Let X be a normed space with finite dimension and T: X --> IR^n linear.Show that T is continuous.?

Answer 1

HOMEWORK!

Outline:By considering components, it is enought to assume that n=1. Since X is finite dimensional, we may take a basis. If p is the number of things in the basis, the closed graph theorem will show that X is continuously isomorphic to R^p. Since T is determined by the values on the basis, we have an explicit form for T which can be shown to be continuous on R^p and hence on X.

Answer 2

enable the oblong container be placed with 3 of its adjoining edges alongside x-, y- and z-axis and one vertex on the inspiration. The coordinates of four vertices are (0,0,0), (a,0,0), (0,b,0) and (0,0,c) enable the distances of a aspect P(p,q,r) be a million, 2, 3 and four from those vertices. Then, p^2 + q^2 + r^2 = a million (p-a)^2 + q^2 + r^2 = 4 p^2 + (q-b)^2 + r^2 = 9 and p^2 + q^2 + (r-c)^2 = 16 => (p-a)^2 - p^2 = 3 (q-b)^2 - q^2 = 8 and (r-c)^2 - r^2 = 15 including, (p-a)^2 + (q-b)^2 + (r-c)^2 - (p2 + q^2 + r^2) = 26 => (p-a)^2 + (q-b)^2 + (r-c)^2 - a million = 26 => (p-a)^2 + (q-b)^2 + (r-c)^2 = 27 => distance between P(p,q,r) and the vertex (a,b,c) = ?(27).