It is usual to introduce Fréchet and Cakes derivatives in Banach spaces. In this context, Taylor's familiar expansion with the rest is also at hand, as you can see in the image below taken from this reference. Now, I would like to know if there is an analogous theorem for locally convex spaces and where should I go to learn it if this is the case. Do we also need Fréchet and Cakes derivatives? Any comments would be helpful. Thanks in advance!

# Tag: convex

## About the definition of the convex from Rudin's analysis

This is from Rudin's mathematical analysis:

We call a set $ E subset R ^ k $ convex if $ lambda x + (1- lambda) y in E $ anytime $ x in E $, $ y in E $, and $ 0 < lambda <1 $.

What is the meaning of this definition? And how we come to the equation $ lambda x + (1- lambda) y in E $?

## Find if the convex cone of a vector list is zero

Given a list of vectors, I want to know if there is a vector such that its dot product with those of the list is all (semi) positive, or at least above some small negative value .

## convex optimization – Conditions for $ min_x max_yf (x, y) = min_y max_x f (x, y) $?

Let $ f (x, y) $ be a real function of variables $ x, y $ (which can be real vectors). Under what conditions do we have the following equality:

$$ min_x max_yf (x, y) = min_y max_x f (x, y) $$

For example, this equality is true if $ f (x, y) = xy $ and $ x, y $ are real scalars.

Note that this is **do not** the same as Von Neumann's minimax theorem (https://en.wikipedia.org/wiki/Minimax_theorem), because here the role of variables is exchanged (*for example.*, $ x $ is minimized on the left side, but it is maximized on the right side).

Although I do not know if the convexity / concavity of $ f (x, y) $ with regard to one or the other of the arguments plays a role here (as it is the case for the minimon of Von Neumann), I use the `convex`

related tags here since this is the context where I saw related questions. Likewise, I mark `game-theory`

, but I'm not sure if it's directly applicable.

## digital geometry – Convex shell of unit circles

(Also posted on Math StackExchange, but I don't know if it belongs here or here).

I know that if we are trying to get the convex hull of the unit circles, we can just shrink the circles to their centers and consider the convex hull of their centers, but I'm trying to prove a few steps intermediaries towards that.

a) Show that the limit of the convex shell consists only of straight line segments and parts of circles.

b) Show that each circle appears at most once within the limit of the convex shell.

(This is from Berg's book of Computational Geometry.)

I kind of have an intuition of why this is true, but my problem is that every time I try to find a solution, I end up looking at a lot of cases, and I have the impression that it is not rigorous or elegant enough because of so many cases. Is there a good way to prove it?

(Note: I know this has already been published in https://math.stackexchange.com/questions/122867/convex-hull-algorithms, but I am not satisfied with the answers there, and I am trying to go for something more rigorous.)

## Collision detection – Can I test the confinement of convex polyhedra with the separation axis theorem?

I would like to refine the output of the separation axis test to not only tell me if the convex polyhedron A intersects the convex polyhedron B, but also if A is completely inside B. Is it safe to say that if the projections of the angle points of A (on the separating the axes) are still inside the projections of the angle points of B, A is inside from B?

My reasoning is

- If all the corner points of A are inside B, A is completely inside B
- A point is inside a convex polyhedron if it is on the right side of each face
- If any point, projected on a normal face of B, is inside the interval formed by the angle points of B projected on the same normal face, it is on the right side with respect to this face
- The face normals are part of the separation axes

but I'm not sure if it's correct, especially 1 and 3.

## real analysis – Closed convex sets

Let $ X $ to be a finite dimension Hilbet space. Consider the functions

$ g_1, cdots, g_p: X longrightarrow mathbb {R} $ , $ h_1, cdots, h_q: X longrightarrow mathbb {R} $ and the whole $ I (x) = {i in {1, cdots, p }: g_i (x) = 0 } $

Let $ bar {x} in mathbb {R} $ .

how can i prove that the sets

$$ A = left { sum_ {j = 1} ^ q lambda_j nabla h_j ( bar {x}): lambda_1, cdots, lambda_q in mathbb {R} right } $ $

$$ B = left { sum_ {i in I ( bar {x})} mu_i nabla g_i ( bar {x}): sum_ {i in I ( bar {x}) } mu_i = 1, mu_i ge0, i in I ( bar {x}) right } $$

are closed sets? even more, is B compact?

## real analysis – If $ f_ {11} geq 0 $, $ f_ {22} geq 0 $, $ f_ {11} f_ {22} geq f_ {12} ^ 2 $ then f is convex.

Yes $ f_ {11} geq 0 $, $ f_ {22} geq 0 $, $ f_ {11} f_ {22} geq f_ {12} ^ 2 $ then f is convex.

Let $ (x, y), (x & # 39 ;, y & # 39;) $ and $ t in (0,1) $. We define $ g (t) = f (t (x & # 39 ;, y & # 39;) + (1-t) (x, y)) $, if $ g (t) $ is then convex $ f $ is convex but I don't understand, could you give this reason?

## $ nabla f (x ^ *) ^ T (x-x ^ *) ge0 quad forall x $ implies a global minimum at $ x ^ * $ for a convex function on a convex set.

I want to show that for a convex, 1-smooth function $ f $ on a non-empty convex set $ X $ he maintains that: $ x ^ * $ is an overall minimum of $ f $ sure $ X $ if and only if for everyone $ x in X $: $ nabla f (x ^ *) ^ T (x-x ^ *) ge0 $.

i got $ implied by $: As $ f $ is convex on a convex set $ X $, for everyone $ x_1, x_2 in X $ we know that

$$ f (x_2) ge f (x_1) + (x_2-x_1) ^ T nabla f (x_1) $$

so obviously if $ (x_2-x_1) ^ T nabla f (x_1) ge 0 $ then $ x_1 le x_2 $.

I can not $ implies $ however, I don't think it stems from the same equation. Can anyone make a suggestion?

## mg.metric geometry – Cut a given triangle into 2 mutually congruent convex pieces which together cover most of the triangle

Two planar regions are congruent if one can coincide perfectly with the other by translation, rotation or reflection (flipping).

**The problem:** Given a triangular region T, how are we going to divide it into 2 mutually congruent convex pieces that together cover the highest fraction of the area of T? Let's call such a partition, the best 2-congruent partition of T.

And what is the specific shape of the triangle T so that the largest fraction of its area remains ("wasted") under its best partition with 2 congruents?

**Note:** We can generalize from T and ask for a general convex region C which “wastes” the largest fraction of its area when it receives its best congruent partition 2.

An attempt on this issue has been published here: https: //arxiv.org/pdf/1012.3106.pdf.

Please note that the document only deals with partitions in convex pieces. I am not aware of conclusive results on this issue.