monotone real analytic function

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Suppose the function $f:[a,b]rightarrow mathbbR$ is real analytic and monotonically increasing, i.e. for all $x,yin [a,b]$ such that $x<y$ we have $f(x)leq f(y)$. Further, suppose that $f$ is non constant (for example suppose $f(x)<f(b)$ for all $x<b$). I am wondering if the function has to be strictly increasing due to the assumptions. First I tried to find a counterexample, but all examples I could think of failed to be analytic at some point.

I would be very happy if someone can provide me a counterexample, or can help me proving that $f$ has to be strictly increasing.

real-analysis analysis analytic-functions

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asked Jul 16 at 17:12

Oliver Watt

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Suppose the function $f:[a,b]rightarrow mathbbR$ is real analytic and monotonically increasing, i.e. for all $x,yin [a,b]$ such that $x<y$ we have $f(x)leq f(y)$. Further, suppose that $f$ is non constant (for example suppose $f(x)<f(b)$ for all $x<b$). I am wondering if the function has to be strictly increasing due to the assumptions. First I tried to find a counterexample, but all examples I could think of failed to be analytic at some point.

I would be very happy if someone can provide me a counterexample, or can help me proving that $f$ has to be strictly increasing.

real-analysis analysis analytic-functions

share|cite|improve this question

asked Jul 16 at 17:12

Oliver Watt

30217

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose the function $f:[a,b]rightarrow mathbbR$ is real analytic and monotonically increasing, i.e. for all $x,yin [a,b]$ such that $x<y$ we have $f(x)leq f(y)$. Further, suppose that $f$ is non constant (for example suppose $f(x)<f(b)$ for all $x<b$). I am wondering if the function has to be strictly increasing due to the assumptions. First I tried to find a counterexample, but all examples I could think of failed to be analytic at some point.

I would be very happy if someone can provide me a counterexample, or can help me proving that $f$ has to be strictly increasing.

real-analysis analysis analytic-functions

share|cite|improve this question

asked Jul 16 at 17:12

Oliver Watt

30217

Suppose the function $f:[a,b]rightarrow mathbbR$ is real analytic and monotonically increasing, i.e. for all $x,yin [a,b]$ such that $x<y$ we have $f(x)leq f(y)$. Further, suppose that $f$ is non constant (for example suppose $f(x)<f(b)$ for all $x<b$). I am wondering if the function has to be strictly increasing due to the assumptions. First I tried to find a counterexample, but all examples I could think of failed to be analytic at some point.

I would be very happy if someone can provide me a counterexample, or can help me proving that $f$ has to be strictly increasing.

real-analysis analysis analytic-functions

share|cite|improve this question

asked Jul 16 at 17:12

Oliver Watt

30217

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 17:12

Oliver Watt

30217

asked Jul 16 at 17:12

Oliver Watt

30217

asked Jul 16 at 17:12

Oliver Watt

30217

30217

add a comment | 

add a comment | 

1 Answer
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Yes, $f$ must be strictly increasing. If $x<y$ and $f(x)=f(y)$ then $f$ is constant on $[x,y]$ and hence $f$ is constant, since it's analytic.

Details for readers new to real-analytic functions: You can get that last bit in either of two ways. (i) note that $f$ extends to a function holomorphic in some neighborhood of $(a,b)$ in $Bbb C$, (ii) repeat the proof for holomorphic functions: Let $A$ be the set of points in $[a,b]$ where every derivative of $f$ vanishes; show that $A$ is open and closed, hence $A=(a,b)$ or $A=emptyset$..

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edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Yes, $f$ must be strictly increasing. If $x<y$ and $f(x)=f(y)$ then $f$ is constant on $[x,y]$ and hence $f$ is constant, since it's analytic.

Details for readers new to real-analytic functions: You can get that last bit in either of two ways. (i) note that $f$ extends to a function holomorphic in some neighborhood of $(a,b)$ in $Bbb C$, (ii) repeat the proof for holomorphic functions: Let $A$ be the set of points in $[a,b]$ where every derivative of $f$ vanishes; show that $A$ is open and closed, hence $A=(a,b)$ or $A=emptyset$..

share|cite|improve this answer

edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

add a comment | 

up vote
2
down vote



accepted

Yes, $f$ must be strictly increasing. If $x<y$ and $f(x)=f(y)$ then $f$ is constant on $[x,y]$ and hence $f$ is constant, since it's analytic.

Details for readers new to real-analytic functions: You can get that last bit in either of two ways. (i) note that $f$ extends to a function holomorphic in some neighborhood of $(a,b)$ in $Bbb C$, (ii) repeat the proof for holomorphic functions: Let $A$ be the set of points in $[a,b]$ where every derivative of $f$ vanishes; show that $A$ is open and closed, hence $A=(a,b)$ or $A=emptyset$..

share|cite|improve this answer

edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Yes, $f$ must be strictly increasing. If $x<y$ and $f(x)=f(y)$ then $f$ is constant on $[x,y]$ and hence $f$ is constant, since it's analytic.

Details for readers new to real-analytic functions: You can get that last bit in either of two ways. (i) note that $f$ extends to a function holomorphic in some neighborhood of $(a,b)$ in $Bbb C$, (ii) repeat the proof for holomorphic functions: Let $A$ be the set of points in $[a,b]$ where every derivative of $f$ vanishes; show that $A$ is open and closed, hence $A=(a,b)$ or $A=emptyset$..

share|cite|improve this answer

edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

Yes, $f$ must be strictly increasing. If $x<y$ and $f(x)=f(y)$ then $f$ is constant on $[x,y]$ and hence $f$ is constant, since it's analytic.

Details for readers new to real-analytic functions: You can get that last bit in either of two ways. (i) note that $f$ extends to a function holomorphic in some neighborhood of $(a,b)$ in $Bbb C$, (ii) repeat the proof for holomorphic functions: Let $A$ be the set of points in $[a,b]$ where every derivative of $f$ vanishes; show that $A$ is open and closed, hence $A=(a,b)$ or $A=emptyset$..

share|cite|improve this answer

edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 17:28

edited Jul 16 at 17:28

edited Jul 16 at 17:28

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

answered Jul 16 at 17:17

David C. Ullrich

54.3k33583

54.3k33583

add a comment | 

add a comment | 

 

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