Mixing
continuous-sup property and equicontinuous set.
$begingroup$
Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.
(a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.
(b) Show that (a) fails without equicontinuity.
(c) Show that this continuous-sup property does not imply equicontinuity.
(d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.
My attempt.
(a) We have that
$$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
then
$$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
and
$$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
that is,
$$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
$$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$
(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.
(c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
$$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.
(d) By Mindlack's observation, the same example of (c) works in (d).
Can someone help me?
real-analysis equicontinuity
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
$endgroup$
add a comment |
$begingroup$
Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.
(a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.
(b) Show that (a) fails without equicontinuity.
(c) Show that this continuous-sup property does not imply equicontinuity.
(d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.
My attempt.
(a) We have that
$$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
then
$$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
and
$$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
that is,
$$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
$$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$
(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.
(c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
$$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.
(d) By Mindlack's observation, the same example of (c) works in (d).
Can someone help me?
real-analysis equicontinuity
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
$endgroup$
add a comment |
$begingroup$
Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.
(a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.
(b) Show that (a) fails without equicontinuity.
(c) Show that this continuous-sup property does not imply equicontinuity.
(d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.
My attempt.
(a) We have that
$$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
then
$$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
and
$$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
that is,
$$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
$$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$
(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.
(c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
$$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.
(d) By Mindlack's observation, the same example of (c) works in (d).
Can someone help me?
real-analysis equicontinuity
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
$endgroup$
Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.
(a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.
(b) Show that (a) fails without equicontinuity.
(c) Show that this continuous-sup property does not imply equicontinuity.
(d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.
My attempt.
(a) We have that
$$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
then
$$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
and
$$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
that is,
$$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
$$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$
(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.
(c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
$$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.
(d) By Mindlack's observation, the same example of (c) works in (d).
Can someone help me?
real-analysis equicontinuity
real-analysis equicontinuity
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
edited Jan 10 at 17:28
Lucas Corrêa
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
asked Jan 10 at 17:11
Lucas CorrêaLucas Corrêa
1,6151321
1,6151321
add a comment |
add a comment |
1 Answer
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$begingroup$
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?
answered Jan 10 at 17:16
MindlackMindlack
3,61517
$endgroup$
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?
answered Jan 10 at 17:16
MindlackMindlack
3,61517
$endgroup$
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
add a comment |
$begingroup$
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?
answered Jan 10 at 17:16
MindlackMindlack
3,61517
$endgroup$
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
add a comment |
$begingroup$
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?
answered Jan 10 at 17:16
MindlackMindlack
3,61517
$endgroup$
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?
answered Jan 10 at 17:16
MindlackMindlack
3,61517
edited Jan 10 at 17:25
answered Jan 10 at 17:16
MindlackMindlack
3,61517
answered Jan 10 at 17:16
MindlackMindlack
3,61517
answered Jan 10 at 17:16
MindlackMindlack
3,61517
3,61517
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
add a comment |
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
Of course! Seems to me that (c) works in (d), thank you!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:21
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
$begingroup$
About (a), what would be the problem?
$endgroup$
– Lucas Corrêa
Jan 10 at 17:22
1
1
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
Nothing, it works.
$endgroup$
– Mindlack
Jan 10 at 17:24
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
$begingroup$
All right. I'll try your suggestion for (b). Thanks!
$endgroup$
– Lucas Corrêa
Jan 10 at 17:26
add a comment |
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