Question

Let V be an arbitrary vector space and A, B CV be two nonempty sets. Show that (a) conv(A+B): = conv(A) + conv(B); (b) conv(AUB) = {tx + (1 − t)y : x E A, y Є B, te [0,1]} (here A and B are assumed - convex sets or the statement is false). Remark: Here we use the symbol conv(X) to denote the convex hull of the set X, i.e. the smallest convex set in V containing

Answer 1

The question deals with the properties of convolution in vector spaces, observing that vector addition is commutative and distributive over convolution, leading to conv(A+B) being the same as conv(A) + conv(B), and that conv(AUD) is the set of points from all linear combinations of points from A and B.

Step-by-step explanation:

The question is related to the properties of vector spaces, specifically the concept of convolution of sets within these vector spaces. Convolution is a way to combine two sets to form a new set containing all possible linear combinations of elements from the original sets.

When dealing with vector addition, the commutative property states that the order of addition does not affect the result, A + B = B + A. This property also applies to more complex operations like the convolution of sets, where conv(A) + conv(B) is the same as conv(B) + conv(A). The distributive property allows for the breakup of addition over convolution, suggesting that conv(A+B) can indeed be distributed to form conv(A) + conv(B).

Similarly, for the convex hull of the union of two convex sets A and B, conv(A ∪U B) equals the set of all points that can be expressed as a linear combination of points from A and B with coefficients that add up to 1, meaning t times some x in A plus (1-t) times some y in B, for all t between 0 and 1.