Let $Ω$ be an open set in $R^n$, and let $a$ be a measurable complex-valued function on $Ω$. The (maximal) operator of multiplication by $a$ on $L_p(Ω) \ (1 ≤ p < ∞)$ is defined by

$M_au = au, u ∈ dom(M_a)$

$dom(M_a) = \{u ∈ Lp(Ω):\ au ∈ L_p(Ω)\}.$

The multiplication operator $M_a$ defined on the whole space $L_p(Ω)$ if and only if $a$ is essential bounded, that is, $a ∈ L_∞(Ω)$.


$||M_au||_p ≤ ||a||_∞||u||_p,\ u ∈ Lp(Ω)$

Therefore, $M_a$ is a bounded operator on $L_p(Ω)$. In fact, we have $||M_a|| =||a||_∞$.

I saw a lot of similar posts, but still I have questions.

I think there should be also information about our measure (should be finite or $\sigma$- finite).

I found the inequality. $||M_a|| \le ||a||_\infty$ but I don't know how to show that.

My calculation:

$$||M_af||_p=||af||_p \le ||a||_\infty (\mu (\Omega))^{1/p}||f||_p$$

There I don't know why this should imply the above inequality.

I know how to show $\ge$.

After that I will get implication to the left.


Suppose $a$ isn't essential bounded. $\Omega = \bigcup_{n=1}^\infty A_n$

$B_m=\{x: \ |a(x)|>m\}$

$f_{n,m}:= \chi _{A_n \cap B_m}$ and $\exists n$ such that $X=A_n \cap B_m$ has positive measure.

$$||M_af_{n,m}||_{p}^p=||af_{n,m}||_p^p=\int_X |a(x)|^pdx>\int_X m^pdx=m^p \mu(X)\xrightarrow{m \to \infty}\infty$$ and this is contradiction, becouse $M_a$ is bounded.


1 Answer 1


If $\Omega \subset \mathbb{R}$ the space $L^{p}(\Omega)$ is assumed to have the standard Lebesgue measure. Also we have

$$ \lvert \lvert M_{a}f \rvert \rvert^{p} = \int_{\Omega} \lvert af \rvert^{p} dx \leq \lvert \lvert a \rvert \rvert_{L^{\infty}}^{p} \int_{\Omega} \lvert f \rvert^{p} dx = \lvert \lvert a \rvert \rvert_{L^{\infty}}^{p} \lvert \lvert f \rvert \rvert_{L^{p}}^{p}. $$

Which proves the inequality $\lvert \lvert M_{a} \rvert \rvert \leq \lvert \lvert a \rvert\rvert_{\infty}$.


The second part of your proof does not show $\lvert \lvert M_{a} \rvert \rvert \geq \lvert \lvert a \rvert \rvert_{\infty}$. You have instead shown that if $a$ is not essentially bounded then $M_{a}$ is not bounded.

I will denote the Lebesgue measure by $m$. For the right implication it is enough to construct a sequence $(f_{n}) \subset L^{2}$, with $\lvert \lvert f_{n} \rvert \rvert = 1$, such that $\lvert \lvert M_{a}f_{n} \rvert \rvert \geq \lvert \lvert a \rvert \rvert_{\infty} - \frac{1}{n}$. To this end note that for each $n$ the set $$ E_{n} = \left\{ x \in \Omega : \lvert a(x) \rvert > \lvert \lvert a \rvert \rvert_{\infty} - \frac{1}{n} \right\} $$ has positive Lebesgue measure, $m(E_{n}) > 0$. We can also assume the measure is finite (otherwise intersect it with a set of finite measure, such that the intersection has positive measure). Now set $f_{n} = (\frac{1}{m(E_{n})})^{1/p}\chi_{E_{n}}$, where $\chi$ is the characteristic function. We have $\lvert \lvert f_{n} \rvert \rvert_{L^{p}} = 1$, for all $n$, also

$$ \lvert \lvert M_{a}f_{n} \rvert \rvert_{p} = \left(\int_{\Omega} \lvert af_{n} \rvert^{p} dx\right)^{1/p} > \left( \lvert \lvert a \rvert \rvert_{\infty} - \frac{1}{n} \right)\lvert \lvert f_{n} \rvert \rvert_{L^{p}} = \lvert \lvert a \rvert \rvert_{\infty} - \frac{1}{n} $$

Thus we have $\lvert \lvert M_{a} \rvert \rvert \geq \lvert \lvert a \rvert \rvert_{\infty}$


As you mention we have similar results for an arbitrary sigma finite measure space.

  • $\begingroup$ Oh, I get it now. I looked at it from the wrong side. Implication to the right is correct? $\endgroup$
    – KrzysztofB
    Commented May 11, 2020 at 15:08
  • $\begingroup$ No. You started by assuming a was not essentially bounded, however, in this case the operator $M_{a}$ is not bounded (so talking about its norm makes no sense). I've added the missing argument. $\endgroup$
    – Alex
    Commented May 11, 2020 at 18:58
  • $\begingroup$ Now I'm confused. You showed that if $a$ is ess. bounded then $M_a$ is defined in whole space, yes? I think that I showed that if $M_a$ is defined then $a$ is ess. bounded. If not then what about the implications in the right? $\endgroup$
    – KrzysztofB
    Commented May 11, 2020 at 19:39
  • $\begingroup$ Ah, I am sorry. I thought your question was how to prove $\lvert \lvert M_{a} \rvert \rvert \geq \lvert \lvert a \rvert \rvert_{\infty}$. Yes, your right implication proof is correct. $\endgroup$
    – Alex
    Commented May 11, 2020 at 20:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .