Mixing
Prove or disprove that there exists $K$ such that $|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1],$ edited...
$begingroup$
Let $f$ be a function on $[0,1]$ into $Bbb{R}$. Suppose that if $xin[0,1],$ there exists $K_x$ such that begin{align}|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].end{align}
Prove or disprove that there exists $K$ such that begin{align}|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1].end{align}
DISPROOF
Consider the function begin{align} f:[0&,1]to
Bbb{R}, \&xmapsto sqrt{x} end{align}
Let $x=0$ and $yin (0,1]$ be fixed. Then,
begin{align} left| f(0)-f(y) right|&=left|0-sqrt{y} right| end{align}
Take $y=1/(4n^2)$ for all $n.$ Then,
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=left|0-dfrac{1}{sqrt{4n^2}} right| \&=2n^{3/2}left|dfrac{1}{4n^2} -0 right| end{align}
By assumption, there exists $K_0$ such that
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=2n^{3/2}left|dfrac{1}{4n^2} -0 right|leq K_0left|dfrac{1}{4n^2} -0 right| end{align}
Sending $ntoinfty,$ we have
begin{align} infty leq K_0<infty,;;text{contradiction}. end{align}
Hence, the function $f$ is not Lipschitz in $[0,1]$.
QUESTION: Is my disproof correct?
real-analysis analysis lipschitz-functions
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
$endgroup$
|
show 4 more comments
$begingroup$
Let $f$ be a function on $[0,1]$ into $Bbb{R}$. Suppose that if $xin[0,1],$ there exists $K_x$ such that begin{align}|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].end{align}
Prove or disprove that there exists $K$ such that begin{align}|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1].end{align}
DISPROOF
Consider the function begin{align} f:[0&,1]to
Bbb{R}, \&xmapsto sqrt{x} end{align}
Let $x=0$ and $yin (0,1]$ be fixed. Then,
begin{align} left| f(0)-f(y) right|&=left|0-sqrt{y} right| end{align}
Take $y=1/(4n^2)$ for all $n.$ Then,
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=left|0-dfrac{1}{sqrt{4n^2}} right| \&=2n^{3/2}left|dfrac{1}{4n^2} -0 right| end{align}
By assumption, there exists $K_0$ such that
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=2n^{3/2}left|dfrac{1}{4n^2} -0 right|leq K_0left|dfrac{1}{4n^2} -0 right| end{align}
Sending $ntoinfty,$ we have
begin{align} infty leq K_0<infty,;;text{contradiction}. end{align}
Hence, the function $f$ is not Lipschitz in $[0,1]$.
QUESTION: Is my disproof correct?
real-analysis analysis lipschitz-functions
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
$endgroup$
1
$begingroup$
I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 15:21
1
$begingroup$
It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
$endgroup$
– Cameron Buie
Jan 5 at 15:24
1
$begingroup$
Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
$endgroup$
– 6005
Jan 5 at 15:26
$begingroup$
Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
$endgroup$
– 6005
Jan 5 at 15:31
$begingroup$
related
$endgroup$
– Omnomnomnom
Jan 5 at 16:42
|
show 4 more comments
$begingroup$
Let $f$ be a function on $[0,1]$ into $Bbb{R}$. Suppose that if $xin[0,1],$ there exists $K_x$ such that begin{align}|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].end{align}
Prove or disprove that there exists $K$ such that begin{align}|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1].end{align}
DISPROOF
Consider the function begin{align} f:[0&,1]to
Bbb{R}, \&xmapsto sqrt{x} end{align}
Let $x=0$ and $yin (0,1]$ be fixed. Then,
begin{align} left| f(0)-f(y) right|&=left|0-sqrt{y} right| end{align}
Take $y=1/(4n^2)$ for all $n.$ Then,
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=left|0-dfrac{1}{sqrt{4n^2}} right| \&=2n^{3/2}left|dfrac{1}{4n^2} -0 right| end{align}
By assumption, there exists $K_0$ such that
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=2n^{3/2}left|dfrac{1}{4n^2} -0 right|leq K_0left|dfrac{1}{4n^2} -0 right| end{align}
Sending $ntoinfty,$ we have
begin{align} infty leq K_0<infty,;;text{contradiction}. end{align}
Hence, the function $f$ is not Lipschitz in $[0,1]$.
QUESTION: Is my disproof correct?
real-analysis analysis lipschitz-functions
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
$endgroup$
Let $f$ be a function on $[0,1]$ into $Bbb{R}$. Suppose that if $xin[0,1],$ there exists $K_x$ such that begin{align}|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].end{align}
Prove or disprove that there exists $K$ such that begin{align}|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1].end{align}
DISPROOF
Consider the function begin{align} f:[0&,1]to
Bbb{R}, \&xmapsto sqrt{x} end{align}
Let $x=0$ and $yin (0,1]$ be fixed. Then,
begin{align} left| f(0)-f(y) right|&=left|0-sqrt{y} right| end{align}
Take $y=1/(4n^2)$ for all $n.$ Then,
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=left|0-dfrac{1}{sqrt{4n^2}} right| \&=2n^{3/2}left|dfrac{1}{4n^2} -0 right| end{align}
By assumption, there exists $K_0$ such that
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=2n^{3/2}left|dfrac{1}{4n^2} -0 right|leq K_0left|dfrac{1}{4n^2} -0 right| end{align}
Sending $ntoinfty,$ we have
begin{align} infty leq K_0<infty,;;text{contradiction}. end{align}
Hence, the function $f$ is not Lipschitz in $[0,1]$.
QUESTION: Is my disproof correct?
real-analysis analysis lipschitz-functions
real-analysis analysis lipschitz-functions
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
edited Jan 7 at 9:34
Omojola Micheal
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 15:11
Omojola MichealOmojola Micheal
1,831324
1,831324
1
$begingroup$
I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 15:21
1
$begingroup$
It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
$endgroup$
– Cameron Buie
Jan 5 at 15:24
1
$begingroup$
Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
$endgroup$
– 6005
Jan 5 at 15:26
$begingroup$
Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
$endgroup$
– 6005
Jan 5 at 15:31
$begingroup$
related
$endgroup$
– Omnomnomnom
Jan 5 at 16:42
|
show 4 more comments
1
$begingroup$
I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 15:21
1
$begingroup$
It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
$endgroup$
– Cameron Buie
Jan 5 at 15:24
1
$begingroup$
Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
$endgroup$
– 6005
Jan 5 at 15:26
$begingroup$
Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
$endgroup$
– 6005
Jan 5 at 15:31
$begingroup$
related
$endgroup$
– Omnomnomnom
Jan 5 at 16:42
1
1
$begingroup$
I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 15:21
$begingroup$
I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 15:21
1
1
$begingroup$
It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
$endgroup$
– Cameron Buie
Jan 5 at 15:24
$begingroup$
It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
$endgroup$
– Cameron Buie
Jan 5 at 15:24
1
1
$begingroup$
Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
$endgroup$
– 6005
Jan 5 at 15:26
$begingroup$
Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
$endgroup$
– 6005
Jan 5 at 15:26
$begingroup$
Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
$endgroup$
– 6005
Jan 5 at 15:31
$begingroup$
Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
$endgroup$
– 6005
Jan 5 at 15:31
$begingroup$
related
$endgroup$
– Omnomnomnom
Jan 5 at 16:42
$begingroup$
related
$endgroup$
– Omnomnomnom
Jan 5 at 16:42
|
show 4 more comments
2 Answers
2
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oldest
votes
$begingroup$
You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.
On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
$endgroup$
$begingroup$
@Mike I added a few lines. Your disproof is incorrect.
$endgroup$
– Robert Z
Jan 5 at 16:48
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
$endgroup$
– David C. Ullrich
Jan 5 at 23:06
add a comment |
$begingroup$
This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.
The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.
Better still just to do an explicit calculation, as the other fellow did.
answered Jan 5 at 15:20
Ben WBen W
2,234615
$endgroup$
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.
On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
$endgroup$
$begingroup$
@Mike I added a few lines. Your disproof is incorrect.
$endgroup$
– Robert Z
Jan 5 at 16:48
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
$endgroup$
– David C. Ullrich
Jan 5 at 23:06
add a comment |
$begingroup$
You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.
On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
$endgroup$
$begingroup$
@Mike I added a few lines. Your disproof is incorrect.
$endgroup$
– Robert Z
Jan 5 at 16:48
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
$endgroup$
– David C. Ullrich
Jan 5 at 23:06
add a comment |
$begingroup$
You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.
On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
$endgroup$
You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.
On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
edited Jan 6 at 9:47
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
answered Jan 5 at 15:22
Robert ZRobert Z
95.3k1063134
95.3k1063134
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@Mike I added a few lines. Your disproof is incorrect.
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– Robert Z
Jan 5 at 16:48
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You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
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– David C. Ullrich
Jan 5 at 23:06
add a comment |
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@Mike I added a few lines. Your disproof is incorrect.
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– Robert Z
Jan 5 at 16:48
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
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– David C. Ullrich
Jan 5 at 23:06
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@Mike I added a few lines. Your disproof is incorrect.
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– Robert Z
Jan 5 at 16:48
$begingroup$
@Mike I added a few lines. Your disproof is incorrect.
$endgroup$
– Robert Z
Jan 5 at 16:48
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
$endgroup$
– David C. Ullrich
Jan 5 at 23:06
$begingroup$
You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...)
$endgroup$
– David C. Ullrich
Jan 5 at 23:06
add a comment |
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This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.
The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.
Better still just to do an explicit calculation, as the other fellow did.
answered Jan 5 at 15:20
Ben WBen W
2,234615
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Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
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– Omojola Micheal
Jan 5 at 16:36
add a comment |
$begingroup$
This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.
The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.
Better still just to do an explicit calculation, as the other fellow did.
answered Jan 5 at 15:20
Ben WBen W
2,234615
$endgroup$
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
add a comment |
$begingroup$
This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.
The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.
Better still just to do an explicit calculation, as the other fellow did.
answered Jan 5 at 15:20
Ben WBen W
2,234615
$endgroup$
This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.
The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.
Better still just to do an explicit calculation, as the other fellow did.
answered Jan 5 at 15:20
Ben WBen W
2,234615
edited Jan 5 at 15:25
answered Jan 5 at 15:20
Ben WBen W
2,234615
answered Jan 5 at 15:20
Ben WBen W
2,234615
answered Jan 5 at 15:20
Ben WBen W
2,234615
2,234615
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
add a comment |
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
$begingroup$
Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1)
$endgroup$
– Omojola Micheal
Jan 5 at 16:36
add a comment |
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I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose.
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– Andrés E. Caicedo
Jan 5 at 15:21
1
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It would be much simpler to just define $f:[0,1]toBbb R$ by $f(x)=sqrt{x}.$ There is no need for a piecewise definition.
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– Cameron Buie
Jan 5 at 15:24
1
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Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| le K |x - y|$ for all $x, y$, or else find a counterexample $f$.
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– 6005
Jan 5 at 15:26
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Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either.
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– 6005
Jan 5 at 15:31
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related
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– Omnomnomnom
Jan 5 at 16:42