Convexity of a Function Implies Function Lies Above Any Tangent Line












1












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I’m trying to use the following definition to show the result listed below.



Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+frac{f(b)-f(a)}{b-a}(x-a)ge f(x)$$ for each $xin(a,b)$.



Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain.



Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $xin(x_1,x_2)$, $$f(x_1)+frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)ge f(x)$$



Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $xin(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$



The Mean Value Theorem guarantees the existence of $cin(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$



Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$



So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$



This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously.










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$endgroup$












  • $begingroup$
    If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
    $endgroup$
    – RRL
    Nov 29 '18 at 23:10
















1












$begingroup$


I’m trying to use the following definition to show the result listed below.



Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+frac{f(b)-f(a)}{b-a}(x-a)ge f(x)$$ for each $xin(a,b)$.



Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain.



Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $xin(x_1,x_2)$, $$f(x_1)+frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)ge f(x)$$



Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $xin(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$



The Mean Value Theorem guarantees the existence of $cin(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$



Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$



So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$



This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously.










share|cite|improve this question









$endgroup$












  • $begingroup$
    If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
    $endgroup$
    – RRL
    Nov 29 '18 at 23:10














1












1








1





$begingroup$


I’m trying to use the following definition to show the result listed below.



Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+frac{f(b)-f(a)}{b-a}(x-a)ge f(x)$$ for each $xin(a,b)$.



Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain.



Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $xin(x_1,x_2)$, $$f(x_1)+frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)ge f(x)$$



Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $xin(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$



The Mean Value Theorem guarantees the existence of $cin(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$



Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$



So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$



This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously.










share|cite|improve this question









$endgroup$




I’m trying to use the following definition to show the result listed below.



Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+frac{f(b)-f(a)}{b-a}(x-a)ge f(x)$$ for each $xin(a,b)$.



Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain.



Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $xin(x_1,x_2)$, $$f(x_1)+frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)ge f(x)$$



Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $xin(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$



The Mean Value Theorem guarantees the existence of $cin(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$



Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$



So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$



This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously.







real-analysis functions convex-analysis






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asked Nov 29 '18 at 22:39









TreeTree

1046




1046












  • $begingroup$
    If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
    $endgroup$
    – RRL
    Nov 29 '18 at 23:10


















  • $begingroup$
    If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
    $endgroup$
    – RRL
    Nov 29 '18 at 23:10
















$begingroup$
If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
$endgroup$
– RRL
Nov 29 '18 at 23:10




$begingroup$
If $f$ is twice differentiable, then it is easy to prove using Taylor's theorem: $f(x) = f(a) + f'(a)(x-a) + frac{1}{2} f''(xi)(x-a)^2 > f(a) + f'(a)(x-a)$ since $f'' > 0$ for a convex function
$endgroup$
– RRL
Nov 29 '18 at 23:10










2 Answers
2






active

oldest

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2












$begingroup$

Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have



$$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$



A convex function on an open interval always has one-sided derivatives, so



$$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$



giving you half of what you need.



If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that



$$f(z) geqslant f(x) + f'(x)(z-x)$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider



    $f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$



    First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities



    $frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$



    Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.



    $f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2












      $begingroup$

      Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have



      $$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$



      A convex function on an open interval always has one-sided derivatives, so



      $$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$



      giving you half of what you need.



      If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that



      $$f(z) geqslant f(x) + f'(x)(z-x)$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have



        $$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$



        A convex function on an open interval always has one-sided derivatives, so



        $$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$



        giving you half of what you need.



        If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that



        $$f(z) geqslant f(x) + f'(x)(z-x)$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have



          $$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$



          A convex function on an open interval always has one-sided derivatives, so



          $$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$



          giving you half of what you need.



          If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that



          $$f(z) geqslant f(x) + f'(x)(z-x)$$






          share|cite|improve this answer











          $endgroup$



          Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have



          $$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$



          A convex function on an open interval always has one-sided derivatives, so



          $$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$



          giving you half of what you need.



          If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that



          $$f(z) geqslant f(x) + f'(x)(z-x)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 '18 at 23:12

























          answered Nov 29 '18 at 23:05









          RRLRRL

          50.9k42573




          50.9k42573























              1












              $begingroup$

              Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider



              $f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$



              First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities



              $frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$



              Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.



              $f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider



                $f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$



                First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities



                $frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$



                Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.



                $f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider



                  $f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$



                  First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities



                  $frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$



                  Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.



                  $f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.






                  share|cite|improve this answer











                  $endgroup$



                  Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider



                  $f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$



                  First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities



                  $frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$



                  Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.



                  $f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 5:21

























                  answered Nov 30 '18 at 1:02









                  MatematletaMatematleta

                  10.7k2918




                  10.7k2918






























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