Mixing
Is equicontinuity needed in theorem 7.25 part (b) from Rudin's “Principles”
$begingroup$
About Rudin's theorem 7.25:
7.25 Theorem If $K$ is compact, if $f_n in mathscr{C}(K)$ for $n=1,2,3,dots,$ and if ${f_n}$ is pointwise bounded and equicontinuous on $K$, then
${f_n}$ is uniformly bounded on $K$,
${f_n}$ contains a uniformly convergent subsequence.
The equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
The proof is the following:
Proof
(1) Let $varepsilon > 0$ be given and choose $delta > 0$, in accordance with Definition 7.22, so that
begin{equation}
|f_n(x) - f_n(y)| < varepsilon tag{44}
end{equation}
for all $n$, provided that $d(x,y) < delta$.
Since $K$ is compact, there are finitely many points $p_1, dots, p_r$ in $K$ such that to every $x in K$ corresponds at least one $p_i$ with $d(x, p_i) < delta$. Since ${f_n}$ is pointwise bounded, there exist $M_i < infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = max(M_1, dots, M_r)$, then $|f_n(x)| < M + varepsilon$ for every $x in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that ${f_n}$ has a subsequence ${f_{n_i}}$ such that ${f_{n_i}(x)}$ converges for every $x in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that ${g_i}$ converrges uniformly on $K$.
Let $varepsilon > 0$, and pick $delta > 0$ as in the beginning of this proof. Let $V(x,delta)$ be the set of all $y in K$ with $d(x,y) < delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, dots, x_m$ in $E$ such that
$$
K subset V(x_1, delta) cup dots cup V(x_m, delta). tag{45}
label{45}
$$
Since ${g_i(x)}$ converges for every $x in E$, there is an integer $N$ such that
$$
|g_i(x_s) - g_j(x_s)| < varepsilon tag{46}
label{46}
$$
whenever $i geq N$, $j geq N$, $1 leq s leq m$.
If $x in K$, $(ref{45})$ shows that $x in V(x_s, delta)$ for some $s$, so that
$$
|g_i(x) - g_i(x_s) | < varepsilon
$$
for every $i$. If $i geq N$ and $j geq N$, it follows from $(ref{46})$ that
begin{align}
|g_i(x) - g_j(x)| &leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \
&< 3varepsilon.
end{align}
This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
analysis uniform-convergence equicontinuity
edited Feb 1 at 13:50
David Kraemer
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
$endgroup$
add a comment |
$begingroup$
About Rudin's theorem 7.25:
7.25 Theorem If $K$ is compact, if $f_n in mathscr{C}(K)$ for $n=1,2,3,dots,$ and if ${f_n}$ is pointwise bounded and equicontinuous on $K$, then
${f_n}$ is uniformly bounded on $K$,
${f_n}$ contains a uniformly convergent subsequence.
The equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
The proof is the following:
Proof
(1) Let $varepsilon > 0$ be given and choose $delta > 0$, in accordance with Definition 7.22, so that
begin{equation}
|f_n(x) - f_n(y)| < varepsilon tag{44}
end{equation}
for all $n$, provided that $d(x,y) < delta$.
Since $K$ is compact, there are finitely many points $p_1, dots, p_r$ in $K$ such that to every $x in K$ corresponds at least one $p_i$ with $d(x, p_i) < delta$. Since ${f_n}$ is pointwise bounded, there exist $M_i < infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = max(M_1, dots, M_r)$, then $|f_n(x)| < M + varepsilon$ for every $x in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that ${f_n}$ has a subsequence ${f_{n_i}}$ such that ${f_{n_i}(x)}$ converges for every $x in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that ${g_i}$ converrges uniformly on $K$.
Let $varepsilon > 0$, and pick $delta > 0$ as in the beginning of this proof. Let $V(x,delta)$ be the set of all $y in K$ with $d(x,y) < delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, dots, x_m$ in $E$ such that
$$
K subset V(x_1, delta) cup dots cup V(x_m, delta). tag{45}
label{45}
$$
Since ${g_i(x)}$ converges for every $x in E$, there is an integer $N$ such that
$$
|g_i(x_s) - g_j(x_s)| < varepsilon tag{46}
label{46}
$$
whenever $i geq N$, $j geq N$, $1 leq s leq m$.
If $x in K$, $(ref{45})$ shows that $x in V(x_s, delta)$ for some $s$, so that
$$
|g_i(x) - g_i(x_s) | < varepsilon
$$
for every $i$. If $i geq N$ and $j geq N$, it follows from $(ref{46})$ that
begin{align}
|g_i(x) - g_j(x)| &leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \
&< 3varepsilon.
end{align}
This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
analysis uniform-convergence equicontinuity
edited Feb 1 at 13:50
David Kraemer
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
$endgroup$
add a comment |
$begingroup$
About Rudin's theorem 7.25:
7.25 Theorem If $K$ is compact, if $f_n in mathscr{C}(K)$ for $n=1,2,3,dots,$ and if ${f_n}$ is pointwise bounded and equicontinuous on $K$, then
${f_n}$ is uniformly bounded on $K$,
${f_n}$ contains a uniformly convergent subsequence.
The equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
The proof is the following:
Proof
(1) Let $varepsilon > 0$ be given and choose $delta > 0$, in accordance with Definition 7.22, so that
begin{equation}
|f_n(x) - f_n(y)| < varepsilon tag{44}
end{equation}
for all $n$, provided that $d(x,y) < delta$.
Since $K$ is compact, there are finitely many points $p_1, dots, p_r$ in $K$ such that to every $x in K$ corresponds at least one $p_i$ with $d(x, p_i) < delta$. Since ${f_n}$ is pointwise bounded, there exist $M_i < infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = max(M_1, dots, M_r)$, then $|f_n(x)| < M + varepsilon$ for every $x in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that ${f_n}$ has a subsequence ${f_{n_i}}$ such that ${f_{n_i}(x)}$ converges for every $x in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that ${g_i}$ converrges uniformly on $K$.
Let $varepsilon > 0$, and pick $delta > 0$ as in the beginning of this proof. Let $V(x,delta)$ be the set of all $y in K$ with $d(x,y) < delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, dots, x_m$ in $E$ such that
$$
K subset V(x_1, delta) cup dots cup V(x_m, delta). tag{45}
label{45}
$$
Since ${g_i(x)}$ converges for every $x in E$, there is an integer $N$ such that
$$
|g_i(x_s) - g_j(x_s)| < varepsilon tag{46}
label{46}
$$
whenever $i geq N$, $j geq N$, $1 leq s leq m$.
If $x in K$, $(ref{45})$ shows that $x in V(x_s, delta)$ for some $s$, so that
$$
|g_i(x) - g_i(x_s) | < varepsilon
$$
for every $i$. If $i geq N$ and $j geq N$, it follows from $(ref{46})$ that
begin{align}
|g_i(x) - g_j(x)| &leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \
&< 3varepsilon.
end{align}
This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
analysis uniform-convergence equicontinuity
edited Feb 1 at 13:50
David Kraemer
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
$endgroup$
About Rudin's theorem 7.25:
7.25 Theorem If $K$ is compact, if $f_n in mathscr{C}(K)$ for $n=1,2,3,dots,$ and if ${f_n}$ is pointwise bounded and equicontinuous on $K$, then
${f_n}$ is uniformly bounded on $K$,
${f_n}$ contains a uniformly convergent subsequence.
The equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
The proof is the following:
Proof
(1) Let $varepsilon > 0$ be given and choose $delta > 0$, in accordance with Definition 7.22, so that
begin{equation}
|f_n(x) - f_n(y)| < varepsilon tag{44}
end{equation}
for all $n$, provided that $d(x,y) < delta$.
Since $K$ is compact, there are finitely many points $p_1, dots, p_r$ in $K$ such that to every $x in K$ corresponds at least one $p_i$ with $d(x, p_i) < delta$. Since ${f_n}$ is pointwise bounded, there exist $M_i < infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = max(M_1, dots, M_r)$, then $|f_n(x)| < M + varepsilon$ for every $x in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that ${f_n}$ has a subsequence ${f_{n_i}}$ such that ${f_{n_i}(x)}$ converges for every $x in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that ${g_i}$ converrges uniformly on $K$.
Let $varepsilon > 0$, and pick $delta > 0$ as in the beginning of this proof. Let $V(x,delta)$ be the set of all $y in K$ with $d(x,y) < delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, dots, x_m$ in $E$ such that
$$
K subset V(x_1, delta) cup dots cup V(x_m, delta). tag{45}
label{45}
$$
Since ${g_i(x)}$ converges for every $x in E$, there is an integer $N$ such that
$$
|g_i(x_s) - g_j(x_s)| < varepsilon tag{46}
label{46}
$$
whenever $i geq N$, $j geq N$, $1 leq s leq m$.
If $x in K$, $(ref{45})$ shows that $x in V(x_s, delta)$ for some $s$, so that
$$
|g_i(x) - g_i(x_s) | < varepsilon
$$
for every $i$. If $i geq N$ and $j geq N$, it follows from $(ref{46})$ that
begin{align}
|g_i(x) - g_j(x)| &leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \
&< 3varepsilon.
end{align}
This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
analysis uniform-convergence equicontinuity
analysis uniform-convergence equicontinuity
edited Feb 1 at 13:50
David Kraemer
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
edited Feb 1 at 13:50
David Kraemer
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
edited Feb 1 at 13:50
David Kraemer
39017
edited Feb 1 at 13:50
David Kraemer
39017
edited Feb 1 at 13:50
David Kraemer
39017
39017
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
asked Feb 1 at 13:07
Antonio Horta RibeiroAntonio Horta Ribeiro
15611
15611
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= begin{cases} 1-nx, & x in [0,1/n), \0, & text{else.} end{cases}$$
The way $delta$ is chosen in the proof above makes use of equicontinuity.
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= begin{cases} 1-nx, & x in [0,1/n), \0, & text{else.} end{cases}$$
The way $delta$ is chosen in the proof above makes use of equicontinuity.
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
add a comment |
$begingroup$
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= begin{cases} 1-nx, & x in [0,1/n), \0, & text{else.} end{cases}$$
The way $delta$ is chosen in the proof above makes use of equicontinuity.
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
add a comment |
$begingroup$
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= begin{cases} 1-nx, & x in [0,1/n), \0, & text{else.} end{cases}$$
The way $delta$ is chosen in the proof above makes use of equicontinuity.
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
$endgroup$
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= begin{cases} 1-nx, & x in [0,1/n), \0, & text{else.} end{cases}$$
The way $delta$ is chosen in the proof above makes use of equicontinuity.
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 13:16
Mars PlasticMars Plastic
1,455122
1,455122
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
add a comment |
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
$begingroup$
Ohh, ok, you need the equicontinuity for: $|g_i(x)- g_i(x_s)| < epsilon$ to hold for every $i$ and every $x$, with the same $delta$. I missed that....Thanks!
$endgroup$
– Antonio Horta Ribeiro
Feb 1 at 13:46
add a comment |
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