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 $\psi \in L^2$ , \begin{eqnarray*} \| M_\phi \psi \|_{L^2} &=& \sqrt{\int \psi(x)^2 \phi(x)^2 \,d\mu(x)} \\ &\le& \sqrt{ (\mathrm{ess } \sup\, \phi^2) \int \psi(x)^2 \,d\mu(x)} \\ &=& (\mathrm{ess }\sup\, |\phi|)\,\| \psi\|_{L^2} \\ \end{eqnarray*}and, hence $$ \| M_\phi \|_\mathrm{op} = \sup {\| M_\phi \psi \|_{L^2} \over \| \psi \|_{L^2}} \le (\mathrm{ess }\sup\, \phi|) $$

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 $$ A = \{ x \mid\,|\phi(x)| \ge \mathrm{ess }\sup\, \phi - \epsilon \} $$ 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 \begin{eqnarray*} \| M_\phi \chi_B \|_{L^2} &=& \sqrt{\int \phi(x)^2 \chi_B(x)^2 \,d\mu(x)} \\ &=& \sqrt{\int_B \phi(x)^2 \,d\mu(x)} \\ &\ge& \mu(B) (\mathrm{ess }\sup\, \phi - \epsilon)\\ \end{eqnarray*}and, hence $$ \| M_\phi \|_\mathrm{op} = \sup {\| M_\phi \psi \|_{L^2} \over \| \psi \|_{L^2}} \ge {\| M_\phi \chi_B \|_{L^2} \over \| \chi_B \|_{L^2}} = \mathrm{ess }\sup\, \phi - \epsilon $$ Since this is true for every $\epsilon > 0$ , we must have $$ \| M_\phi \|_\mathrm{op} \ge \mathrm{ess }\sup\, \phi $$ 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 $\mu (\{ x \mid\,\phi (x)| \ge R\}) > 0$ for all $R>0$ . Furthermore, for any $R>0$ , we can find $N>R$ such that $\mu(A) > 0$ , where $$ A = \{ x \mid N+1 \ge |\phi (x)| \ge N\} $$ 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 \begin{eqnarray*} \| M_\phi \chi_B \|_{L^2} &=& \sqrt{\int \phi(x)^2 \chi_B(x)^2 \,d\mu(x)} \\ &=& \sqrt{\int_B \phi(x)^2 \,d\mu(x)} \\ &\ge& \mu(B) N\\ \end{eqnarray*}and, hence $$ \| M_\phi \|_\mathrm{op} = \sup {\| M_\phi \psi \|_{L^2} \over \| \psi \|_{L^2}} \ge {\| M_\phi \chi_B \|_{L^2} \over \| \chi_B \|_{L^2}} = N \ge R $$ Since this is true for every $R$ , we see that the operator norm is infinite, i.e. the operator is unbounded.