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Four vertex theorem 
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#1
Apr1511, 10:35 PM

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I just began to read Do Carmo's Differential Geometry book and having breezed through most of the chapter 1, I am finding it difficult to see how the four vertex theorem for a convex figure is proved?
Could someone please provide a simple and easy to understand proof? 


#2
Apr1611, 09:07 AM

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#3
Apr1611, 01:14 PM

P: 3

By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is tangent to C at the three points p, q, and r. But then the tangent to a point near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 129(b)). This implies that k = 0 at p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction. image host 


#4
Apr1711, 09:14 AM

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Four vertex theorem
I am also having trouble with this convexity argument.
also it seems wrong. I was trying to draw some pictures and am stumped by this example. Take a heart shape and round out the pointy vertex so the the curve is smooth. Make sure that it is narrow towards the top so the the maximum curvature occurs at the two apexes of the two halves of the heart. The minimum curvature occurs at the rounded out vertex since the curvature is maximum negative at this point. I see no reason why the line connecting the max and min curvature points separates the heart into exactly two pieces. This would not happen if the curve was convex. Or maybe he means the absolute value of the curvature. Anyway. still thinking about it. 


#5
Apr1711, 09:28 AM

P: 3

Lavinia, just to make sure that both of us are talking about the same thing, let me state the theorem from Do Carmo:
The FourVertex Theorem: A simple closed convex curve has at least four vertices. Actually, I can kind of see that for a convex curve this line L should divide the curve in two pieces, since if there was an extra Sshaped piece as shown in fig. a, then a tangent at p would have parts of the curve on both of its side, which violates convexity. So by contradiction, L should divide the curve in two distinct pieces. What I don't understand is this: By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is tangent to C at the three points p, q, and r. But then the tangent to a point near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 129(b)). This implies that k = 0 at p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction. 


#6
Apr1711, 09:50 AM

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It's just that a straight line connecting two points must pass entirely through the interior region that the curve encloses. If the line did not separate the curve into two pieces lying in opposite half planes then it would have to be tangent to prevent it from connecting p to q by passing outside of the interior region. That is what the picture is trying to illustrate. It seems that any line connecting two points of a convex curve must pass through the interior region and then out into the outer region. So there could never be a third point on the line because it could only be reached by returning from the outside region which would contradict convexity. But I guess he doesn't want to assume this and so gives us this ' it must be tangent' argument. I was a little confused because I thought a vertex was just a local extremum of the curvature. With this definition the theorem is still true. Interestingly, if you allow the curve to intersect itself then you can have only two local extrema even though the curvature is always positive. 


#7
Apr1711, 10:16 AM

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Pardon my ignorance,
I guess if convex means that the curvature is always positive then this tangent argument is necessary. what I think he is saying is that for the curve to cross and then come back the curvature would have to go negative. 


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