Question

Let (X,∥⋅∥ X) and (Y,∥⋅∥Y) be two normed spaces. In the lectures we proved the following: Let A:X→Y be a linear mapping. If A is bounded then A is continuous. Proof. Assume that A is bounded. Then for any x 0∈X and for any sequence (xn)n∈N⊂X such that x n→x0∈X, we find ∥Ax n−Ax0∥Y=∥A(xn−x0)∥Y<=(∗)M∥x n−x0∥X →0. This means that A is continuous at x 0. Explain the step labelled

Answer 1

The provided step uses the boundedness of the linear operator A to show that A is continuous. Specifically, it shows that the operator norm of the difference sequence is bounded by M times the norm of the difference sequence in X, which converges to zero, thus establishing continuity.

Step-by-step explanation:

The question pertains to the subject of linear algebra within Mathematics and asks for an explanation of a particular step in a proof that demonstrates the continuity of a bounded linear operator between normed spaces. Specifically, the proof establishes that for a bounded operator A from X to Y, and a sequence (xn) converging to x0 in X, the sequence (A(xn)) converges to A(x0) in Y.

To clarify the step marked with an asterisk (*), assume that A is bounded, which means there exists a constant M such that for all x in X, we have ||A(x)||Y ≤ M||x||X. Using this property, the inequality ||A(xn - x0)||Y ≤ M||xn - x0||X holds. Since xn converges to x0, we know ||xn - x0||X goes to zero, and thus ||A(xn) - A(x0)||Y also approaches zero, showing that A is continuous at x0.