I am trying to see the implication that if $g: [a,b] \to \mathbb{R}$ satisfies $fg \in L_{2}[a,b]$ for all $f \in L_{2}[a,b]$ then $\lVert g \rVert_{\infty} < \infty$.
My attempt for this is to consider the function $f \equiv 1\in L_{2}[a,b]$ and see that $fg = g \in L_{2}[a,b]$. Then, any $g^{n} \in L_{2}[a,b]$ for all $n$. But no other idea emerge to show that $\lVert g\rVert_{\infty} < \infty$. Any help would be appreciated.
Diego.
By the assumption, the operator $M_g:f\mapsto fg$ is everywhere defined. It is also closed, so it is bounded by the closed graph theorem.
Then, letting $\{ x\in [a,b]:|g(x)|\geq n, n\in \mathbb{N}\}=:\Omega_n$ we have $$ n^2\mu(\Omega_n)\leq \int_a^b\mathbb{1}_{\Omega_n}(x)|g(x)|^2d\mu(x)\leq \|M_g\|^2\mu(\Omega_n) $$ forcing $\mu(\Omega_n)=0$ for large enough $n\in \mathbb{N}$ (otherwise, we may divide through by $\mu(\Omega_n)$ and find a contradiction). Thus, $g\in L^\infty$.
edit: to see $M_g$ is closed, let $f_n\to f$ and $gf_n\to h$ in $L^2[a,b]$. Then, along a subsequence $n'$, $f_n'(x)\to f(x)$ for a.e. $x$. Thus, $g(x)f_n'(x)\to g(x)f(x)$ a.e., and $h=gf$.
A more straightforward way to prove without invoking anything in functional analysis: Prove by contradiction. Suppose that contrary that $||g||_{\infty}=\infty$. For each $n\in\mathbb{N}$, let $A_{n}=\{x\in[a,b]\mid|g(x)|>n\}$. Note that $\lambda(A_{n})>0$ (otherwise, $n$ is an essential upper bound of $g$). It is clear that $A_{1}\supseteq A_{2}\supseteq\ldots$ and $\cap_{n}A_{n}=\{x\mid|g(x)|=\infty\}=\emptyset$. By continuity of measure, $\lim_{n}\lambda(A_{n})=\lambda(\cap_{n}A_{n})=0$. Define $B_{n}=A_{n}\setminus A_{n+1}=\{x\mid|g(x)|\in(n,n+1]\}$. Note that $\lambda(B_{n})>0$ for infintely many $n$. (For, if there exists $N$ such that $\lambda(B_{n})=0$ whenever $n\geq N$, then $\lambda(A_{N})=\lambda(A_{N+1})=\lambda(A_{N+2})=\ldots$. It follows that $\lim_{n}\lambda(A_{n})=\lambda(A_{N})>0$, which is a contradiction.). Also note that $B_{1},B_{2},\ldots$ are pairwisely disjoint.
Choose a sequence of integers $(n_{k})$ such that $1\leq n_{1}<n_{2}\leq\ldots$ and $\lambda(B_{n_{k}})>0$ for each $k$. Define $f=\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{\sqrt{\lambda(B_{n_k})}}1_{B_{n_{k}}}.$ Observe that $\int f^{2}d\lambda=\sum_{k=1}^{\infty}\frac{1}{k^{2}}<\infty$. Therefore $f\in L^{2}[a,b]$. By assumption, $fg\in L^{2}$. However, by direct calculation, \begin{eqnarray*} \int|fg|^{2}d\lambda & = & \sum_{k=1}^{\infty}\int_{B_{n_{k}}}\frac{1}{k^{2}\lambda(B_{n_{k}})}g^{2}(x)d\lambda(x)\\ & \geq & \sum_{k=1}^{\infty}\int_{B_{n_{k}}}\frac{1}{k^{2}\lambda(B_{n_{k}})}\cdot(n_{k})^{2}d\lambda(x)\\ & \geq & \sum_{k=1}^{\infty}\int_{B_{n_{k}}}\frac{d\lambda(x)}{\lambda(B_{n_{k}})}\\ & = & \sum_{k=1}^{\infty}1\\ & = & \infty, \end{eqnarray*} which is a contradiction.
