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(Someone may please change the title if they can think of a better one)

We have a Hilbert Space $\mathcal{H}$ that consists of all functions $\psi(x)$ such that

$\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $

Now I have to show that there are functions in $\mathcal{H}$ such that $Q\phi(x)=x\psi(x)$ is not in $\mathcal{H}$ . Does it suffice if I think of a counterexample.. but even then, the only two square integrables i can think of are the Gaussian and the dirac-delta "function". For both the cases, the corresponding $x\psi(x)$ is also L2. Could someone give me more examples of functions that are square integrable over the entire real axis?

Furthermore, if I now consider the function space $\Omega$ with the infinite set of conditions,

$\int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty$ for $n=0,1,2,\cdots$

I have to show now, that for any $\psi(x)$ in $\Omega$ the function $Q\phi(x)=x\psi(x)$ is also in $\Omega$. I can think of a weak argument like

$\int_{-\infty}^{\infty} |Q\psi(x)|^2 (1+|x|^n)dx = \int_{-\infty}^{\infty} |\psi(x)|^2 (x^2+|x|^{n+2})dx \approx \int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty $

The last step because we would be concerned only with the behaviour of $x$ when it is very large, as the function would have to go to zero at large $x$ for it to be in $\Omega$.

I am not so sure about this, and face a basic conceptual difficulty. Can we conclude if $\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $ then $\psi(x)$ should fall off as $1/x^2$?

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    @Jonas I don't know but I somehow felt I'm cheating the problem by defining my own piecewise function. But perhaps you are right.2011-03-08

2 Answers 2

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Hint: Think along lines of Cauchy distribution function. Say $\psi(x) = \frac{1}{\sqrt{1+x^2}}$

For the second part of your question, you could define your function space $\Omega$ to be functions such that $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx < \infty$, for $n \in \mathbb{N}$

(Your function space and the function space defined in the above line are the same. Since if $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx < \infty$, for $n \in \mathbb{N}$, then $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 dx < \infty$ by taking $n=0$ and adding the two gives your definition and also the function satisfying your definition also fits in the definition given in this answer since $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx \leq \displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 (1 + \left|x \right|^n) dx < \infty$)

The conclusion now is trivial since $2 \in \mathbb{N}$ and if $n \in \mathbb{N}$, then so does $n+2$

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    @Jonas: I tend to use $0 \in \mathbb{N}$2011-03-08
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Also I know I can construct piecewise functions, but is there something better than that?

I don't know what's wrong with piecewise functions, but how about $f(x)=1/(1+|x|)$?

As Akhil mentioned, the Dirac delta function is not in $L^2$. A densely defined linear functional on a space is typically a very different object from an element of the space.

To show that $Q(\Omega)\subseteq\Omega$, note that $x^2+|x|^{n+2}\leq 1+|x|^{n+3}$ outside of a bounded interval, and $x^2+|x|^{n+2}$ is bounded on bounded intervals.

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    @Approximist: Yeah, I guess better than my bounds above would be to just compute it, $\int_{-\infty}^\infty f^2(x)dx=2\int_0^\infty1/(1+x)^2dx=2(-1/(1+x))\vert_0^\infty=2$.2011-03-08