# Is this function in Hilbert space?

1. May 21, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

(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 $\frac{d}{dx}f(x)$?

2. Relevant equations

3. 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?

2. May 21, 2015

### 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)

3. May 21, 2015

### 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?

4. May 21, 2015

### 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...

5. May 21, 2015

### 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 $<\nu|\nu>$ ...