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

Looking for a function $f$ that is $n$-differentiable, but $f^(n)$ is not continuousWays for a continuous function to not be smoothContinuous versus differentiableUniformly continuous and differentiable.Relation between differentiable,continuous and integrable functions.Differentiable function with bounded derivative, yet not uniformly continuousHow can we show that the functions are differentiable?Construct a real function with is exactly $C^2$ such that its first derivative does not vanish everywhereConstruct a real function with is exactly C^2 such that its first derivative does not vanish everywhereDifferentiable bijection $f:mathbbR to mathbbR$ with nonzero derivative whose inverse is not differentiableanti-derivative not differentiable at any point

2

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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

41k546166

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

$endgroup$

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  $begingroup$
  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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  $begingroup$
  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
  

  
  
  
  


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  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

41k546166

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

41k546166

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

$endgroup$

share|cite|improve this question

edited Mar 24 at 16:54

user21820

41k546166

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

share|cite|improve this question

edited Mar 24 at 16:54

user21820

41k546166

asked Mar 24 at 15:03

jmarvin_jmarvin_

134

edited Mar 24 at 16:54

user21820

41k546166

edited Mar 24 at 16:54

user21820

41k546166

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

189k24146262

$endgroup$

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  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

2,437213

$endgroup$

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  $begingroup$
  @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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  @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.

76.2k43477

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

189k24146262

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

189k24146262

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

189k24146262

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

189k24146262

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

189k24146262

answered Mar 24 at 15:06

José Carlos SantosJosé Carlos Santos

189k24146262

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

2,437213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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

2,437213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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

2,437213

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:05

MachineLearnerMachineLearner

2,437213

answered Mar 24 at 15:05

MachineLearnerMachineLearner

2,437213

answered Mar 24 at 15:05

MachineLearnerMachineLearner

2,437213

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.

76.2k43477

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

76.2k43477

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

76.2k43477

$endgroup$

share|cite|improve this answer

answered Mar 24 at 17:00

zhw.zhw.

76.2k43477

answered Mar 24 at 17:00

zhw.zhw.

76.2k43477

answered Mar 24 at 17:00

zhw.zhw.

76.2k43477