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Everywhere continuous and differentiable $f : mathbbR → mathbbR$ that is not smooth?

2

$begingroup$

I can't seem to find any counterexamples to the statement "all functions that are continuous and differentiable at every point of the reals are smooth," nor can I find anyone asserting or proving this statement. Are there known functions that are continuous and differentiable at every point (with no holes / discontinuities / bounded domain) but are not smooth, that is, after some number of derivatives the derivative function is no longer fully differentiable?

real-analysis functions derivatives real-numbers

share|cite|improve this question

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Forgot to specify that I am talking about real functions only, if that wasn't clear - complex functions that are discontinuous off the real axis are not interesting counterexamples!
  $endgroup$
  – jmarvin_
  Mar 24 at 15:04
  

  
  
  
  


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  What do you mean by "smooth" --- continuous derivative or infinitely differentiable (or something else)? FYI, both uses of "smooth" occur quite often here, and nearly always without the questioner saying what "smooth" means until asked.
  $endgroup$
  – Dave L. Renfro
  Mar 24 at 15:46
  

  
  
  
  


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  $begingroup$
  Dave: infinitely differentiable was the one I was using, since that's the limit of "smoothness" and thus what I thought was the unequivocal meaning of "smooth" (I specified this in the question at the end - it is "not smooth" if it is no longer totally differentiable after some derivative)
  $endgroup$
  – jmarvin_
  Mar 24 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, when I saw the part about totally differentiable, I thought you were talking about this, although I did wonder why you used the term in a non-multivariable setting (but not enough to think carefully about what you might have intended, since your question already had several answers). Incidentally, this answer might be of interest.
  $endgroup$
  – Dave L. Renfro
  Mar 24 at 18:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, yes, I was using the word "totally" as a non-technical synonym for "completely." My bad - it's been years since I took multivariable calc.
  $endgroup$
  – jmarvin_
  Mar 25 at 19:03
  

  
  
  
  

add a comment | 

2

$begingroup$

I can't seem to find any counterexamples to the statement "all functions that are continuous and differentiable at every point of the reals are smooth," nor can I find anyone asserting or proving this statement. Are there known functions that are continuous and differentiable at every point (with no holes / discontinuities / bounded domain) but are not smooth, that is, after some number of derivatives the derivative function is no longer fully differentiable?

real-analysis functions derivatives real-numbers

share|cite|improve this question

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Forgot to specify that I am talking about real functions only, if that wasn't clear - complex functions that are discontinuous off the real axis are not interesting counterexamples!
  $endgroup$
  – jmarvin_
  Mar 24 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "smooth" --- continuous derivative or infinitely differentiable (or something else)? FYI, both uses of "smooth" occur quite often here, and nearly always without the questioner saying what "smooth" means until asked.
  $endgroup$
  – Dave L. Renfro
  Mar 24 at 15:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dave: infinitely differentiable was the one I was using, since that's the limit of "smoothness" and thus what I thought was the unequivocal meaning of "smooth" (I specified this in the question at the end - it is "not smooth" if it is no longer totally differentiable after some derivative)
  $endgroup$
  – jmarvin_
  Mar 24 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, when I saw the part about totally differentiable, I thought you were talking about this, although I did wonder why you used the term in a non-multivariable setting (but not enough to think carefully about what you might have intended, since your question already had several answers). Incidentally, this answer might be of interest.
  $endgroup$
  – Dave L. Renfro
  Mar 24 at 18:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, yes, I was using the word "totally" as a non-technical synonym for "completely." My bad - it's been years since I took multivariable calc.
  $endgroup$
  – jmarvin_
  Mar 25 at 19:03
  

  
  
  
  

add a comment | 

$begingroup$

I can't seem to find any counterexamples to the statement "all functions that are continuous and differentiable at every point of the reals are smooth," nor can I find anyone asserting or proving this statement. Are there known functions that are continuous and differentiable at every point (with no holes / discontinuities / bounded domain) but are not smooth, that is, after some number of derivatives the derivative function is no longer fully differentiable?

real-analysis functions derivatives real-numbers

share|cite|improve this question

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

$endgroup$

share|cite|improve this question

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

share|cite|improve this question

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

edited Mar 24 at 16:54

user21820

40.4k544163

edited Mar 24 at 16:54

user21820

40.4k544163

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

3 Answers
3

active

oldest

votes

3

$begingroup$

Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - I wasn't thinking about just composing a piecewise function like this.
  $endgroup$
  – jmarvin_
  Mar 24 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 18:21
  

  
  
  
  

add a comment | 

3

$begingroup$

An example is

$$f(x) = begincases0 & textfor x<0\x^2 & textfor xgeq 0 endcases.$$

It is clear that the function is continuous and differentiable for all $xin mathbbR$. But $f'(x)$ is not differentiable at $x=0$.

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TomislavOstojich I think you didn't notice the prime. I said that $f(x)$ differentiable. And $f'(x)$ (the derivative of f(x)$ is not differentiable.
  $endgroup$
  – MachineLearner
  Mar 24 at 15:27
  
  

  
  
  
  


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  $begingroup$
  @TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
  $endgroup$
  – coffeemath
  Mar 24 at 15:27
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $W$ be the continuous, nowhere differentiable Weierstrass function. Then $f(x)=int_0^x W(t),dt$ is continuously differentiable on $mathbb R,$ but $f''(x)$ fails to exist for every $x.$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting answer, probably the most extreme one easily available. Thanks!
  $endgroup$
  – jmarvin_
  Mar 24 at 18:19
  

  
  
  
  

add a comment | 

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3

$begingroup$

Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - I wasn't thinking about just composing a piecewise function like this.
  $endgroup$
  – jmarvin_
  Mar 24 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 18:21
  

  
  
  
  

add a comment | 

3

$begingroup$

Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks - I wasn't thinking about just composing a piecewise function like this.
  $endgroup$
  – jmarvin_
  Mar 24 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 18:21
  

  
  
  
  

add a comment | 

$begingroup$

Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

176k24137247

3

$begingroup$

An example is

$$f(x) = begincases0 & textfor x<0\x^2 & textfor xgeq 0 endcases.$$

It is clear that the function is continuous and differentiable for all $xin mathbbR$. But $f'(x)$ is not differentiable at $x=0$.

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TomislavOstojich I think you didn't notice the prime. I said that $f(x)$ differentiable. And $f'(x)$ (the derivative of f(x)$ is not differentiable.
  $endgroup$
  – MachineLearner
  Mar 24 at 15:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
  $endgroup$
  – coffeemath
  Mar 24 at 15:27
  

  
  
  
  

add a comment | 

3

$begingroup$

An example is

$$f(x) = begincases0 & textfor x<0\x^2 & textfor xgeq 0 endcases.$$

It is clear that the function is continuous and differentiable for all $xin mathbbR$. But $f'(x)$ is not differentiable at $x=0$.

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TomislavOstojich I think you didn't notice the prime. I said that $f(x)$ differentiable. And $f'(x)$ (the derivative of f(x)$ is not differentiable.
  $endgroup$
  – MachineLearner
  Mar 24 at 15:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
  $endgroup$
  – coffeemath
  Mar 24 at 15:27
  

  
  
  
  

add a comment | 

$begingroup$

An example is

$$f(x) = begincases0 & textfor x<0\x^2 & textfor xgeq 0 endcases.$$

It is clear that the function is continuous and differentiable for all $xin mathbbR$. But $f'(x)$ is not differentiable at $x=0$.

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

answered Mar 24 at 15:05

MachineLearnerMachineLearner

1,539212

1

$begingroup$

Let $W$ be the continuous, nowhere differentiable Weierstrass function. Then $f(x)=int_0^x W(t),dt$ is continuously differentiable on $mathbb R,$ but $f''(x)$ fails to exist for every $x.$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting answer, probably the most extreme one easily available. Thanks!
  $endgroup$
  – jmarvin_
  Mar 24 at 18:19
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $W$ be the continuous, nowhere differentiable Weierstrass function. Then $f(x)=int_0^x W(t),dt$ is continuously differentiable on $mathbb R,$ but $f''(x)$ fails to exist for every $x.$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting answer, probably the most extreme one easily available. Thanks!
  $endgroup$
  – jmarvin_
  Mar 24 at 18:19
  

  
  
  
  

add a comment | 

$begingroup$

Let $W$ be the continuous, nowhere differentiable Weierstrass function. Then $f(x)=int_0^x W(t),dt$ is continuously differentiable on $mathbb R,$ but $f''(x)$ fails to exist for every $x.$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

$endgroup$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275

answered Mar 24 at 17:00

zhw.zhw.

75.2k43275