(a) For what range of $\mathbf{v}$is the function $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}$in Hilbert space, on the interval $\left(0.1\right)$? Assume $\mathit{v}$is real, but not necessarily positive.

(b) For the specific case $\mathbf{v}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{2}$, is $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$in this Hilbert space? What about$\mathbf{xf}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$? How about $\mathbf{(}\mathbf{d}\mathbf{/}\mathbf{dx}\mathbf{)}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$?

Hilbert space:

The collection of all functions of $\mathrm{x}$constitutes a vector space, but for our purposes it is much too large. The wave function $\mathrm{\Psi }$ must be normalised to represent a hypothetical physical state:

$\int |\mathrm{\Psi }{|}^2\mathrm{dx}=1$

The set of all square integrable function, on a specified interval,

$\mathrm{f}\left(\mathrm{x}\right)\text{suchthat}{\int }_{\mathrm{a}}^{\mathrm{b}}\left|\mathrm{f}\right(\mathrm{x}){|}^2\mathrm{dx}<\mathrm{\infty }$

(a)

Given the function: Hilbert space is the vector space of all square-integrable functions.

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{\nu }}$

Apply the normalization we get:

Now, $⟨f\mid f⟩$ is finite only if $0^{2\nu +1}$ is finite, the value of$0^{2\nu +1}$ is zero, excepts if $2\nu +1$is less than zero, the function become infinite. Provided,$(2\mathrm{\nu }+1)>0$

The function$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{v}}$ in Hilbert space, thus, the range is,$\mathrm{\nu }>-\frac{1}{2}$.

b)

From part (a), for $\mathrm{v}=\frac{1}{2}>-\frac{1}{2}$, $\mathrm{f}\left(\mathrm{x}\right)$is in Hilbert space, and $\mathrm{xf}\left(\mathrm{x}\right)$is Hilbert space also, but $\mathrm{f}\text{'}\left(\mathrm{x}\right)$is not.

For $\mathrm{xf}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{v}+1}$, we have:

which is in Hilbert space for $\nu =\frac{1}{2}$. Now for ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)={\mathrm{vx}}^{\mathrm{\nu }-1}$, we have:

For $\nu =\frac{1}{2}$the lower limit of the integration gives usdata-custom-editor="chemistry" $0°$ which is infinite, therefore $f^{\text{'}}\left(x\right)$is not in the Hilbert space.