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.










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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
















0












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 5:13









blue boyblue boy

1,236613




1,236613








  • 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




    $begingroup$
    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










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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 $x$.






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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 $x$.






    share|cite|improve this answer









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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 $x$.






      share|cite|improve this answer









      $endgroup$
















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        1





        $begingroup$

        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 $x$.






        share|cite|improve this answer









        $endgroup$



        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 $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 5:32









        John BJohn B

        1766




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