Convexity of a Function Implies Function Lies Above Any Tangent Line












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












$begingroup$


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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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

votes


















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











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019331%2fconvexity-of-a-function-implies-function-lies-above-any-tangent-line%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019331%2fconvexity-of-a-function-implies-function-lies-above-any-tangent-line%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      How to change which sound is reproduced for terminal bell?

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents