Mixing
$f$ continuous, monotone, what do we know about differentiability?
6
$begingroup$
I am interested in knowing what we can say in general about when a continuous function $f:mathbb{R} to mathbb{R}$ is differentiable.
To my mind, there are various ways a continuous function can fail to be differentiable. It could have a corner (i.e. its left-derivative is not equal to its right-derivative, but both exist). It could oscillate wildly, like $xsin dfrac{1}{x}$ at $x=0$. I'm not really sure if there are other options.
For instance, suppose we eliminate the oscillation by saying that $f$ is monotone on $[a,b]$. Can we then say for instance that its left-derivative exists almost everywhere?
real-analysis derivatives
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
$endgroup$
add a comment |
6
$begingroup$
I am interested in knowing what we can say in general about when a continuous function $f:mathbb{R} to mathbb{R}$ is differentiable.
To my mind, there are various ways a continuous function can fail to be differentiable. It could have a corner (i.e. its left-derivative is not equal to its right-derivative, but both exist). It could oscillate wildly, like $xsin dfrac{1}{x}$ at $x=0$. I'm not really sure if there are other options.
For instance, suppose we eliminate the oscillation by saying that $f$ is monotone on $[a,b]$. Can we then say for instance that its left-derivative exists almost everywhere?
real-analysis derivatives
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
$endgroup$
$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46
add a comment |
6
6
6
3
$begingroup$
I am interested in knowing what we can say in general about when a continuous function $f:mathbb{R} to mathbb{R}$ is differentiable.
To my mind, there are various ways a continuous function can fail to be differentiable. It could have a corner (i.e. its left-derivative is not equal to its right-derivative, but both exist). It could oscillate wildly, like $xsin dfrac{1}{x}$ at $x=0$. I'm not really sure if there are other options.
For instance, suppose we eliminate the oscillation by saying that $f$ is monotone on $[a,b]$. Can we then say for instance that its left-derivative exists almost everywhere?
real-analysis derivatives
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
$endgroup$
I am interested in knowing what we can say in general about when a continuous function $f:mathbb{R} to mathbb{R}$ is differentiable.
To my mind, there are various ways a continuous function can fail to be differentiable. It could have a corner (i.e. its left-derivative is not equal to its right-derivative, but both exist). It could oscillate wildly, like $xsin dfrac{1}{x}$ at $x=0$. I'm not really sure if there are other options.
For instance, suppose we eliminate the oscillation by saying that $f$ is monotone on $[a,b]$. Can we then say for instance that its left-derivative exists almost everywhere?
real-analysis derivatives
real-analysis derivatives
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
asked Mar 15 '14 at 9:52
Eric AuldEric Auld
13.1k431112
13.1k431112
$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46
add a comment |
$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46
$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46
$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46
add a comment |
1 Answer
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$begingroup$
A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
$endgroup$
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
add a comment |
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1 Answer
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1 Answer
1
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oldest
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0
$begingroup$
A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
$endgroup$
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
add a comment |
0
$begingroup$
A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
$endgroup$
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
add a comment |
0
0
0
$begingroup$
A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
$endgroup$
A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
edited Jan 1 '18 at 17:27
user99914
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
answered Mar 15 '14 at 10:01
Michael Greinecker♦Michael Greinecker
24.7k453104
24.7k453104
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
add a comment |
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing?
$endgroup$
– Eric Auld
Mar 15 '14 at 14:31
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
@EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere.
$endgroup$
– user127096
Mar 16 '14 at 19:40
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips?
$endgroup$
– Eric Auld
Apr 8 '14 at 23:12
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
A followup question would be whether the function can fail to be differentiable uncountably often.
$endgroup$
– Solomonoff's Secret
Jan 1 '18 at 17:34
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
$begingroup$
@Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions.
$endgroup$
– Dave L. Renfro
Jan 1 '18 at 18:42
add a comment |
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$begingroup$
Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$.
$endgroup$
– David Mitra
Mar 15 '14 at 10:46