measure theory - Finding the norm of the operator $M_g : L_p to L_p$ where $g in L_{infty}$

Suppose $(X,\,\cal{A},\,\mu)$ is a measure space where $\mu$ is a finite measure, and $p \in (1,\,\infty)$. Say that we take some $g \in L_{\infty}$, and we define $M_g : L_p \to L_p$ by $$M_g (f) = fg.$$ Then I'd like to show that $||M_g|| = ||g||_{\infty}$. I'm not absolutely sure this is true, but I think it is. I've shown that $||M_g|| \leq ||g||_{\infty}$, which is pretty simple, but now I'd like to find a function $f\in L_p$ such that $$\frac{||M_g (f)||_p}{||f||_p} = ||g||_{\infty}.$$ Here's what I've got so far:

Let $M=||g||_{\infty}$, the essential supremum of $|g|$. Then let $N$ be the subset of $X$ on which $|g(x)| < \frac{M}{2}$. Now $0<\mu (N)<\mu(X)<\infty$ and $0<\mu(X\setminus N)<\infty$ too. These things follow from the definition of $M$ and the fact that $\mu$ is a finite measure. So define $f:X \to \mathbb{R}$ as
$$f(x) = \begin{cases} 0 & x\in N \\ \frac{M}{|g(x)|(\mu(X \setminus N))^{\frac{1}{p}}} & x \in X \setminus N \end{cases} .$$

Then $$\begin{align} ||M_g (f)||_p ^p &= \int_{X\setminus N} \frac{M^p}{\mu(X \setminus N)} d\mu\\ &= M^p ,\,\text{ so} \\ ||M_g (f)||_p &= M. \end{align}$$

And we can easily show that $f \in L_p$ because $\frac{M}{|g(x)|}$ is not going to blow up when $|g(x)| \geq \frac{M}{2}$.

However, we don't have $||f||_p = 1$, so we only have $||M_g (f)||_p = M$, not $\frac{||M_g (f)||_p}{||f||_p} = M$ as we require.

So the question is, can this example be tweaked easily to get a unit-normed $f$ satisfying $||M_g (f)||_p = M$, or is the whole thing off track?

Answer

With some small modifications, your proof can be converted into a working proof.

Instead of using $|g(x)|<\frac M2$, we use
$|g(x)|<\frac Ma$, where $a>1$ is arbitrary.
Then, the rest of the proof can be repeated as you did.

We have to calculate $\|f\|_p$:
$$\begin{aligned} \|f\|_p^p & = \int_{X\setminus N} \frac{M^p}{|g(x)|^p \mu(X\setminus N)} \mathrm d\mu \\ & \leq \int_{X\setminus N} \frac{M^p}{(M/a)^p \mu(X\setminus N)} \mathrm d\mu \\ & \leq \mu(X\setminus N) \frac{M^p}{(M/a)^p \mu(X\setminus N)} \\ & = a^p \end{aligned}$$

So $\|f\|_p\leq a$.

Then we have
$$\frac{\|M_g(f)\|_p}{\|f\|_p} = \frac{M}{\|f\|_p} \geq \frac{M}{a}.$$

Because $a$ can be arbitrarily close to $1$, this completes the proof.

Note that because of the $\sup$ in the definition of the operator norm, you do not need to show that

$\frac{\|M_g(f)\|_p}{\|f\|_p} = M$,
only that you can find functions $f$ such that
$\frac{\|M_g(f)\|_p}{\|f\|_p}$
is arbitrarily close to $M$.