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Let the position operator be $$ \begin{array}{cccc} \hat{x}: & L_2[a,b] & \to & L_2[a,b] \\ & f(x) & \mapsto & xf(x) \end{array}\ . $$

We can prove that this operator is bounded: $$ ||\hat{x}f(x)||_2 = ||xf(x)||_2 = \sqrt{\int_a^b |xf(x)|^2\ dx} = \sqrt{\int_a^b |x|^2|f(x)|^2\ dx}\ , $$ now define $M = \max\{|a|,|b|\}$ to bound that integral, and $$ ||\hat{x}f(x)||_2 \leq \sqrt{M^2 \int_a^b |f(x)|^2\ dx} = M ||f(x)||_2\ , $$ proving that $\hat{x}$ is a bounded operator.

However, this also makes a restriction on the norm of the operator, such that $$ ||\hat{x}|| \leq \max\{|a|,|b|\}\ . $$ But I don't know how to calculate the norm from here. I know that three expressions can be used, $$ ||\hat{x}|| = \sup_{||f(x)||\leq 1} ||\hat{x}f(x)|| = \sup_{||f(x)|| = 1} ||\hat{x}f(x)|| = \sup_{||f(x)||\neq 0} \dfrac{||\hat{x}f(x)||}{||f(x)||}\ , $$ so I tried finding a $g(x)$ such that $||g(x)|| = 1$, and use it as a lower bound: $$ ||\hat{x}|| = \sup_{||f(x)|| = 1} ||\hat{x}f(x)|| \geq ||\hat{x}g(x)||\ , $$ but I didn't have too much success on this.

I would appreciate any help on this problem.

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1 Answer 1

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Let $f_n=\sqrt{n}1_{[b-1/n,b]}$ and note that for $n\geq\frac{1}{b-a}$, $f_n\in L^2([a,b])$ and $\|f_n\|_{L^2}=1$. Furthermore, we have $$ \int_a^b x^2f_n(x)^2\textrm{d}x=n\int_{b-1/n}^b x^2\textrm{d}x=\frac{n}{3}(b^3-(b-1/n)^3)=\frac{n}{3}\left( \frac{1}{n^3}-3\frac{b}{n^2}+3\frac{b^2}{n}\right), $$ which converges to $b^2$ as $n\to\infty$. Accordingly, $\|\hat{x}\|\geq |b|.$ Similarly, we can get $\|\hat{x}\|\geq |a|$ and thus, you get that $\|\hat{x}\|=\max\{|a|,|b|\}$.

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  • $\begingroup$ Thanks! Sorry for thew question but, what do you mean by the number 1 of the first definition? A function which is identical to 1 on all the given interval? $\endgroup$
    – Jaime_mc2
    Jan 7, 2020 at 11:26
  • $\begingroup$ Yes, $1_A(x)$ is the function which is $1$ if $x\in A$ and $0$ else. It's called the indicator function or characteristic function of $A$. $\endgroup$ Jan 7, 2020 at 11:31
  • $\begingroup$ Ok, perfect. Thank you again! $\endgroup$
    – Jaime_mc2
    Jan 7, 2020 at 11:52

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