$begingroup$
Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
$endgroup$
– Guido A.
Dec 14 '18 at 1:10
$begingroup$
Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
$endgroup$
– Guido A.
Dec 14 '18 at 1:10
add a comment |
1 Answer
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Fix $d in D$. Now, it is always true that
$$
begin{align}
|f(d) - f(y)| &leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \ &leq 2d(f_n,f) + |f_n(d)-f_n(y)|. tag{1}
end{align}
$$
You also know that $f_n to f$, that is $d(f_n,f) to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.
Let $varepsilon > 0$. Now, there exists $m gg 1$ such that $2d(f_n,f) < varepsilon/2$ when $n geq m$. Since $f_m$ is continuous, there exists $delta > 0$ such that $|d-y| < delta $ implies $|f_m(d)-f_m(y)| < varepsilon /2$. Thus, if $|d-y| < delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < varepsilon$, and so $f$ is continuous.
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fix $d in D$. Now, it is always true that
$$
begin{align}
|f(d) - f(y)| &leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \ &leq 2d(f_n,f) + |f_n(d)-f_n(y)|. tag{1}
end{align}
$$
You also know that $f_n to f$, that is $d(f_n,f) to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.
Let $varepsilon > 0$. Now, there exists $m gg 1$ such that $2d(f_n,f) < varepsilon/2$ when $n geq m$. Since $f_m$ is continuous, there exists $delta > 0$ such that $|d-y| < delta $ implies $|f_m(d)-f_m(y)| < varepsilon /2$. Thus, if $|d-y| < delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < varepsilon$, and so $f$ is continuous.
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
$endgroup$
add a comment |
$begingroup$
Fix $d in D$. Now, it is always true that
$$
begin{align}
|f(d) - f(y)| &leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \ &leq 2d(f_n,f) + |f_n(d)-f_n(y)|. tag{1}
end{align}
$$
You also know that $f_n to f$, that is $d(f_n,f) to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.
Let $varepsilon > 0$. Now, there exists $m gg 1$ such that $2d(f_n,f) < varepsilon/2$ when $n geq m$. Since $f_m$ is continuous, there exists $delta > 0$ such that $|d-y| < delta $ implies $|f_m(d)-f_m(y)| < varepsilon /2$. Thus, if $|d-y| < delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < varepsilon$, and so $f$ is continuous.
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
$endgroup$
add a comment |
$begingroup$
Fix $d in D$. Now, it is always true that
$$
begin{align}
|f(d) - f(y)| &leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \ &leq 2d(f_n,f) + |f_n(d)-f_n(y)|. tag{1}
end{align}
$$
You also know that $f_n to f$, that is $d(f_n,f) to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.
Let $varepsilon > 0$. Now, there exists $m gg 1$ such that $2d(f_n,f) < varepsilon/2$ when $n geq m$. Since $f_m$ is continuous, there exists $delta > 0$ such that $|d-y| < delta $ implies $|f_m(d)-f_m(y)| < varepsilon /2$. Thus, if $|d-y| < delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < varepsilon$, and so $f$ is continuous.
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
$endgroup$
Fix $d in D$. Now, it is always true that
$$
begin{align}
|f(d) - f(y)| &leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \ &leq 2d(f_n,f) + |f_n(d)-f_n(y)|. tag{1}
end{align}
$$
You also know that $f_n to f$, that is $d(f_n,f) to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.
Let $varepsilon > 0$. Now, there exists $m gg 1$ such that $2d(f_n,f) < varepsilon/2$ when $n geq m$. Since $f_m$ is continuous, there exists $delta > 0$ such that $|d-y| < delta $ implies $|f_m(d)-f_m(y)| < varepsilon /2$. Thus, if $|d-y| < delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < varepsilon$, and so $f$ is continuous.
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
answered Dec 14 '18 at 1:14
Guido A.Guido A.
7,7231730
7,7231730
add a comment |
add a comment |
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$begingroup$
Yes, it's a classical theorem in early Analysis/Topology courses.
$endgroup$
– Saucy O'Path
Dec 14 '18 at 1:06
$begingroup$
Is $D$ a metric space?
$endgroup$
– Guido A.
Dec 14 '18 at 1:08
$begingroup$
I don't know what a metric space is. $D$ is a set. $D subseteq mathbb{R}$
$endgroup$
– joseph
Dec 14 '18 at 1:09
$begingroup$
Okay, great. Your question may seem a lot more general, so you may want to include that in the post.
$endgroup$
– Guido A.
Dec 14 '18 at 1:10