About multiplication operator on $L^p(X, \mu)$ where X is sigma finite Let X be sigma finite measure space and $\phi$ $\in$ $L^{\infty}(X, \Omega)$ and $M_\phi:L^p(X, \Omega)$ $\to$ $L^p(X, \Omega)$ multiplication operator then show that $\| M_\phi \|=\| \phi\|_{\infty}$.
My attempt:
I could prove that $\| M_\phi \| \leq \| \phi \|_ {\infty}$ 
And then to prove the reverse inequality, I tried constructing a sequence of functions $f_n$ in $L^p$ such that $\| \phi f_n \|_p$ converges to $\| \phi \|_ \infty$ in the field. If we get such a sequence then that will prove the result by property of sup. Can we construct such a sequence?
As X is sigma finite so $X=\cup_{i=1}^{\infty} X_i$ where $\mu(X_i) < \infty$
I tried defining $f_n(x)=1 $ if $x \in \cup_{i=1}^n X_i$ and $0$ otherwise.
But then the norm of $M_\phi(f_n)$ converges to $\int \phi^p$
Can i modify this? Or is there any other way?

You might first try the (special) case with vectors, i.e., equip $\mathbb{R}^n$ with the $\ell^p$-norm, i.e., $\|x\|_p^p = \sum_{i=1}^n |x_i|^p$. How do you have to choose $x$ to obtain
$$\| x \, \phi \|_p = \|x\|_p \, \|\phi\|_\infty\qquad?$$
