Mixing
Question on one-sided derivatives
$begingroup$
Assume we have a function $f$, say on $mathbb{R}$, such that $f$ is continuously differentiable in all $x$ smaller than some given $x_0 in mathbb{R}$.
I am a bit confused about the connections of $lim_{x to x_0, x < x_0} f'(x)$ and the left derivative $f'_-(x_0)$ at $x_0$, since I could neither prove nor find a counterexample on these questions:
1) If $lim_{x to x_0, x < x_0} f'(x)$ exists, does so the left derivative at $x_0$ and are they equal?
2) If the left derivative exists, does the other limit exist and are they equal?
real-analysis derivatives
asked Jan 27 '16 at 17:39
agbagb
1,069415
$endgroup$
add a comment |
$begingroup$
Assume we have a function $f$, say on $mathbb{R}$, such that $f$ is continuously differentiable in all $x$ smaller than some given $x_0 in mathbb{R}$.
I am a bit confused about the connections of $lim_{x to x_0, x < x_0} f'(x)$ and the left derivative $f'_-(x_0)$ at $x_0$, since I could neither prove nor find a counterexample on these questions:
1) If $lim_{x to x_0, x < x_0} f'(x)$ exists, does so the left derivative at $x_0$ and are they equal?
2) If the left derivative exists, does the other limit exist and are they equal?
real-analysis derivatives
asked Jan 27 '16 at 17:39
agbagb
1,069415
$endgroup$
$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48
add a comment |
$begingroup$
Assume we have a function $f$, say on $mathbb{R}$, such that $f$ is continuously differentiable in all $x$ smaller than some given $x_0 in mathbb{R}$.
I am a bit confused about the connections of $lim_{x to x_0, x < x_0} f'(x)$ and the left derivative $f'_-(x_0)$ at $x_0$, since I could neither prove nor find a counterexample on these questions:
1) If $lim_{x to x_0, x < x_0} f'(x)$ exists, does so the left derivative at $x_0$ and are they equal?
2) If the left derivative exists, does the other limit exist and are they equal?
real-analysis derivatives
asked Jan 27 '16 at 17:39
agbagb
1,069415
$endgroup$
Assume we have a function $f$, say on $mathbb{R}$, such that $f$ is continuously differentiable in all $x$ smaller than some given $x_0 in mathbb{R}$.
I am a bit confused about the connections of $lim_{x to x_0, x < x_0} f'(x)$ and the left derivative $f'_-(x_0)$ at $x_0$, since I could neither prove nor find a counterexample on these questions:
1) If $lim_{x to x_0, x < x_0} f'(x)$ exists, does so the left derivative at $x_0$ and are they equal?
2) If the left derivative exists, does the other limit exist and are they equal?
real-analysis derivatives
real-analysis derivatives
asked Jan 27 '16 at 17:39
agbagb
1,069415
asked Jan 27 '16 at 17:39
agbagb
1,069415
asked Jan 27 '16 at 17:39
agbagb
1,069415
asked Jan 27 '16 at 17:39
agbagb
1,069415
asked Jan 27 '16 at 17:39
agbagb
1,069415
1,069415
$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48
add a comment |
$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48
$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48
$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the following I assume that $f$ is continuous.
The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough
$$
frac{f(x_0-h)-f(x_0)}{-h}=f'(xi_h),quad x_0-h<xi_h<x_0.
$$
Letting $hto0$ we get hat the left derivative exists at $x_0$ and equals $lim_{xto x_0^-}f'(x)$.
For 2) consider $f(x)=(x-x_0)^alphasinfrac{1}{x-x_0}$ if $xne x_0$, $f(x_0)=0$ with $1<alphale2$. Then $f$ is differentiable on $mathbb{R}$, $C^infty$ on $mathbb{R}setminus{x_0}$ and $lim_{xto x_0^pm}f'(x)$ does not exist.
edited Jan 19 at 14:56
Medo
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In the following I assume that $f$ is continuous.
The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough
$$
frac{f(x_0-h)-f(x_0)}{-h}=f'(xi_h),quad x_0-h<xi_h<x_0.
$$
Letting $hto0$ we get hat the left derivative exists at $x_0$ and equals $lim_{xto x_0^-}f'(x)$.
For 2) consider $f(x)=(x-x_0)^alphasinfrac{1}{x-x_0}$ if $xne x_0$, $f(x_0)=0$ with $1<alphale2$. Then $f$ is differentiable on $mathbb{R}$, $C^infty$ on $mathbb{R}setminus{x_0}$ and $lim_{xto x_0^pm}f'(x)$ does not exist.
edited Jan 19 at 14:56
Medo
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
add a comment |
$begingroup$
In the following I assume that $f$ is continuous.
The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough
$$
frac{f(x_0-h)-f(x_0)}{-h}=f'(xi_h),quad x_0-h<xi_h<x_0.
$$
Letting $hto0$ we get hat the left derivative exists at $x_0$ and equals $lim_{xto x_0^-}f'(x)$.
For 2) consider $f(x)=(x-x_0)^alphasinfrac{1}{x-x_0}$ if $xne x_0$, $f(x_0)=0$ with $1<alphale2$. Then $f$ is differentiable on $mathbb{R}$, $C^infty$ on $mathbb{R}setminus{x_0}$ and $lim_{xto x_0^pm}f'(x)$ does not exist.
edited Jan 19 at 14:56
Medo
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
add a comment |
$begingroup$
In the following I assume that $f$ is continuous.
The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough
$$
frac{f(x_0-h)-f(x_0)}{-h}=f'(xi_h),quad x_0-h<xi_h<x_0.
$$
Letting $hto0$ we get hat the left derivative exists at $x_0$ and equals $lim_{xto x_0^-}f'(x)$.
For 2) consider $f(x)=(x-x_0)^alphasinfrac{1}{x-x_0}$ if $xne x_0$, $f(x_0)=0$ with $1<alphale2$. Then $f$ is differentiable on $mathbb{R}$, $C^infty$ on $mathbb{R}setminus{x_0}$ and $lim_{xto x_0^pm}f'(x)$ does not exist.
edited Jan 19 at 14:56
Medo
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
$endgroup$
In the following I assume that $f$ is continuous.
The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough
$$
frac{f(x_0-h)-f(x_0)}{-h}=f'(xi_h),quad x_0-h<xi_h<x_0.
$$
Letting $hto0$ we get hat the left derivative exists at $x_0$ and equals $lim_{xto x_0^-}f'(x)$.
For 2) consider $f(x)=(x-x_0)^alphasinfrac{1}{x-x_0}$ if $xne x_0$, $f(x_0)=0$ with $1<alphale2$. Then $f$ is differentiable on $mathbb{R}$, $C^infty$ on $mathbb{R}setminus{x_0}$ and $lim_{xto x_0^pm}f'(x)$ does not exist.
edited Jan 19 at 14:56
Medo
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
edited Jan 19 at 14:56
Medo
642214
edited Jan 19 at 14:56
Medo
642214
edited Jan 19 at 14:56
Medo
642214
642214
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
answered Jan 27 '16 at 17:59
Julián AguirreJulián Aguirre
69.1k24096
69.1k24096
add a comment |
add a comment |
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$begingroup$
At a minimum, you need continuity at $x_0$ for (1) to be true.
$endgroup$
– Thomas Andrews
Jan 27 '16 at 17:48