# Norm of multiplication operator in $L_p$

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, $$a ∈ L_∞(Ω)$$.

Moreover,

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

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.

$$\Rightarrow$$

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.

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 $$\lvert \lvert M_{a} \rvert \rvert \leq \lvert \lvert a \rvert\rvert_{\infty}$$.

(EDIT)

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 $$\lvert \lvert M_{a} \rvert \rvert \geq \lvert \lvert a \rvert \rvert_{\infty}$$

(EDIT)

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

• Oh, I get it now. I looked at it from the wrong side. Implication to the right is correct? Commented May 11, 2020 at 15:08
• 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.
– Alex
Commented May 11, 2020 at 18:58
• 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? Commented May 11, 2020 at 19:39
• 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.
– Alex
Commented May 11, 2020 at 20:02