Prove that position operator is bounded

Question

I have to prove that the operator $$ \begin{array}{cccc} \hat{x}: & L_2(A) & \to & L_2(A) \\ & f(x) & \mapsto & xf(x) \end{array} $$ is bounded if $A=[a,b]$ and it's not bounded if $A=\mathbb{R}$. Then, calculate $||\hat{x}||$ when $A=[a,b]$.

I have tried the usual procedure, that is to find that $||\hat{x}f(x)||\leq M ||f(x)||$. If $A=[a,b]$, the left hand-side of the inequality is $$ ||\hat{x}f(x)|| = ||xf(x)|| = \int_a^b |xf(x)|^2\ dx\ . $$ I tried to use the property $$ \left|\int_a^b f(x)\ dx\right| \leq \int_a^b |f(x)|\ dx\ , $$ but I realized that this does not make any sense, probably. So I don't know how to start solving this.

A hint is given to solve the problem: Use the vectors $\chi_{[c,d]},\ c,d\in[a,b]$, where $$ \chi_{[c,d]}(x) \left\{\begin{array}{ll}1 & x\in[c,d]\\0 & x\notin[c,d]\end{array}\right.\ . $$ But I don't get how this would be useful. I would appreciate any kind of help about the problem itself or about use of $\chi$.

Answer 1

Well you have to say something about $x$. Note that

$$ \min\{a^2,b^2\}\int_a^b |f(x)|^2\textrm{d}x \leq \int_a^b |xf(x)|^2\textrm{d}x\leq \max\{a^2,b^2\}\int_a^b |f(x)|^2\textrm{d}x, $$ The right-hand inequality explicitly shows that your operator is bounded on $L^2(A)$ with norm bounded by $\max\{|a|,|b|\}$. You can get equality by considering $f=1_{[a,b]}$. Conversely, the left-hand inequality shows, by considering $f_n=1_{[n,n+1]}$ that the operator is unbounded on $L^2(\mathbb{R})$.

Answer 2

Let $M$ be the larger of $|a|$ and $|b|$, and notice that all points $x$ in the interval $[a,b]$ satisfy $|x|\leq M$. So $$ \int_a^b|xf(x)|^2\,dx\leq M^2\!\!\int_a^b|f(x)|^2\,dx, $$ which implies $\Vert\hat x\Vert\leq M$.

For the converse inequality, see what happens when $f$ is the $\chi_{[c,d]}$ suggested in the hint and the interval $[c.d]$ is very short and adjacent to one of the endpoints of $[a,b]$.