Proving characterization of convexity as “first order Taylor approximation is a global underestimate”











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From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,yin dom f$,



$$f(y)ge f(x)+nabla f(x)^T(y-x).$$



I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:



They start out by picking $thetain[0,1]$, and $x,yin dom f$. If $f$ is a single-variable function, the assumed equality tells us that $theta f(x)ge theta f(z)+theta f'(z)(x-z)$ and $(1-theta)f(y)ge (1-theta) f(z)+(1-theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $theta f(x)+(1-theta) f(y)ge f(z)$. So it must be the case that $theta f'(z)(x-z)+(1-theta)f'(z)(y-z)ge 0$. I'm not able to show this.



Any help is greatly appreciated. Thanks in advance.






real-analysis convex-optimization







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    up vote
    1
    down vote

    favorite












    From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,yin dom f$,



    $$f(y)ge f(x)+nabla f(x)^T(y-x).$$



    I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:



    They start out by picking $thetain[0,1]$, and $x,yin dom f$. If $f$ is a single-variable function, the assumed equality tells us that $theta f(x)ge theta f(z)+theta f'(z)(x-z)$ and $(1-theta)f(y)ge (1-theta) f(z)+(1-theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $theta f(x)+(1-theta) f(y)ge f(z)$. So it must be the case that $theta f'(z)(x-z)+(1-theta)f'(z)(y-z)ge 0$. I'm not able to show this.



    Any help is greatly appreciated. Thanks in advance.






    real-analysis convex-optimization







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,yin dom f$,



      $$f(y)ge f(x)+nabla f(x)^T(y-x).$$



      I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:



      They start out by picking $thetain[0,1]$, and $x,yin dom f$. If $f$ is a single-variable function, the assumed equality tells us that $theta f(x)ge theta f(z)+theta f'(z)(x-z)$ and $(1-theta)f(y)ge (1-theta) f(z)+(1-theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $theta f(x)+(1-theta) f(y)ge f(z)$. So it must be the case that $theta f'(z)(x-z)+(1-theta)f'(z)(y-z)ge 0$. I'm not able to show this.



      Any help is greatly appreciated. Thanks in advance.






      real-analysis convex-optimization







      share|cite|improve this question













      From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,yin dom f$,



      $$f(y)ge f(x)+nabla f(x)^T(y-x).$$



      I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:



      They start out by picking $thetain[0,1]$, and $x,yin dom f$. If $f$ is a single-variable function, the assumed equality tells us that $theta f(x)ge theta f(z)+theta f'(z)(x-z)$ and $(1-theta)f(y)ge (1-theta) f(z)+(1-theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $theta f(x)+(1-theta) f(y)ge f(z)$. So it must be the case that $theta f'(z)(x-z)+(1-theta)f'(z)(y-z)ge 0$. I'm not able to show this.



      Any help is greatly appreciated. Thanks in advance.






      real-analysis convex-optimization




      real-analysis convex-optimization






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      asked 2 days ago









      manofbear

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          The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
          Adding the inequalities gives
          $$
          theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
          $$






          share|cite|improve this answer





















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            active

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            up vote
            2
            down vote



            accepted










            The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
            Adding the inequalities gives
            $$
            theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
            $$






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
              Adding the inequalities gives
              $$
              theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
              $$






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
                Adding the inequalities gives
                $$
                theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
                $$






                share|cite|improve this answer












                The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
                Adding the inequalities gives
                $$
                theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Martin R

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