If ‎$‎f$ ‎is a ‎concave function ‎then ‎$‎f$ ‎is ‎increasing‎ The Next CEO...

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If ‎$‎f$ ‎is a ‎concave function ‎then ‎$‎f$ ‎is ‎increasing‎



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1












$begingroup$


My question is how to prove the following assertion:




If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.




I know that if ‎$‎‎f$ ‎is ‎differentiable ‎then the ‎mentioned ‎hypothesis ($f'$ is decreasing) and using M.V.T ‎imply ‎that firstly ‎‎$limlimits_{xtoinfty}f'‎(x)=0$ also ‎‎$‎f'‎geq0$ and $‎‎f$‎ ‎is ‎increasing. ‎but ‎here ‎we ‎do not ‎have ‎the differentiability.



‎Anyone ‎can ‎help ‎me. Thanks.‎






calculus convex-analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    My question is how to prove the following assertion:




    If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.




    I know that if ‎$‎‎f$ ‎is ‎differentiable ‎then the ‎mentioned ‎hypothesis ($f'$ is decreasing) and using M.V.T ‎imply ‎that firstly ‎‎$limlimits_{xtoinfty}f'‎(x)=0$ also ‎‎$‎f'‎geq0$ and $‎‎f$‎ ‎is ‎increasing. ‎but ‎here ‎we ‎do not ‎have ‎the differentiability.



    ‎Anyone ‎can ‎help ‎me. Thanks.‎






    calculus convex-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      My question is how to prove the following assertion:




      If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.




      I know that if ‎$‎‎f$ ‎is ‎differentiable ‎then the ‎mentioned ‎hypothesis ($f'$ is decreasing) and using M.V.T ‎imply ‎that firstly ‎‎$limlimits_{xtoinfty}f'‎(x)=0$ also ‎‎$‎f'‎geq0$ and $‎‎f$‎ ‎is ‎increasing. ‎but ‎here ‎we ‎do not ‎have ‎the differentiability.



      ‎Anyone ‎can ‎help ‎me. Thanks.‎






      calculus convex-analysis







      share|cite|improve this question











      $endgroup$




      My question is how to prove the following assertion:




      If $f:mathbb{R}tomathbb{R}$ is a concave function such that $limlimits_{xtoinfty}(f(x)-f(x-1))=0$ then $f$ increasing.




      I know that if ‎$‎‎f$ ‎is ‎differentiable ‎then the ‎mentioned ‎hypothesis ($f'$ is decreasing) and using M.V.T ‎imply ‎that firstly ‎‎$limlimits_{xtoinfty}f'‎(x)=0$ also ‎‎$‎f'‎geq0$ and $‎‎f$‎ ‎is ‎increasing. ‎but ‎here ‎we ‎do not ‎have ‎the differentiability.



      ‎Anyone ‎can ‎help ‎me. Thanks.‎






      calculus convex-analysis




      calculus convex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 16 at 7:52









      Saad

      20.2k92352




      20.2k92352










      asked Mar 16 at 7:33









      soodehMehboodisoodehMehboodi

      64238




      64238






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
          $$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
          is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.



          Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
          $$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
          where $k$ is any positive integer such that $x_0+k>x_1$.



          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 8:26












          • $begingroup$
            @soodehMehboodi See my edit.
            $endgroup$
            – Robert Z
            Mar 16 at 9:16










          • $begingroup$
            @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 10:00












          • $begingroup$
            One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
            $endgroup$
            – Robert Z
            Mar 16 at 10:48










          • $begingroup$
            @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
            $endgroup$
            – soodehMehboodi
            Mar 16 at 14:47














          Your Answer





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          2












          $begingroup$

          Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
          $$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
          is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.



          Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
          $$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
          where $k$ is any positive integer such that $x_0+k>x_1$.



          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 8:26












          • $begingroup$
            @soodehMehboodi See my edit.
            $endgroup$
            – Robert Z
            Mar 16 at 9:16










          • $begingroup$
            @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 10:00












          • $begingroup$
            One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
            $endgroup$
            – Robert Z
            Mar 16 at 10:48










          • $begingroup$
            @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
            $endgroup$
            – soodehMehboodi
            Mar 16 at 14:47


















          2












          $begingroup$

          Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
          $$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
          is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.



          Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
          $$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
          where $k$ is any positive integer such that $x_0+k>x_1$.



          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 8:26












          • $begingroup$
            @soodehMehboodi See my edit.
            $endgroup$
            – Robert Z
            Mar 16 at 9:16










          • $begingroup$
            @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 10:00












          • $begingroup$
            One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
            $endgroup$
            – Robert Z
            Mar 16 at 10:48










          • $begingroup$
            @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
            $endgroup$
            – soodehMehboodi
            Mar 16 at 14:47
















          2












          2








          2





          $begingroup$

          Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
          $$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
          is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.



          Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
          $$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
          where $k$ is any positive integer such that $x_0+k>x_1$.



          Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          Hint. Since $f$ is concave it follows that for any $x_0in mathbb{R}$ the function
          $$xto R(x,x_0):=frac{f(x)-f(x_0)}{x-x_0}$$
          is decreasing in $mathbb{R}setminus{x_0}$. Here it is a reference for the convex function $-f$.



          Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
          $$0>R(x_1,x_0)geq R(x_0+k+1,x_0)geq R(x_0+k+1,x_0+k)$$
          where $k$ is any positive integer such that $x_0+k>x_1$.



          Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 16 at 9:15

























          answered Mar 16 at 7:46









          Robert ZRobert Z

          101k1070143




          101k1070143












          • $begingroup$
            @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 8:26












          • $begingroup$
            @soodehMehboodi See my edit.
            $endgroup$
            – Robert Z
            Mar 16 at 9:16










          • $begingroup$
            @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 10:00












          • $begingroup$
            One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
            $endgroup$
            – Robert Z
            Mar 16 at 10:48










          • $begingroup$
            @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
            $endgroup$
            – soodehMehboodi
            Mar 16 at 14:47




















          • $begingroup$
            @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 8:26












          • $begingroup$
            @soodehMehboodi See my edit.
            $endgroup$
            – Robert Z
            Mar 16 at 9:16










          • $begingroup$
            @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
            $endgroup$
            – soodehMehboodi
            Mar 16 at 10:00












          • $begingroup$
            One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
            $endgroup$
            – Robert Z
            Mar 16 at 10:48










          • $begingroup$
            @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
            $endgroup$
            – soodehMehboodi
            Mar 16 at 14:47


















          $begingroup$
          @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
          $endgroup$
          – soodehMehboodi
          Mar 16 at 8:26






          $begingroup$
          @ Robert Z, Would you explain more that what happens for $f$ And how limit condition helps?
          $endgroup$
          – soodehMehboodi
          Mar 16 at 8:26














          $begingroup$
          @soodehMehboodi See my edit.
          $endgroup$
          – Robert Z
          Mar 16 at 9:16




          $begingroup$
          @soodehMehboodi See my edit.
          $endgroup$
          – Robert Z
          Mar 16 at 9:16












          $begingroup$
          @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
          $endgroup$
          – soodehMehboodi
          Mar 16 at 10:00






          $begingroup$
          @ Robert Z, You mean that it made contradiction with be decreasing of $R(x)$ but I want to know that where is $f(x)-f(x-1)to0$ used?
          $endgroup$
          – soodehMehboodi
          Mar 16 at 10:00














          $begingroup$
          One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
          $endgroup$
          – Robert Z
          Mar 16 at 10:48




          $begingroup$
          One single step is missing and I did not use the limit of $f(x)-f(x-1)=R(x-1,x)$ yet.
          $endgroup$
          – Robert Z
          Mar 16 at 10:48












          $begingroup$
          @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
          $endgroup$
          – soodehMehboodi
          Mar 16 at 14:47






          $begingroup$
          @ Robert Z, I think you proved the assertion "$f$ is increasing from a number on," but it seems the limit condition makes it hold for all $xin R$.
          $endgroup$
          – soodehMehboodi
          Mar 16 at 14:47




















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