Existence of a function under certain given conditions.
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Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
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add a comment |
$begingroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
$endgroup$
1
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Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
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– John B
Dec 7 '18 at 5:16
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$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
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Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
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I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
$begingroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
$endgroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,243613
1,243613
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
1
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
1 Answer
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Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
3287
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add a comment |
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$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
3287
$endgroup$
add a comment |
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
3287
$endgroup$
add a comment |
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
3287
$endgroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
3287
answered Dec 7 '18 at 5:32
John BJohn B
3287
answered Dec 7 '18 at 5:32
John BJohn B
3287
answered Dec 7 '18 at 5:32
John BJohn B
3287
3287
add a comment |
add a comment |
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1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25