Proving that a specific value is taken by a continuous function

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Let $f:xlongmapsto dfrac 1n displaystylesum_k=1^n |x - a_k|$ where $a_k$ are elements of the interval $[0,1]$ satisfying $a_1<a_2<cdots <a_n$.



The question asks to prove that $f$ takes the value $dfraca_1+a_n2$



I tried using IVT by proving that $dfraca_1+a_n2$ lies between two known taken values of $f$ but I didn't achieve any progress.



Some remarks :
$$f(0) + f(1) = 1$$
$$f(a_1)=f(0)-a_1$$
$$f(a_n)=a_n-f(0)$$



Any hints would be appreciated.






continuity absolute-value







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  • 1




    Selection $a_k$s are random?
    – Nosrati
    1 hour ago










  • yes, in principle
    – Oussama Sarih
    1 hour ago






  • 1




    Is $f$ defined on $mathbbR$?
    – Xiangxiang Xu
    1 hour ago










  • @XiangxiangXu yes
    – Oussama Sarih
    1 hour ago














up vote
1
down vote

favorite
1












Let $f:xlongmapsto dfrac 1n displaystylesum_k=1^n |x - a_k|$ where $a_k$ are elements of the interval $[0,1]$ satisfying $a_1<a_2<cdots <a_n$.



The question asks to prove that $f$ takes the value $dfraca_1+a_n2$



I tried using IVT by proving that $dfraca_1+a_n2$ lies between two known taken values of $f$ but I didn't achieve any progress.



Some remarks :
$$f(0) + f(1) = 1$$
$$f(a_1)=f(0)-a_1$$
$$f(a_n)=a_n-f(0)$$



Any hints would be appreciated.






continuity absolute-value







share|cite|improve this question



















  • 1




    Selection $a_k$s are random?
    – Nosrati
    1 hour ago










  • yes, in principle
    – Oussama Sarih
    1 hour ago






  • 1




    Is $f$ defined on $mathbbR$?
    – Xiangxiang Xu
    1 hour ago










  • @XiangxiangXu yes
    – Oussama Sarih
    1 hour ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let $f:xlongmapsto dfrac 1n displaystylesum_k=1^n |x - a_k|$ where $a_k$ are elements of the interval $[0,1]$ satisfying $a_1<a_2<cdots <a_n$.



The question asks to prove that $f$ takes the value $dfraca_1+a_n2$



I tried using IVT by proving that $dfraca_1+a_n2$ lies between two known taken values of $f$ but I didn't achieve any progress.



Some remarks :
$$f(0) + f(1) = 1$$
$$f(a_1)=f(0)-a_1$$
$$f(a_n)=a_n-f(0)$$



Any hints would be appreciated.






continuity absolute-value







share|cite|improve this question















Let $f:xlongmapsto dfrac 1n displaystylesum_k=1^n |x - a_k|$ where $a_k$ are elements of the interval $[0,1]$ satisfying $a_1<a_2<cdots <a_n$.



The question asks to prove that $f$ takes the value $dfraca_1+a_n2$



I tried using IVT by proving that $dfraca_1+a_n2$ lies between two known taken values of $f$ but I didn't achieve any progress.



Some remarks :
$$f(0) + f(1) = 1$$
$$f(a_1)=f(0)-a_1$$
$$f(a_n)=a_n-f(0)$$



Any hints would be appreciated.






continuity absolute-value




continuity absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago

























asked 1 hour ago









Oussama Sarih

27617




27617







  • 1




    Selection $a_k$s are random?
    – Nosrati
    1 hour ago










  • yes, in principle
    – Oussama Sarih
    1 hour ago






  • 1




    Is $f$ defined on $mathbbR$?
    – Xiangxiang Xu
    1 hour ago










  • @XiangxiangXu yes
    – Oussama Sarih
    1 hour ago












  • 1




    Selection $a_k$s are random?
    – Nosrati
    1 hour ago










  • yes, in principle
    – Oussama Sarih
    1 hour ago






  • 1




    Is $f$ defined on $mathbbR$?
    – Xiangxiang Xu
    1 hour ago










  • @XiangxiangXu yes
    – Oussama Sarih
    1 hour ago







1




1




Selection $a_k$s are random?
– Nosrati
1 hour ago




Selection $a_k$s are random?
– Nosrati
1 hour ago












yes, in principle
– Oussama Sarih
1 hour ago




yes, in principle
– Oussama Sarih
1 hour ago




1




1




Is $f$ defined on $mathbbR$?
– Xiangxiang Xu
1 hour ago




Is $f$ defined on $mathbbR$?
– Xiangxiang Xu
1 hour ago












@XiangxiangXu yes
– Oussama Sarih
1 hour ago




@XiangxiangXu yes
– Oussama Sarih
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










Note that $f(x)$ is convex, hence
$$
fleft(fraca_1 + a_n2right) leq fracf(a_1) + f(a_n)2 = fraca_n - a_12 leq fraca_1 + a_n2.
$$
Also, we have
$$
f(2) geq 1 geq fraca_1 + a_n2.
$$
Then from IVT there exists $xi in left[fraca_1 + a_n2, 2right]$ such that $f(xi) = fraca_1 + a_n2$.






share|cite|improve this answer



























    up vote
    3
    down vote













    Yes, IVT is the right tool here. Note that $f$ is continuous and convex function in $mathbbR$ because it is a convex combination of the continuous and convex functions $xto |x-a_k|$ for $k=1,2,dots,n$. Hence, by your remarks,
    $$fleft(fraca_1+a_n2right)leq fracf(a_1)+f(a_n)2=fraca_n-a_12leq fraca_n+a_12.$$
    Now, in order to apply IVT, it suffices to say that
    $$fraca_n+a_12< sup_xin mathbbR f(x)=+infty.$$
    P.S. It is not always true that
    $$fraca_n+a_12leq max_xin [0,1] f(x)=maxf(0),f(1).$$
    Take for example $a_1=1/3$, $a_2=1/2$, and $a_3=3/4$. It follows that the value $fraca_1+a_n2$ could be attained by $f$ outside the interval $[0,1]$.






    share|cite|improve this answer






















    • Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
      – Oussama Sarih
      1 hour ago







    • 1




      Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
      – Robert Z
      1 hour ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Note that $f(x)$ is convex, hence
    $$
    fleft(fraca_1 + a_n2right) leq fracf(a_1) + f(a_n)2 = fraca_n - a_12 leq fraca_1 + a_n2.
    $$
    Also, we have
    $$
    f(2) geq 1 geq fraca_1 + a_n2.
    $$
    Then from IVT there exists $xi in left[fraca_1 + a_n2, 2right]$ such that $f(xi) = fraca_1 + a_n2$.






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      Note that $f(x)$ is convex, hence
      $$
      fleft(fraca_1 + a_n2right) leq fracf(a_1) + f(a_n)2 = fraca_n - a_12 leq fraca_1 + a_n2.
      $$
      Also, we have
      $$
      f(2) geq 1 geq fraca_1 + a_n2.
      $$
      Then from IVT there exists $xi in left[fraca_1 + a_n2, 2right]$ such that $f(xi) = fraca_1 + a_n2$.






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Note that $f(x)$ is convex, hence
        $$
        fleft(fraca_1 + a_n2right) leq fracf(a_1) + f(a_n)2 = fraca_n - a_12 leq fraca_1 + a_n2.
        $$
        Also, we have
        $$
        f(2) geq 1 geq fraca_1 + a_n2.
        $$
        Then from IVT there exists $xi in left[fraca_1 + a_n2, 2right]$ such that $f(xi) = fraca_1 + a_n2$.






        share|cite|improve this answer












        Note that $f(x)$ is convex, hence
        $$
        fleft(fraca_1 + a_n2right) leq fracf(a_1) + f(a_n)2 = fraca_n - a_12 leq fraca_1 + a_n2.
        $$
        Also, we have
        $$
        f(2) geq 1 geq fraca_1 + a_n2.
        $$
        Then from IVT there exists $xi in left[fraca_1 + a_n2, 2right]$ such that $f(xi) = fraca_1 + a_n2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Xiangxiang Xu

        89948




        89948




















            up vote
            3
            down vote













            Yes, IVT is the right tool here. Note that $f$ is continuous and convex function in $mathbbR$ because it is a convex combination of the continuous and convex functions $xto |x-a_k|$ for $k=1,2,dots,n$. Hence, by your remarks,
            $$fleft(fraca_1+a_n2right)leq fracf(a_1)+f(a_n)2=fraca_n-a_12leq fraca_n+a_12.$$
            Now, in order to apply IVT, it suffices to say that
            $$fraca_n+a_12< sup_xin mathbbR f(x)=+infty.$$
            P.S. It is not always true that
            $$fraca_n+a_12leq max_xin [0,1] f(x)=maxf(0),f(1).$$
            Take for example $a_1=1/3$, $a_2=1/2$, and $a_3=3/4$. It follows that the value $fraca_1+a_n2$ could be attained by $f$ outside the interval $[0,1]$.






            share|cite|improve this answer






















            • Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
              – Oussama Sarih
              1 hour ago







            • 1




              Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
              – Robert Z
              1 hour ago















            up vote
            3
            down vote













            Yes, IVT is the right tool here. Note that $f$ is continuous and convex function in $mathbbR$ because it is a convex combination of the continuous and convex functions $xto |x-a_k|$ for $k=1,2,dots,n$. Hence, by your remarks,
            $$fleft(fraca_1+a_n2right)leq fracf(a_1)+f(a_n)2=fraca_n-a_12leq fraca_n+a_12.$$
            Now, in order to apply IVT, it suffices to say that
            $$fraca_n+a_12< sup_xin mathbbR f(x)=+infty.$$
            P.S. It is not always true that
            $$fraca_n+a_12leq max_xin [0,1] f(x)=maxf(0),f(1).$$
            Take for example $a_1=1/3$, $a_2=1/2$, and $a_3=3/4$. It follows that the value $fraca_1+a_n2$ could be attained by $f$ outside the interval $[0,1]$.






            share|cite|improve this answer






















            • Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
              – Oussama Sarih
              1 hour ago







            • 1




              Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
              – Robert Z
              1 hour ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            Yes, IVT is the right tool here. Note that $f$ is continuous and convex function in $mathbbR$ because it is a convex combination of the continuous and convex functions $xto |x-a_k|$ for $k=1,2,dots,n$. Hence, by your remarks,
            $$fleft(fraca_1+a_n2right)leq fracf(a_1)+f(a_n)2=fraca_n-a_12leq fraca_n+a_12.$$
            Now, in order to apply IVT, it suffices to say that
            $$fraca_n+a_12< sup_xin mathbbR f(x)=+infty.$$
            P.S. It is not always true that
            $$fraca_n+a_12leq max_xin [0,1] f(x)=maxf(0),f(1).$$
            Take for example $a_1=1/3$, $a_2=1/2$, and $a_3=3/4$. It follows that the value $fraca_1+a_n2$ could be attained by $f$ outside the interval $[0,1]$.






            share|cite|improve this answer














            Yes, IVT is the right tool here. Note that $f$ is continuous and convex function in $mathbbR$ because it is a convex combination of the continuous and convex functions $xto |x-a_k|$ for $k=1,2,dots,n$. Hence, by your remarks,
            $$fleft(fraca_1+a_n2right)leq fracf(a_1)+f(a_n)2=fraca_n-a_12leq fraca_n+a_12.$$
            Now, in order to apply IVT, it suffices to say that
            $$fraca_n+a_12< sup_xin mathbbR f(x)=+infty.$$
            P.S. It is not always true that
            $$fraca_n+a_12leq max_xin [0,1] f(x)=maxf(0),f(1).$$
            Take for example $a_1=1/3$, $a_2=1/2$, and $a_3=3/4$. It follows that the value $fraca_1+a_n2$ could be attained by $f$ outside the interval $[0,1]$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 16 mins ago

























            answered 1 hour ago









            Robert Z

            85.9k1056124




            85.9k1056124











            • Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
              – Oussama Sarih
              1 hour ago







            • 1




              Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
              – Robert Z
              1 hour ago

















            • Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
              – Oussama Sarih
              1 hour ago







            • 1




              Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
              – Robert Z
              1 hour ago
















            Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
            – Oussama Sarih
            1 hour ago





            Isn't it rather $dfraca_1+a_n2 ge min f(x) $ (min taken over $mathbbR$) as the sup is $+infty$ ?
            – Oussama Sarih
            1 hour ago





            1




            1




            Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
            – Robert Z
            1 hour ago





            Yes, my first line shows that $dfraca_1+a_n2 ge min f(x)$. The sup is of course $+infty$.
            – Robert Z
            1 hour ago


















             

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