# operator norm of multiplication operator on $L^{2}$

The operator norm of the multiplication operator $M_{\phi}$ is the essential supremum of the absolute value of $\phi$. (This may be expressed as $\|M_{\phi}\|_{\mathrm{op}}=\|\phi\|_{L^{\infty}}$.) In particular, if $\phi$ is essentially unbounded, the multiplication operator is unbounded.

For the time being, assume that $\phi$ is essentially bounded.

On the one hand, the operator norm is bounded by the essential supremum of the absolute value because, for any $\psi\in L^{2}$,

 $\displaystyle\|M_{\phi}\psi\|_{L^{2}}$ $\displaystyle=$ $\displaystyle\sqrt{\int\psi(x)^{2}\phi(x)^{2}\,d\mu(x)}$ $\displaystyle\leq$ $\displaystyle\sqrt{(\mathrm{ess}\sup\,\phi^{2})\int\psi(x)^{2}\,d\mu(x)}$ $\displaystyle=$ $\displaystyle(\mathrm{ess}\sup\,|\phi|)\,\|\psi\|_{L^{2}}$

and, hence

 $\|M_{\phi}\|_{\mathrm{op}}=\sup{\|M_{\phi}\psi\|_{L^{2}}\over\|\psi\|_{L^{2}}}% \leq(\mathrm{ess}\sup\,|\phi|).$

On the other hand, the operator norm bounds by the essential supremum of the absolute value . For any $\epsilon>0$, the measure of the set

 $A=\{x\mid\,|\phi(x)|\geq\mathrm{ess}\sup\,|\phi|-\epsilon\}$

is greater than zero. If $\mu(A)<\infty$, set $B=A$, otherwise let $B$ be a subset of $A$ whose measure is finite. Then, if $\chi_{B}$ is the characteristic function of $B$, we have

 $\displaystyle\|M_{\phi}\chi_{B}\|_{L^{2}}$ $\displaystyle=$ $\displaystyle\sqrt{\int\phi(x)^{2}\chi_{B}(x)^{2}\,d\mu(x)}$ $\displaystyle=$ $\displaystyle\sqrt{\int_{B}\phi(x)^{2}\,d\mu(x)}$ $\displaystyle\geq$ $\displaystyle\mu(B)(\mathrm{ess}\sup\,|\phi|-\epsilon)$

and, hence

 $\|M_{\phi}\|_{\mathrm{op}}=\sup{\|M_{\phi}\psi\|_{L^{2}}\over\|\psi\|_{L^{2}}}% \geq{\|M_{\phi}\chi_{B}\|_{L^{2}}\over\|\chi_{B}\|_{L^{2}}}=\mathrm{ess}\sup\,% |\phi|-\epsilon.$

Since this is true for every $\epsilon>0$, we must have

 $\|M_{\phi}\|_{\mathrm{op}}\geq\mathrm{ess}\sup\,|\phi|.$

Combining with the inequality in the opposite direction,

 $\|M_{\phi}\|_{\mathrm{op}}=\mathrm{ess}\sup\,|\phi|.$

It remains to consider the case where $|\phi|$ is essentially unbounded. This can be dealt with by a variation on the preceeding argument.

If $\phi$ is unbounded, then $\mu(\{x\mid\,|\phi(x)|\geq R\})>0$ for all $R>0$. Furthermore, for any $R>0$, we can find $N>R$ such that $\mu(A)>0$, where

 $A=\{x\mid N+1\geq|\phi(x)|\geq N\}.$

If $\mu(A)<\infty$, set $B=A$, otherwise let $B$ be a subset of $A$ whose measure is finite. Then, if $\chi_{B}$ is the characteristic function of $B$, we have

 $\displaystyle\|M_{\phi}\chi_{B}\|_{L^{2}}$ $\displaystyle=$ $\displaystyle\sqrt{\int\phi(x)^{2}\chi_{B}(x)^{2}\,d\mu(x)}$ $\displaystyle=$ $\displaystyle\sqrt{\int_{B}\phi(x)^{2}\,d\mu(x)}$ $\displaystyle\geq$ $\displaystyle\mu(B)N$

and, hence

 $\|M_{\phi}\|_{\mathrm{op}}=\sup{\|M_{\phi}\psi\|_{L^{2}}\over\|\psi\|_{L^{2}}}% \geq{\|M_{\phi}\chi_{B}\|_{L^{2}}\over\|\chi_{B}\|_{L^{2}}}=N\geq R.$

Since this is true for every $R$, we see that the operator norm is infinite, i.e. the operator is unbounded.

Title operator norm of multiplication operator on $L^{2}$ OperatorNormOfMultiplicationOperatorOnL2 2013-04-06 22:14:23 2013-04-06 22:14:23 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Theorem msc 47B38