Everywhere continuous and differentiable $f : mathbb{R} → mathbb{R}$ 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
edited Mar 24 at 16:54
user21820
39.8k544158
asked Mar 24 at 15:03
jmarvin_jmarvin_
134
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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
edited Mar 24 at 16:54
user21820
39.8k544158
asked Mar 24 at 15:03
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_
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 |
2
2
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
edited Mar 24 at 16:54
user21820
39.8k544158
asked Mar 24 at 15:03
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 Mar 24 at 16:54
user21820
39.8k544158
asked Mar 24 at 15:03
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 Mar 24 at 16:54
user21820
39.8k544158
asked Mar 24 at 15:03
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 Mar 24 at 16:54
user21820
39.8k544158
edited Mar 24 at 16:54
user21820
39.8k544158
edited Mar 24 at 16:54
user21820
39.8k544158
39.8k544158
asked Mar 24 at 15:03
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 Mar 24 at 15:03
jmarvin_jmarvin_
134
asked Mar 24 at 15:03
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.
jmarvin_ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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_
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
3 Answers
3
active
oldest
votes
3
$begingroup$
Sure. Take, for instance$$f(x)=begin{cases}x^2&text{ if }xgeqslant0\-x^2&text{ otherwise.}end{cases}$$
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
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
$$f(x) = begin{cases}0 & text{for } x<0\x^2 & text{for } xgeq 0 end{cases}.$$
It is clear that the function is continuous and differentiable for all $xin mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
$endgroup$
$begingroup$
@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 |
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 $x.$
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
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 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Sure. Take, for instance$$f(x)=begin{cases}x^2&text{ if }xgeqslant0\-x^2&text{ otherwise.}end{cases}$$
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
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$$f(x)=begin{cases}x^2&text{ if }xgeqslant0\-x^2&text{ otherwise.}end{cases}$$
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
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
3
3
$begingroup$
Sure. Take, for instance$$f(x)=begin{cases}x^2&text{ if }xgeqslant0\-x^2&text{ otherwise.}end{cases}$$
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Sure. Take, for instance$$f(x)=begin{cases}x^2&text{ if }xgeqslant0\-x^2&text{ otherwise.}end{cases}$$
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 24 at 15:06
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
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 |
$begingroup$
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
$begingroup$
Thanks - I wasn't thinking about just composing a piecewise function like this.
$endgroup$
– jmarvin_
Mar 24 at 18:18
$begingroup$
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
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Mar 24 at 18:21
add a comment |
3
$begingroup$
An example is
$$f(x) = begin{cases}0 & text{for } x<0\x^2 & text{for } xgeq 0 end{cases}.$$
It is clear that the function is continuous and differentiable for all $xin mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
$endgroup$
$begingroup$
@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
$$f(x) = begin{cases}0 & text{for } x<0\x^2 & text{for } xgeq 0 end{cases}.$$
It is clear that the function is continuous and differentiable for all $xin mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
$endgroup$
$begingroup$
@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
3
3
$begingroup$
An example is
$$f(x) = begin{cases}0 & text{for } x<0\x^2 & text{for } xgeq 0 end{cases}.$$
It is clear that the function is continuous and differentiable for all $xin mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
$endgroup$
An example is
$$f(x) = begin{cases}0 & text{for } x<0\x^2 & text{for } xgeq 0 end{cases}.$$
It is clear that the function is continuous and differentiable for all $xin mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
answered Mar 24 at 15:05
MachineLearnerMachineLearner
1,319112
1,319112
$begingroup$
@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 |
$begingroup$
@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
$begingroup$
@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 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
$begingroup$
@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 $x.$
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
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 $x.$
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
Mar 24 at 18:19
add a comment |
1
1
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 $x.$
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
$endgroup$
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 Mar 24 at 17:00
zhw.zhw.
74.8k43175
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
answered Mar 24 at 17:00
zhw.zhw.
74.8k43175
74.8k43175
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
Mar 24 at 18:19
add a comment |
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
Mar 24 at 18:19
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
Mar 24 at 18:19
$begingroup$
This is an interesting answer, probably the most extreme one easily available. Thanks!
$endgroup$
– jmarvin_
Mar 24 at 18:19
add a comment |
jmarvin_ is a new contributor. Be nice, and check out our Code of Conduct.
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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_
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.
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– 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)
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– jmarvin_
Mar 24 at 18:07
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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
Mar 24 at 18:37
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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_
Mar 25 at 19:03