Mixing
Differentiability implies Lipschitz continuity
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Let $f:[0,1]tomathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0in [0,1]$. Is it true that there exists $L>0$ such that $lvert f(x)-f(x_0)lvertleq Llvert x-x_0lvert$?
I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?
real-analysis functional-analysis
asked May 19 '13 at 11:21
JulianJulian
1,04611025
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]tomathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0in [0,1]$. Is it true that there exists $L>0$ such that $lvert f(x)-f(x_0)lvertleq Llvert x-x_0lvert$?
I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?
real-analysis functional-analysis
asked May 19 '13 at 11:21
JulianJulian
1,04611025
$endgroup$
1
$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43
add a comment |
$begingroup$
Let $f:[0,1]tomathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0in [0,1]$. Is it true that there exists $L>0$ such that $lvert f(x)-f(x_0)lvertleq Llvert x-x_0lvert$?
I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?
real-analysis functional-analysis
asked May 19 '13 at 11:21
JulianJulian
1,04611025
$endgroup$
Let $f:[0,1]tomathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0in [0,1]$. Is it true that there exists $L>0$ such that $lvert f(x)-f(x_0)lvertleq Llvert x-x_0lvert$?
I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?
real-analysis functional-analysis
real-analysis functional-analysis
asked May 19 '13 at 11:21
JulianJulian
1,04611025
asked May 19 '13 at 11:21
JulianJulian
1,04611025
asked May 19 '13 at 11:21
JulianJulian
1,04611025
asked May 19 '13 at 11:21
JulianJulian
1,04611025
asked May 19 '13 at 11:21
JulianJulian
1,04611025
1,04611025
1
$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43
add a comment |
1
$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43
1
1
$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43
$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43
add a comment |
2 Answers
2
active
oldest
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$begingroup$
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |leq L_1| x-x_0|$ for $|x-x_0| < delta$. Since $f$ is continuous on a compact set, $|f(x)|_{infty} < infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |leq L_2| x-x_0|$ for $|x-x_0| geq delta$. Take $L = max {L_1, L_2}$.
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
$endgroup$
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
add a comment |
$begingroup$
It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:
Let $f:[0,1]tomathbb{R},, f(x)=x^{frac{3}{2}}sin(frac{1}{x}),,x>0,, f(0)=0$. Then it's derivate outside $0$ is $f^`(x)=frac{3}{2}x^{frac{1}{2}}sin(frac{1}{x})-x^{-frac{1}{2}}cos(frac{1}{x})$ and note that $limsup_{xdownarrow 0}|f^´(x)|=infty$.
Furthermore $f$ is differentiable in $0$ since
begin{equation}
lim_{hdownarrow 0}leftlvertfrac{h^{frac{3}{2}}sin(frac{1}{h})}{h}rightrvertleqlim_{hdownarrow 0}leftlvert h^{frac{1}{2}}rightrvert=0 , ,
end{equation}
so $f$ actually meets your requirements.
But the unbounded derivative outside $0$ gives us that there is no $epsilon > 0$ and $Lgeq0$ such that $|f(x)-f(y)|leq L|x-y|, forall, x,,yin[0,epsilon]$.
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |leq L_1| x-x_0|$ for $|x-x_0| < delta$. Since $f$ is continuous on a compact set, $|f(x)|_{infty} < infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |leq L_2| x-x_0|$ for $|x-x_0| geq delta$. Take $L = max {L_1, L_2}$.
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
$endgroup$
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
add a comment |
$begingroup$
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |leq L_1| x-x_0|$ for $|x-x_0| < delta$. Since $f$ is continuous on a compact set, $|f(x)|_{infty} < infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |leq L_2| x-x_0|$ for $|x-x_0| geq delta$. Take $L = max {L_1, L_2}$.
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
$endgroup$
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
add a comment |
$begingroup$
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |leq L_1| x-x_0|$ for $|x-x_0| < delta$. Since $f$ is continuous on a compact set, $|f(x)|_{infty} < infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |leq L_2| x-x_0|$ for $|x-x_0| geq delta$. Take $L = max {L_1, L_2}$.
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
$endgroup$
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |leq L_1| x-x_0|$ for $|x-x_0| < delta$. Since $f$ is continuous on a compact set, $|f(x)|_{infty} < infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |leq L_2| x-x_0|$ for $|x-x_0| geq delta$. Take $L = max {L_1, L_2}$.
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
answered May 19 '13 at 12:07
N.U.N.U.
2,54821426
2,54821426
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
add a comment |
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
$begingroup$
Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention).
$endgroup$
– gnometorule
May 19 '13 at 12:12
1
1
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
$begingroup$
@gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point.
$endgroup$
– N.U.
May 19 '13 at 12:15
add a comment |
$begingroup$
It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:
Let $f:[0,1]tomathbb{R},, f(x)=x^{frac{3}{2}}sin(frac{1}{x}),,x>0,, f(0)=0$. Then it's derivate outside $0$ is $f^`(x)=frac{3}{2}x^{frac{1}{2}}sin(frac{1}{x})-x^{-frac{1}{2}}cos(frac{1}{x})$ and note that $limsup_{xdownarrow 0}|f^´(x)|=infty$.
Furthermore $f$ is differentiable in $0$ since
begin{equation}
lim_{hdownarrow 0}leftlvertfrac{h^{frac{3}{2}}sin(frac{1}{h})}{h}rightrvertleqlim_{hdownarrow 0}leftlvert h^{frac{1}{2}}rightrvert=0 , ,
end{equation}
so $f$ actually meets your requirements.
But the unbounded derivative outside $0$ gives us that there is no $epsilon > 0$ and $Lgeq0$ such that $|f(x)-f(y)|leq L|x-y|, forall, x,,yin[0,epsilon]$.
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
$endgroup$
add a comment |
$begingroup$
It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:
Let $f:[0,1]tomathbb{R},, f(x)=x^{frac{3}{2}}sin(frac{1}{x}),,x>0,, f(0)=0$. Then it's derivate outside $0$ is $f^`(x)=frac{3}{2}x^{frac{1}{2}}sin(frac{1}{x})-x^{-frac{1}{2}}cos(frac{1}{x})$ and note that $limsup_{xdownarrow 0}|f^´(x)|=infty$.
Furthermore $f$ is differentiable in $0$ since
begin{equation}
lim_{hdownarrow 0}leftlvertfrac{h^{frac{3}{2}}sin(frac{1}{h})}{h}rightrvertleqlim_{hdownarrow 0}leftlvert h^{frac{1}{2}}rightrvert=0 , ,
end{equation}
so $f$ actually meets your requirements.
But the unbounded derivative outside $0$ gives us that there is no $epsilon > 0$ and $Lgeq0$ such that $|f(x)-f(y)|leq L|x-y|, forall, x,,yin[0,epsilon]$.
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
$endgroup$
add a comment |
$begingroup$
It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:
Let $f:[0,1]tomathbb{R},, f(x)=x^{frac{3}{2}}sin(frac{1}{x}),,x>0,, f(0)=0$. Then it's derivate outside $0$ is $f^`(x)=frac{3}{2}x^{frac{1}{2}}sin(frac{1}{x})-x^{-frac{1}{2}}cos(frac{1}{x})$ and note that $limsup_{xdownarrow 0}|f^´(x)|=infty$.
Furthermore $f$ is differentiable in $0$ since
begin{equation}
lim_{hdownarrow 0}leftlvertfrac{h^{frac{3}{2}}sin(frac{1}{h})}{h}rightrvertleqlim_{hdownarrow 0}leftlvert h^{frac{1}{2}}rightrvert=0 , ,
end{equation}
so $f$ actually meets your requirements.
But the unbounded derivative outside $0$ gives us that there is no $epsilon > 0$ and $Lgeq0$ such that $|f(x)-f(y)|leq L|x-y|, forall, x,,yin[0,epsilon]$.
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
$endgroup$
It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:
Let $f:[0,1]tomathbb{R},, f(x)=x^{frac{3}{2}}sin(frac{1}{x}),,x>0,, f(0)=0$. Then it's derivate outside $0$ is $f^`(x)=frac{3}{2}x^{frac{1}{2}}sin(frac{1}{x})-x^{-frac{1}{2}}cos(frac{1}{x})$ and note that $limsup_{xdownarrow 0}|f^´(x)|=infty$.
Furthermore $f$ is differentiable in $0$ since
begin{equation}
lim_{hdownarrow 0}leftlvertfrac{h^{frac{3}{2}}sin(frac{1}{h})}{h}rightrvertleqlim_{hdownarrow 0}leftlvert h^{frac{1}{2}}rightrvert=0 , ,
end{equation}
so $f$ actually meets your requirements.
But the unbounded derivative outside $0$ gives us that there is no $epsilon > 0$ and $Lgeq0$ such that $|f(x)-f(y)|leq L|x-y|, forall, x,,yin[0,epsilon]$.
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
answered Jan 18 at 17:41
mortenmcfishmortenmcfish
905
905
add a comment |
add a comment |
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$begingroup$
This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness.
$endgroup$
– robjohn♦
May 19 '13 at 17:43