Proof Position Operator Is Dense

Question

This is an exercise from my last homework sheet, proofing that $P$ is unbounded and self-adjoint was clear, however I'm having trouble proofing that $P$ is densely defined.

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How my Instructor solved the problem:

Suppose we have the function $f_n(x)=\begin{cases} 1 & x\in \,[-n,n] \\ 0 & x\notin \,[-n,n] \end{cases}$

It is easy to show that $f_n(x)\in D_P$

If we take an arbitrary function $f(x)\in D_P$ we can show for all $n\in \mathbb{N}$ that:

1. $f(x)f_n(x)\in D_P$

2. $\lim_{n\to\infty}(f(x), f_n(x))=f(x)$

I can proof the individual steps, but I don't understand how this has to do anything with $D_P$ being dense?

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How I tried to approach the problem:

We know that $P$ is densely defined if $(D_P)^\perp=\{0\}$

If we consider an arbitrary function $f(x) \in D_P$, we know that $\langle f(x), 0 \rangle=0$.

But is this the only function that is orthogonal to $D_P$?

For that I'm setting $f(x)=e^{-x^2}$, which is inside of $D_P$ since the Gaussian Integral converges.

We can see that the only function $g(x)$ that fullfills following equation is $0$, since there is no other $g(x)$ that would make the integralkernel equal to $0$ (I'm using the fact that the sgn$(x)$ function is not in $D_P$):

$0\stackrel{!}{=}\langle e^{-x^2}, g(x) \rangle = \int_{-\infty}^{\infty} e^{-x^2} g(x) dx$

Hence $0$ is the only function, orthogonal to every $f(x)\in D_P$.

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I'm not really satisfied with either one of these solutions, and I hope someone could explain what the idea behind the first solution was, if my solution is correct, and if there are alternative\easier solutions. Thanks in advance.

Answer 1

The properties $$f(x)f_n(x)\in D_P\quad \lim_{n\to\infty}\|f(x) f_n(x)-f(x)\|_2=0$$ hold for any $f\in L^2(\Bbb{R})$, not just for $f\in D_P$. Therefore, any $f\in L^2(\Bbb{R})$ is a limit of a sequence of functions in $D_P$, so $D_P$ is dense.

Answer 2

Yes, neither of the solutions is clear.

Let $a>0$ and $f(x)=1$ for $0<x<a$ and $0$ otherwise. Then $f \in L^{2}$ and $\hat f(t)=\frac {e^{iat}-1} {it}$ for $t \neq 0$, $\hat f(0)=a$.

Spilt $\int |\hat f(t)|^{2}dt $ into integral over $|t|<1$ and integral over $|t|\ge 1$ to see that $\hat f \in L^{2}$.

[ Use :$|e^{iat}-1|\leq 2$ in the second part].

Thus, $f \in \mathcal D_P$.

If $g \in \mathcal (D_P)^{\perp}$ it follows that $\int_0^{a} g(x)dx=0$. This is true for every $a>0$ so $g=0$ a.e. on $(0,\infty)$. Use a similar argument for $(-\infty, 0)$.