# Multiplication operator on $L^1$

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?

• Assume $\phi$ is not essentially bounded, i.e. for every $n\in\mathbb{N}$ there exists $E_n$ of positive measure where $|\phi(x)|\ge n$. Take $f_n$ as the $L_1$-normalized indicator function of $E_n$, so $\|M_\phi f_n\|_1\ge n\to +\infty$. Unbounded (contradiction).
– A.Γ.
Commented Jul 29, 2015 at 21:28
• You need the measure to be semifinite to guarantee that you can normalize. Or at least you need to be able to guarantee that $E_n$ has finite measure for infinitely many $n$. Commented Jul 29, 2015 at 21:36
• I tried such things but if we have atoms of infinite measure the argument gets complicated. I also thought to take cases on the measure space to drop the pathological cases but i was hoping to find a faster and more elegant solution. @ A.G. Commented Jul 29, 2015 at 21:50
• @PhilipHoskins Yeah, that's right.
– A.Γ.
Commented Jul 29, 2015 at 22:00

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 $||\phi||_\infty = \infty$.

• Nice and simple! Commented Jul 29, 2015 at 23:04

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.

• .Nice solution but how do we know that there exists such f that is not zero a.e.? What if there is an atom with infinite measure? Commented Jul 29, 2015 at 22:02
• Good point. I'm used to always assuming the measures are at least semifinite. Many functional analysis textbooks use that assumption as well. This proof will work in that case, but I think the statement is false if you drop that assumption. Commented Jul 29, 2015 at 22:58
• @user163644 Such an $f$ cannot always be chosen. See where his proof breaks with the $\mu$ given in my answer. Commented Jul 29, 2015 at 23:00

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.