Why is the inner product of two square integrable function finite?

898 Views Asked by At

consider an infinite dimensional Hilbert space, e.g. the vector space of complex-valued functions with the inner product:

$\langle \psi, \phi \rangle = \int_{-\infty}^\infty \psi^{*}\phi dx \tag{1}$

Since this is a inner product, we need it to be finite. We apparently can guarantee that by forcing $\psi, \phi$ to be square-integrable since for a square integrable functon we know

$\int_{-\infty}^\infty |\psi |^2 dx < \infty \tag{3}$

Now, I don't really see how (3) leads to (1) being finite.

1

There are 1 best solutions below

1
On BEST ANSWER

By Holder's /Cauchy -Schwarz inequality $\int |\psi^{*} \phi |dx \leq \sqrt {\int |\phi|^{2}dx}\sqrt {\int |\psi|^{2}dx} <\infty$.