Does there exist such a monotone function?
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Can we find a monotone function $f$ in $R^1$ satisfying that there is no continuous function $g$ equal to $f$ almost everywhere in every open interval $I$?
We know that a monotone function is differentiable almost everywhere,so I think the property of $f$ can not be too bad.but I don't know if $f$ can be approached by a continuous function for every open interval.I tried to construct some functions but didn't work...
Thank you in advance!
real-analysis
asked Jan 16 at 16:50
MaxwellMaxwell
264
$endgroup$
|
show 2 more comments
$begingroup$
Can we find a monotone function $f$ in $R^1$ satisfying that there is no continuous function $g$ equal to $f$ almost everywhere in every open interval $I$?
We know that a monotone function is differentiable almost everywhere,so I think the property of $f$ can not be too bad.but I don't know if $f$ can be approached by a continuous function for every open interval.I tried to construct some functions but didn't work...
Thank you in advance!
real-analysis
asked Jan 16 at 16:50
MaxwellMaxwell
264
$endgroup$
$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
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@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
1
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
1
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09
|
show 2 more comments
$begingroup$
Can we find a monotone function $f$ in $R^1$ satisfying that there is no continuous function $g$ equal to $f$ almost everywhere in every open interval $I$?
We know that a monotone function is differentiable almost everywhere,so I think the property of $f$ can not be too bad.but I don't know if $f$ can be approached by a continuous function for every open interval.I tried to construct some functions but didn't work...
Thank you in advance!
real-analysis
asked Jan 16 at 16:50
MaxwellMaxwell
264
$endgroup$
Can we find a monotone function $f$ in $R^1$ satisfying that there is no continuous function $g$ equal to $f$ almost everywhere in every open interval $I$?
We know that a monotone function is differentiable almost everywhere,so I think the property of $f$ can not be too bad.but I don't know if $f$ can be approached by a continuous function for every open interval.I tried to construct some functions but didn't work...
Thank you in advance!
real-analysis
real-analysis
asked Jan 16 at 16:50
MaxwellMaxwell
264
asked Jan 16 at 16:50
MaxwellMaxwell
264
edited Jan 16 at 16:59
Maxwell
asked Jan 16 at 16:50
MaxwellMaxwell
264
asked Jan 16 at 16:50
MaxwellMaxwell
264
asked Jan 16 at 16:50
MaxwellMaxwell
264
264
$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
$begingroup$
@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
1
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
1
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09
|
show 2 more comments
$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
$begingroup$
@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
1
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
1
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09
$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
$begingroup$
@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
$begingroup$
@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
1
1
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
1
1
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09
|
show 2 more comments
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$begingroup$
I don't know what you mean by "a monotone function is derivative almost everywhere", but the Cantor function is continuous everywhere but nowhere continuous.
$endgroup$
– Bermudes
Jan 16 at 16:54
$begingroup$
@Bermudes I think the OP means differentiable. See beginning of this question.
$endgroup$
– Dog_69
Jan 16 at 16:58
$begingroup$
@Bermudes thank you,but can these two condition hold at the same time?It seems like a contradiction...continuous and nowhere continuous...
$endgroup$
– Maxwell
Jan 16 at 17:16
1
$begingroup$
@Maxwell I think he was wrong too. According to Wiki's page the Cantor function is continuous (uniformly continuous) but it is not completely continuous.
$endgroup$
– Dog_69
Jan 16 at 17:48
1
$begingroup$
Does a monotone increasing function which is discontinuous at all rational points do the job? The only possible discontinuities are jump discontinuities. I haven’t written a formal argument, though.
$endgroup$
– LinearOperator32
Jan 17 at 6:09