operator norm of multiplication operator on $L^2$

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

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

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

${\parallel M_{\varphi }\psi \parallel }_{L^2}$ $=$ $\sqrt{{\displaystyle \int \psi {(x)}^2\varphi {(x)}^2𝑑\mu (x)}}$

$\le$ $\sqrt{(\mathrm{ess}sup{\varphi }^2){\displaystyle \int \psi {(x)}^2𝑑\mu (x)}}$

$=$ $(\mathrm{ess}sup|\varphi |){\parallel \psi \parallel }_{L^2}$

and, hence

$${\parallel M_{\varphi }\parallel }_{\mathrm{op}}=sup\frac{{\parallel M_{\varphi }\psi \parallel }_{L^2}}{{\parallel \psi \parallel }_{L^2}}\le (\mathrm{ess}sup|\varphi |).$$

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

$$A=\{x\mid |\varphi (x)|\ge \mathrm{ess}sup|\varphi |-ϵ\}$$

is greater than zero. If , 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

${\parallel M_{\varphi }{\chi }_B\parallel }_{L^2}$ $=$ $\sqrt{{\displaystyle \int \varphi {(x)}^2{\chi }_B{(x)}^2𝑑\mu (x)}}$

$=$ $\sqrt{{\displaystyle {\int }_B}\varphi {(x)}^2𝑑\mu (x)}$

$\ge$ $\mu (B)(\mathrm{ess}sup|\varphi |-ϵ)$

and, hence

$${\parallel M_{\varphi }\parallel }_{\mathrm{op}}=sup\frac{{\parallel M_{\varphi }\psi \parallel }_{L^2}}{{\parallel \psi \parallel }_{L^2}}\ge \frac{{\parallel M_{\varphi }{\chi }_B\parallel }_{L^2}}{{\parallel {\chi }_B\parallel }_{L^2}}=\mathrm{ess}sup|\varphi |-ϵ.$$

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

$${\parallel M_{\varphi }\parallel }_{\mathrm{op}}\ge \mathrm{ess}sup|\varphi |.$$

Combining with the inequality in the opposite direction,

$${\parallel M_{\varphi }\parallel }_{\mathrm{op}}=\mathrm{ess}sup|\varphi |.$$

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

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

$$A=\{x\mid N+1\ge |\varphi (x)|\ge N\}.$$

If , 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

${\parallel M_{\varphi }{\chi }_B\parallel }_{L^2}$ $=$ $\sqrt{{\displaystyle \int \varphi {(x)}^2{\chi }_B{(x)}^2𝑑\mu (x)}}$

$=$ $\sqrt{{\displaystyle {\int }_B}\varphi {(x)}^2𝑑\mu (x)}$

$\ge$ $\mu (B)N$

and, hence

$${\parallel M_{\varphi }\parallel }_{\mathrm{op}}=sup\frac{{\parallel M_{\varphi }\psi \parallel }_{L^2}}{{\parallel \psi \parallel }_{L^2}}\ge \frac{{\parallel M_{\varphi }{\chi }_B\parallel }_{L^2}}{{\parallel {\chi }_B\parallel }_{L^2}}=N\ge R.$$

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 $L^2$
Canonical name	OperatorNormOfMultiplicationOperatorOnL2
Date of creation	2013-04-06 22:14:23
Last modified on	2013-04-06 22:14:23
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	13
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 47B38