Is $sqrt x$ locally Lipschitz continuous everywhere? The 2019 Stack Overflow Developer Survey...
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Is $sqrt x$ locally Lipschitz continuous everywhere?
The 2019 Stack Overflow Developer Survey Results Are InLocally Lipschitz and differentiable almost everywhereDifferentiable almost everywhere and LipschitzIs $sqrt{x}$ Lipschitz continuous on $(0,infty)$?Does global lipschitz imply locally lipschitz?Continuous and almost everywhere continuously differentiable with bounded gradient implies Lipschitz?Globally differentiable non-Lipschitz function that is uniformly continuous?Is a continuous and locally Lipschitz function Lipschitz?Concavity implies locally Lipschitz continuity?Property in-between Lipschitz continuous and *locally* Lipschitz continuous?Can a Lipschitz continuous function be linear almost everywhere but not linear everywhere?
$begingroup$
Is $f(x)=sqrt x$ locally Lipschitz continuous everywhere?
It is definitely not (globally) Lipschitz continuous. I wonder if it is locally Lipschitz continuous at $x=0$.
real-analysis calculus functions continuity lipschitz-functions
edited Mar 21 at 5:15
Ruslan
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
$endgroup$
add a comment |
$begingroup$
Is $f(x)=sqrt x$ locally Lipschitz continuous everywhere?
It is definitely not (globally) Lipschitz continuous. I wonder if it is locally Lipschitz continuous at $x=0$.
real-analysis calculus functions continuity lipschitz-functions
edited Mar 21 at 5:15
Ruslan
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
$endgroup$
add a comment |
$begingroup$
Is $f(x)=sqrt x$ locally Lipschitz continuous everywhere?
It is definitely not (globally) Lipschitz continuous. I wonder if it is locally Lipschitz continuous at $x=0$.
real-analysis calculus functions continuity lipschitz-functions
edited Mar 21 at 5:15
Ruslan
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
$endgroup$
Is $f(x)=sqrt x$ locally Lipschitz continuous everywhere?
It is definitely not (globally) Lipschitz continuous. I wonder if it is locally Lipschitz continuous at $x=0$.
real-analysis calculus functions continuity lipschitz-functions
real-analysis calculus functions continuity lipschitz-functions
edited Mar 21 at 5:15
Ruslan
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
edited Mar 21 at 5:15
Ruslan
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
edited Mar 21 at 5:15
Ruslan
3,73221634
edited Mar 21 at 5:15
Ruslan
3,73221634
edited Mar 21 at 5:15
Ruslan
3,73221634
3,73221634
asked Mar 21 at 5:03
High GPAHigh GPA
916422
asked Mar 21 at 5:03
High GPAHigh GPA
916422
asked Mar 21 at 5:03
High GPAHigh GPA
916422
916422
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Consider any interval $[0,varepsilon)$ where $varepsilon > 0$. For $x,yin [0,varepsilon)$, with $x < y$, we have
$$begin{align*}
frac{|f(x)- f(y)|}{|x-y|} &= frac{sqrt{y} - sqrt{x}}{y - x}\
&= frac{1}{sqrt{y}+sqrt{x}}.
end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
add a comment |
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1 Answer
1
active
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1 Answer
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active
oldest
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active
oldest
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active
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votes
$begingroup$
Hint: Consider any interval $[0,varepsilon)$ where $varepsilon > 0$. For $x,yin [0,varepsilon)$, with $x < y$, we have
$$begin{align*}
frac{|f(x)- f(y)|}{|x-y|} &= frac{sqrt{y} - sqrt{x}}{y - x}\
&= frac{1}{sqrt{y}+sqrt{x}}.
end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
add a comment |
$begingroup$
Hint: Consider any interval $[0,varepsilon)$ where $varepsilon > 0$. For $x,yin [0,varepsilon)$, with $x < y$, we have
$$begin{align*}
frac{|f(x)- f(y)|}{|x-y|} &= frac{sqrt{y} - sqrt{x}}{y - x}\
&= frac{1}{sqrt{y}+sqrt{x}}.
end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
add a comment |
$begingroup$
Hint: Consider any interval $[0,varepsilon)$ where $varepsilon > 0$. For $x,yin [0,varepsilon)$, with $x < y$, we have
$$begin{align*}
frac{|f(x)- f(y)|}{|x-y|} &= frac{sqrt{y} - sqrt{x}}{y - x}\
&= frac{1}{sqrt{y}+sqrt{x}}.
end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
Hint: Consider any interval $[0,varepsilon)$ where $varepsilon > 0$. For $x,yin [0,varepsilon)$, with $x < y$, we have
$$begin{align*}
frac{|f(x)- f(y)|}{|x-y|} &= frac{sqrt{y} - sqrt{x}}{y - x}\
&= frac{1}{sqrt{y}+sqrt{x}}.
end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
edited Mar 21 at 5:12
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 21 at 5:09
Minus One-TwelfthMinus One-Twelfth
3,363413
3,363413
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
add a comment |
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
$begingroup$
My guess: if we bound $x,yin[0,a]$ then the answer is "yes", so $f(x)$ is not locally lipschitz. If we bound $x,yin(0,a]$ then the answer is "no."
$endgroup$
– High GPA
Mar 21 at 5:12
1
1
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
$begingroup$
Even if $x,y>0$, we can make $frac{1}{sqrt{y}+sqrt{x}}$ arbitrarily large. For example take $x= frac{1}{N+1}$ and $y = frac{1}{N}$ for some $N$ sufficiently large.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:15
add a comment |
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