Proof: Bounded and Unbounded Operators on Lp Space

Let $\phi:\mathbb{R}\to\mathbb{C}$ be measurable function and $1\leq p\leq \infty$.

• If $\phi\in L^\infty(\mathbb{R})$, then the operator $M_\phi f=f\phi$ for $f\in L^p(\mathbb{R})$ is a bounded operator from $L^p(\mathbb{R})$ to $L^p(\mathbb{R})$, and its operator norm is precisely $\Vert M_\phi\Vert=\Vert\phi\Vert_\infty$.
• Conversely, if $f\phi\in L^p(\mathbb{R})$ for every $f\in L^p(\mathbb{R})$, then $\phi\in L^\infty(\mathbb{R})$.

Proof: Since $\phi\in L^\infty(\mathbb{R})$ and there is some fixed number $K=\Vert\phi\Vert_\infty$, i.e. $\vert\phi\vert\leq K$ a.e., it follows that

$$\Vert M_\phi f\Vert =\Vert f\phi\Vert\leq K\Vert f\Vert=\Vert\phi\Vert_\infty\Vert f\Vert$$

We can get the strict equality because

$$\Vert M_\phi\Vert=\sup_{f\in L^\infty(\mathbb{R})}{\Vert M_\phi f\Vert\over \Vert f\Vert}$$

and for any $\varepsilon>0$, there exists a set $E$ with positive measure, s.t. $\vert\phi\vert>K-\varepsilon$. Let $f=\mathbb{I}_{E\cap[-l,l]}\in L^p(\mathbb{R})$ and we have

$$\Vert M_\phi f\Vert =\Vert f\phi\Vert\geq\Vert(K-\varepsilon)f\Vert=(K-\varepsilon)\Vert f\Vert$$

Therefore, we conclude $\Vert M_\phi\Vert=\Vert\phi\Vert_\infty$.

If $p=\infty$ and for any $f\in L^\infty(\mathbb{R})$ we have $f\phi\in L^\infty(\mathbb{R})$, we claim that $\phi$ must be essentially bounded. Otherwise, $f=\mathbb{I}_\mathbb{R}\in L^\infty(\mathbb{R})$ will give $f\phi=\phi\notin L^\infty(\mathbb{R})$.

If $p\neq\infty$ and $\phi\notin L^\infty(\mathbb{R})$, there are infinitely many $E_k=\lbrace k\leq|\phi|<k+1\rbrace$ with positive measure, each measurable and disjoint. Choose any $E_{n_k}, k\in \mathbb{N}$, all with positive measure and take $E=\cup E_{n_k}$. Define

$$f(x)= \begin{cases} {1\over k^{2/p}\mu(E_{n_k})^{1/p}},\quad x\in E_{n_k} \\ 0,\quad x\notin E \end{cases}$$

Then

$$\int|f|^pd\mu=\sum_{k=1}^\infty\int_{E{n_k}}{1\over k^2\mu(E_{n_k})}=\sum_{k=1}^\infty {1\over k^2}<\infty$$

and

$$\int|f\phi|^pd\mu\geq\sum_{k=1}^\infty\int_{E{n_k}}{k^p\over k^2\mu(E_{n_k})}$$

As $p\geq1$, it is obvious that RHS gives

$$\int|f\phi|^pd\mu\geq\sum_{k=1}^\infty\int_{E{n_k}}{k^p\over k^2\mu(E_{n_k})}\geq \sum_{k=1}^\infty {1\over k}=\infty$$

Contradiction!

$$\tag*{$\Box$}$$