Definition of convexity












1














We know that a function is convex if we have $$lambda f(x_1) + (1-lambda)f(x_2) ge f(lambda x_1 + (1-lambda)x_2)$$
where $0lelambdale1$



But I don't know where is it come from ? Unfortunately , I can't understand it. I searched in the internet many times but it didn't help to me. If someone explain this expression is helpful. 










share|cite|improve this question
























  • Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
    – hardmath
    Jan 14 '17 at 22:15










  • Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:32










  • A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
    – littleO
    Jan 14 '17 at 22:32










  • @hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
    – dxiv
    Jan 14 '17 at 22:34












  • @S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
    – dxiv
    Jan 14 '17 at 22:38


















1














We know that a function is convex if we have $$lambda f(x_1) + (1-lambda)f(x_2) ge f(lambda x_1 + (1-lambda)x_2)$$
where $0lelambdale1$



But I don't know where is it come from ? Unfortunately , I can't understand it. I searched in the internet many times but it didn't help to me. If someone explain this expression is helpful. 










share|cite|improve this question
























  • Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
    – hardmath
    Jan 14 '17 at 22:15










  • Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:32










  • A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
    – littleO
    Jan 14 '17 at 22:32










  • @hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
    – dxiv
    Jan 14 '17 at 22:34












  • @S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
    – dxiv
    Jan 14 '17 at 22:38
















1












1








1


1





We know that a function is convex if we have $$lambda f(x_1) + (1-lambda)f(x_2) ge f(lambda x_1 + (1-lambda)x_2)$$
where $0lelambdale1$



But I don't know where is it come from ? Unfortunately , I can't understand it. I searched in the internet many times but it didn't help to me. If someone explain this expression is helpful. 










share|cite|improve this question















We know that a function is convex if we have $$lambda f(x_1) + (1-lambda)f(x_2) ge f(lambda x_1 + (1-lambda)x_2)$$
where $0lelambdale1$



But I don't know where is it come from ? Unfortunately , I can't understand it. I searched in the internet many times but it didn't help to me. If someone explain this expression is helpful. 







convex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 '17 at 22:12









hardmath

28.6k94994




28.6k94994










asked Jan 14 '17 at 22:09









S.H.W

1,1491922




1,1491922












  • Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
    – hardmath
    Jan 14 '17 at 22:15










  • Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:32










  • A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
    – littleO
    Jan 14 '17 at 22:32










  • @hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
    – dxiv
    Jan 14 '17 at 22:34












  • @S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
    – dxiv
    Jan 14 '17 at 22:38




















  • Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
    – hardmath
    Jan 14 '17 at 22:15










  • Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:32










  • A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
    – littleO
    Jan 14 '17 at 22:32










  • @hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
    – dxiv
    Jan 14 '17 at 22:34












  • @S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
    – dxiv
    Jan 14 '17 at 22:38


















Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
– hardmath
Jan 14 '17 at 22:15




Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is convex essentially amounts to saying the region above the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent.
– hardmath
Jan 14 '17 at 22:15












Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
– S.H.W
Jan 14 '17 at 22:32




Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "?
– S.H.W
Jan 14 '17 at 22:32












A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
– littleO
Jan 14 '17 at 22:32




A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions.
– littleO
Jan 14 '17 at 22:32












@hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
– dxiv
Jan 14 '17 at 22:34






@hardmath A convex curve is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a convex function while the latter by a concave function. The geometrical definition of a convex function is that the set of points above its graph $y ge f(x)$ (a.k.a. its epigraph) is convex .
– dxiv
Jan 14 '17 at 22:34














@S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
– dxiv
Jan 14 '17 at 22:38






@S.H.W If you look at the very first picture on the wikipedia page for convex functions you'll see what the inequality means geometrically (except they use $t$ instead of $lambda$).
– dxiv
Jan 14 '17 at 22:38












2 Answers
2






active

oldest

votes


















3














The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.



enter image description here





Edit. For the proof, it suffices to notice that for all $A,Binmathbb{R}^2$, one has:
$$[AB]={(1-t)A+tB;tin[0,1]}.$$
This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.






share|cite|improve this answer























  • I know it , but can you show it mathematically ?
    – S.H.W
    Jan 14 '17 at 22:13












  • Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
    – Andrew Whelan
    Jan 14 '17 at 22:18










  • @Andrew Whelan Is your purpose slope of it ?
    – S.H.W
    Jan 14 '17 at 22:21












  • @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:29










  • @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
    – C. Falcon
    Jan 14 '17 at 22:41





















0














Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.



The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.



We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.



Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.



The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.






share|cite|improve this answer























    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%2f2098008%2fdefinition-of-convexity%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









    3














    The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.



    enter image description here





    Edit. For the proof, it suffices to notice that for all $A,Binmathbb{R}^2$, one has:
    $$[AB]={(1-t)A+tB;tin[0,1]}.$$
    This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.






    share|cite|improve this answer























    • I know it , but can you show it mathematically ?
      – S.H.W
      Jan 14 '17 at 22:13












    • Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
      – Andrew Whelan
      Jan 14 '17 at 22:18










    • @Andrew Whelan Is your purpose slope of it ?
      – S.H.W
      Jan 14 '17 at 22:21












    • @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
      – S.H.W
      Jan 14 '17 at 22:29










    • @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
      – C. Falcon
      Jan 14 '17 at 22:41


















    3














    The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.



    enter image description here





    Edit. For the proof, it suffices to notice that for all $A,Binmathbb{R}^2$, one has:
    $$[AB]={(1-t)A+tB;tin[0,1]}.$$
    This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.






    share|cite|improve this answer























    • I know it , but can you show it mathematically ?
      – S.H.W
      Jan 14 '17 at 22:13












    • Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
      – Andrew Whelan
      Jan 14 '17 at 22:18










    • @Andrew Whelan Is your purpose slope of it ?
      – S.H.W
      Jan 14 '17 at 22:21












    • @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
      – S.H.W
      Jan 14 '17 at 22:29










    • @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
      – C. Falcon
      Jan 14 '17 at 22:41
















    3












    3








    3






    The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.



    enter image description here





    Edit. For the proof, it suffices to notice that for all $A,Binmathbb{R}^2$, one has:
    $$[AB]={(1-t)A+tB;tin[0,1]}.$$
    This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.






    share|cite|improve this answer














    The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.



    enter image description here





    Edit. For the proof, it suffices to notice that for all $A,Binmathbb{R}^2$, one has:
    $$[AB]={(1-t)A+tB;tin[0,1]}.$$
    This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 14 '17 at 22:16

























    answered Jan 14 '17 at 22:12









    C. Falcon

    14.9k41950




    14.9k41950












    • I know it , but can you show it mathematically ?
      – S.H.W
      Jan 14 '17 at 22:13












    • Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
      – Andrew Whelan
      Jan 14 '17 at 22:18










    • @Andrew Whelan Is your purpose slope of it ?
      – S.H.W
      Jan 14 '17 at 22:21












    • @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
      – S.H.W
      Jan 14 '17 at 22:29










    • @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
      – C. Falcon
      Jan 14 '17 at 22:41




















    • I know it , but can you show it mathematically ?
      – S.H.W
      Jan 14 '17 at 22:13












    • Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
      – Andrew Whelan
      Jan 14 '17 at 22:18










    • @Andrew Whelan Is your purpose slope of it ?
      – S.H.W
      Jan 14 '17 at 22:21












    • @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
      – S.H.W
      Jan 14 '17 at 22:29










    • @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
      – C. Falcon
      Jan 14 '17 at 22:41


















    I know it , but can you show it mathematically ?
    – S.H.W
    Jan 14 '17 at 22:13






    I know it , but can you show it mathematically ?
    – S.H.W
    Jan 14 '17 at 22:13














    Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
    – Andrew Whelan
    Jan 14 '17 at 22:18




    Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$?
    – Andrew Whelan
    Jan 14 '17 at 22:18












    @Andrew Whelan Is your purpose slope of it ?
    – S.H.W
    Jan 14 '17 at 22:21






    @Andrew Whelan Is your purpose slope of it ?
    – S.H.W
    Jan 14 '17 at 22:21














    @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:29




    @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "?
    – S.H.W
    Jan 14 '17 at 22:29












    @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
    – C. Falcon
    Jan 14 '17 at 22:41






    @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $lambdageqslant 0$ and $B$ weight $mugeqslant 0$, with $lambda+mu=1$, then the barycenter of the system ${(A,lambda),(B,mu)}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$.
    – C. Falcon
    Jan 14 '17 at 22:41













    0














    Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.



    The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.



    We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.



    Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.



    The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.






    share|cite|improve this answer




























      0














      Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.



      The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.



      We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.



      Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.



      The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.






      share|cite|improve this answer


























        0












        0








        0






        Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.



        The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.



        We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.



        Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.



        The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.






        share|cite|improve this answer














        Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.



        The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.



        We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.



        Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.



        The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 at 13:16

























        answered Nov 24 at 13:04









        PJTraill

        651518




        651518






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f2098008%2fdefinition-of-convexity%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

            Aardman Animations

            Are they similar matrix

            “minimization” problem in Euclidean space related to orthonormal basis