A convex set is one which contains the line segment between any two points in the set. How can one prove Euclidean balls and ellipsoids are convex sets?
[Math] How to Euclidean balls and Ellipsoids be convex sets
convex optimizationconvex-analysis
Related Solutions
This is almost certainly false. The following animation shows two convex shapes (with outlines shown in red and green) whose Minkowski sum is a disk of radius 3 (with outline shown in blue). The green shape is an ellipse with major and minor radii 1 and 1/2, which uniquely determines the red shape.
I do not have a proof that the red shape is convex, but it shouldn't be too hard to check.
Incidentally, here is the Mathematica code I used to produce this animation:
MyPlot = ParametricPlot[{3*{Cos[t], Sin[t]},
With[{u = ArcTan[-Sin[t], Cos[t]/2]},
3*{Sin[u], -Cos[u]} - {Cos[t], Sin[t]/2}]},
{t, 0, 2 Pi}];
myframes =
Table[With[{u = ArcTan[-Sin[t], Cos[t]/2]},
With[{pt = 3*{Sin[u], -Cos[u]} - {Cos[t], Sin[t]/2}},
Show[MyPlot,
ParametricPlot[pt + {Cos[r], Sin[r]/2}, {r, 0, 2 Pi},
PlotStyle -> Darker[Green]],
Graphics[{PointSize[Large], Point[pt]}]]]], {t, 0, 2 Pi - Pi/20,
Pi/20}]; ListAnimate[myframes]
Edit: Here is a simpler solution using two congruent shapes. The boundary of each shape is the union of two circular arcs, each of which is congruent to 1/4 of the blue circle.
Well, first note that if we only have two points $x_1$ and $x_2$, then all that's being said is whenever $a + b = 1$ the point $a*x_1 + b*x_2$ is in $D$. This is very clear though, because $b = 1-a$ and so the point in question is $a*x_1 + (1-a)*x_2$, which is a point on the line between $x_1$ and $x_2$.
Generally speaking, if we have points $x_1, ..., x_k$, and $\sum_{i=1}^k a_i = 1$, then you can write $a_1 + ... + a_{k-1} = 1 - a_k$ to get that
$\sum_{i=1}^k a_i x_i = a_k x_k + (1-a_k)\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k $
The points $x_k$ and $\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k$ may by induction be assumed to be points in $D$, so this forms the induction step of the proof.
Best Answer
If $a,b\in\mathbb{R}^+$ we have that $f(x,y)=ax^2+by^2$ is a convex function, as a sum of two convex functions. So, if $f(x_1,y_1)=f(x_2,y_2)=r^2$, for any $\lambda\in(0,1)$ we have: $$ f\left((1-\lambda)x_1+\lambda x_2,(1-\lambda)y_1+\lambda y_2\right)<r^2 $$ by Jensen's inequality. But the last line is equivalent to the open segment joining $(x_1,y_1),(x_2,y_2)$ to lie inside the ellipsoid.