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
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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
$begingroup$
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
edited yesterday
user21820
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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
edited yesterday
user21820
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
5 hours ago
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
edited yesterday
user21820
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
real-analysis functions derivatives real-numbers
edited yesterday
user21820
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
user21820
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
39.7k544158
asked yesterday
jmarvin_jmarvin_
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
jmarvin_jmarvin_
134
asked yesterday
jmarvin_jmarvin_
134
134
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
5 hours ago
add a comment |
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
5 hours ago
$begingroup$
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_
yesterday
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
yesterday
$begingroup$
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_
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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_
5 hours ago
$begingroup$
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_
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
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 $x=0$.
answered yesterday
MachineLearnerMachineLearner
1,327112
$endgroup$
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
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 $x.$
answered yesterday
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Sure. Take, for instance$$f(x)=begincasesx^2&text if xgeqslant0\-x^2&text otherwise.endcases$$
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
answered yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
yesterday
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 $x=0$.
answered yesterday
MachineLearnerMachineLearner
1,327112
$endgroup$
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
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 $x=0$.
answered yesterday
MachineLearnerMachineLearner
1,327112
$endgroup$
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
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 $x=0$.
answered yesterday
MachineLearnerMachineLearner
1,327112
$endgroup$
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 $x=0$.
answered yesterday
MachineLearnerMachineLearner
1,327112
answered yesterday
MachineLearnerMachineLearner
1,327112
answered yesterday
MachineLearnerMachineLearner
1,327112
answered yesterday
MachineLearnerMachineLearner
1,327112
1,327112
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
add a comment |
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
$begingroup$
@TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always.
$endgroup$
– coffeemath
yesterday
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 $x.$
answered yesterday
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
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 $x.$
answered yesterday
zhw.zhw.
74.7k43175
$endgroup$
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
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 $x.$
answered yesterday
zhw.zhw.
74.7k43175
$endgroup$
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 $x.$
answered yesterday
zhw.zhw.
74.7k43175
answered yesterday
zhw.zhw.
74.7k43175
answered yesterday
zhw.zhw.
74.7k43175
answered yesterday
zhw.zhw.
74.7k43175
74.7k43175
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
add a comment |
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
yesterday
add a comment |
jmarvin_ is a new contributor. Be nice, and check out our Code of Conduct.
jmarvin_ is a new contributor. Be nice, and check out our Code of Conduct.
jmarvin_ is a new contributor. Be nice, and check out our Code of Conduct.
jmarvin_ is a new contributor. Be nice, and check out our Code of Conduct.
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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!
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– jmarvin_
yesterday
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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.
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– Dave L. Renfro
yesterday
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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)
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– jmarvin_
yesterday
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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.
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– Dave L. Renfro
yesterday
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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.
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– jmarvin_
5 hours ago