Multiplication operator on $L^1$

Question

Let $\phi :X \rightarrow \mathbb{C}$ be measurable with respect to the measure space $(X,\mu)$. Suppose that $\phi f \in L^1(\mu)$ whenever $f \in L^1(\mu)$. Define $M_{\phi}(f)=\phi f$, for $f \in L^1(\mu)$.

Show that $M_{\phi}$ is continuous, that $\phi \in L^{\infty}(\mu)$, and that $\|M_{\phi}\|\leq \|\phi\|_{\infty}$.

I have proved the first part using the closed graph theorem, and if we have the second, the third question is obvious. My initial thought for the second question was to consider the functional $f \rightarrow \int \phi fd\mu$ and use the Riesz representation theorem for the dual of $L^1$ together with uniqeness. Though the measure space is not $\sigma$-finite so the representation doesn't hold. Any help?

Answer 1

This is false. Consider $$\mu = \sum_{n=-\infty}^\infty \infty \delta_n + \lambda$$ where $\infty \delta_n$ is the measure that has infinite point mass at $n$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$. Let $\phi(n) = n,$ for $n\in \mathbb{Z}$ and $\phi(x)=1$ for $x$ not an integer. Then $f \in L^1(\mu)$ implies $f(n) = 0$ for all $n\in \mathbb{Z}$, so $\phi f = f \in L^1(\mu)$. However $\mu(|\phi|> t) = \infty$ for all $t$ so $||\phi||_\infty = \infty$.

Answer 2

Edit Here is a proof that works if $\mu$ is semifinite.

Suppose $\phi \notin L_\infty(X, \mu).$ Then given any $R>0$ there is a set of positive measure $E \subset X$ such that $\vert \phi \vert > R$ on $E.$ Let $f \in L_1(X,\mu)$ such that $f$ vanishes outside of $E$. Taking absolute value, we can assume $f$ is real and non-negative. Then

$$\Vert Mf \Vert_1 = \int_E \vert \phi \vert \vert f \vert d\mu \geq R \Vert f \Vert_1.$$

This shows

$$\Vert M \Vert \Vert f \Vert_1 \geq R \Vert f \Vert_1 $$

which implies $\Vert M \Vert = \infty$, a contradiction.

Answer 3

For convenience, scale $\phi$ so that $\|M_{\phi}\|_{\mathcal{L}(X)}=1$. For $\delta >0$, the $\chi_{\delta}$ be the characteristic function of the set where $|\phi| \ge 1+\delta$. Then, for any $f \in L^{1}$, $$ (1+\delta)\int |f\chi_{\delta}|d\mu \le \int |f\chi_{\delta}| |\phi|d\mu \le \int |f\chi_{\delta}|d\mu. $$ Thus $\|f\chi_{\delta}\|=0$ for all $\delta > 0$. That means that either (a) $E=\{ x : |\phi(x)| > 1 \}$ is of measure $0$ or (b) $E$ has infinite measure and contains no subset of finite measure.

Ruling out $(b)$ requires some assumption on the measure space. For example, if you allow an atom to have infinite measure, then every $f \in L^{1}$ vanishes on that atom and, yet, $\phi$ may be $2$ on that atom, but $\|M_{\phi}\| \le 1$ can still occur because every $f \in L^{1}$ vanishes on that point. If $\mu$ is a finite measure or a sigma-finite measure, then you'll be okay.