Exercise 8, § 1

Topological Vector Spaces, V, p. 61

Let $E$ be a real or complex normed space, having the following property: if $x, y, x', y'$ are four vectors of $E$ such that

\|x\| = \|x'\|,\quad \|y\| = \|y'\|,\quad \|x + y\| = \|x' + y'\|,

then we have $\|x - y\| = \|x' - y'\|$ . Prove that the norm on $E$ is prehilbertian (use Exercise 7).

Solution

No solution has been written yet.