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.