Reading this paper here, about convex optimization (photo page 2-3).

enter image description here

What about if the object misses the corner point? I think it is still convex but I am not sure whether my intuition fails me here. Is it convex or not? Is a ball without the boundary convex? I doubt it because if I select a point, I can always go nearer-and-nearer (never really reaching the point without extended Number system ofc). I am not sure about the general case with limit points like if I have non-dense space like missing points $\mathbb R^{N} \text{ (exclude sign here please) } \mathbb Q^{N}$ from the space $\mathbb C^{N}$.


The boundary of a convex set has special points, called extreme points, which generalize corners. You can remove any of those and the remaining set will be convex. If you remove an ordinary boundary point that is not an extreme point then you may end up with a set that is not convex.

All points in the boundary of a ball are extreme and so you can remove any or all of them. In the wikipedia illustration, the extreme points appear in red; you can remove any or all of them. But if you leave the endpoints of the red segments then you cannot remove any of the ordinary boundary points without breaking convexity.

A related result is the Krein–Milman theorem, which says that a compact convex set is the closure of the convex hull of its extreme points. Informally, you can recover the shape of a compact convex set from its extreme points.


I'm not sure I understand all of what you're saying, but from what I understand it seems that you've misunderstood the definition of convexity. In the slides you link to, convexity is introduced by the condition

$$f_i(\alpha x + \beta y) \le \alpha f_i(x) + \beta f_i(y)$$

for $\alpha+\beta=1$, $\alpha\ge0$, $\beta\ge0$ on the constraint functions $f_i$ in the constraints

$$f_i(x)\le b_i\;.$$

That is, if the constraints are fulfilled for $x$ and $y$, they are also fulfilled for all convex combinations of $x$ and $y$, that is, for all points on the line segment joining $x$ and $y$. Thus, the constraints define a convex set, which is a set containing all convex combinations of its points.

Thus, you don't have to think about whether you can approach or reach the boundary to decide whether a set is convex. A cube without its corners and a ball without its boundary are both convex, since these boundary points don't lie on line segments joining any other points. By contrast, if you were to remove the midpoints of the sides of a cube, it would no longer be convex, since those midpoints do lie on line segments joining other points, e.g. on the face diagonals – this is similar to the third example in the image.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.