Recent content by gherardo

This only shows c) here below. Another argument, besides direct computation, is by standard properties of Fourier transform: a) Fourier transform of a gaussian is a gaussian b) Fourier transform of convolution product = pointwise product of Fourier transforms (up to choice of...

A few points: 1. You cannot extend an unbounded operator on an infinite separable Hilbert space. Unbounded is the same as non continuous at zero, and by linearity non continuous at 0 is the same as non continuous everywhere. 2. Hellinger Toeplitz theorem: everywhere defined symmetric ==>...