Is this function in Hilbert space?

[spaghetti3451]

Homework Statement

(a) For what range of ##\nu## is the function ##f(x) = x^{\nu}## in Hilbert space, on the interval ##(0,1)##. Assume ##\nu## is real, but not necessarily positive.

(b) For the specific case ##\nu = \frac{1}{2}##, is ##f(x)## in Hilbert space? What about ##xf(x)##? What about ##\frac{d}{dx}f(x)##?

Homework Equations

The Attempt at a Solution

If the function ##f(x) = x^{\nu}## belongs in Hilbert space, then ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

(assuming that, by 'Hilbert space', the question means 'the vector space of square-integrable functions,' as that is the only Hilbert space of interest to physicists).

Therefore,

##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

## = \int_{0}^{1} x^{2 \nu} dx < \infty## since ##\nu## is real

## = \int_{0}^{1} e^{{\rm ln} x^{2 \nu}} dx < \infty##

## = \int_{0}^{1} e^{2 \nu {\rm ln} x} dx < \infty##

## = [ \frac{2 \nu}{x} e^{2 \nu {\rm ln} x} ]_{0}^{1} < \infty##

## = [ 2 \nu e^{2 \nu {\rm ln} 1} - \frac{2 \nu}{0} e^{2 \nu {\rm ln} 0} ] < \infty##

The only way for this the last sentence to be valid is for ##\nu## to be 0.

Am I on the right track?

 

[BvU]

Hi,
No, you derail when you do the integration. You can easily check by taking ##\nu = 1## .
There are easier ways to integrate ##x^{2\nu}## (if you differentiate it, you'll probably see it)

 

[spaghetti3451]

Thanks for pointing it out! Don't know why I made my life harder when it was already an easy problem.

So, here's my new attempt:

If the function ##f(x) = x^{\nu}## belongs in Hilbert space, then ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

(assuming that, by 'Hilbert space', the question means 'the vector space of square-integrable functions,' as that is the only Hilbert space of interest to physicists).

Therefore,

##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

## = \int_{0}^{1} x^{2 \nu} dx < \infty## since ##\nu## is real

## = [ \frac{x^{2 \nu + 1}}{2 \nu + 1} ]_{0}^{1} < \infty##

## = \frac{1}{2 \nu + 1} < \infty##

Therefore, ##2 \nu + 1 \neq 0##, so that ##\nu \neq - \frac{1}{2}##.

Therefore, the function ##f(x) = x^{\nu}## is in Hilbert space, on the interval ##(0,1)##, for all real values of ##\nu## except ##-\frac{1}{2}##.

P.S.: I'd assume that for the function ##f(x) = x^{\nu}## to ##\textit{really}## be in Hilbert space (on the interval ##(0,1)##), we must also show that there is an inner product (that converges to a complex number) defined on the Hilbert space, and that the Hilbert space is complete. I have no idea, though, as to how I can form an inner product with only the one function ##f(x) = x^{\nu}##.

(b) ##\int_{0}^{1} |x^{\frac{1}{2}}|^{2} dx##

## = \int_{0}^{1} x dx##

## = [ \frac{x^{2}}{2} ]_{0}^{1}##

## = \frac{1}{2}##

##< \infty##

So, for the specific case ##\nu = \frac{1}{2}##, ##f(x)## in Hilbert space.Am I on the right track?

 

[BvU]

On the right track allright, but overshooting the length of the rails a little. Look again at your primitive. Can you carry out the integration for e.g. ##\nu = -2## ?

And (b) is right, so far, but there are two more things asked there...

 

[BvU]

Re inner product: you can try looking at your integral as the inner product ##f(x) \cdot f(x)##...

And in a fancy notation it would be ##<\nu|\nu>## ...