Riemann integrable functions in \(H^\infty\) (Q2721577)

Let \({\mathbb D}\) be the open unit disk in the complex plane and let \({\mathbb T}\) be the boundary of \({\mathbb D}\). \(C({\mathbb T})\) denotes the algebra of continuous functions on \({\mathbb T}\) and \(L^\infty\) denotes the algebra of essentially bounded measurable functions with respect to the normalized Lebesgue measure \(m\) on \({\mathbb T}\). \(A\) and \(H^\infty\) are the algebras of functions in \(C({\mathbb T})\) and \(L^\infty\) whose Fourier coefficients with negative indices vanish, respectively. The algebra \(A\) is called the disk algebra. NEWLINENEWLINENEWLINE\textit{D. Sarason} [Bull. Am. Math. Soc. 79, 286-299 (1973; Zbl 0257.46079)] considered the problem: When closed subalgebras of \(L^\infty\) contain \(C({\mathbb T})\). He gave an answer in terms of Riemann integrable functions. NEWLINENEWLINENEWLINEIn the present article the author considers Riemann integrable functions in \(H^\infty\) and reconsiders Sarason's results. NEWLINENEWLINENEWLINEFirst the author presents a direct proof of one of Sarason's results:NEWLINENEWLINENEWLINETheorem 1. Let \(B\) be the closed subalgebra of \(L^\infty\) containing \(A\), which contains at least one Riemann integrable function not belonging to \(H^\infty\). Then \(B\) contains \(C({\mathbb T})\). NEWLINENEWLINENEWLINEThen the author obtains the following theorem:NEWLINENEWLINENEWLINETheorem 2. Let \(B\) be the closed subalgebra of \(L^\infty\) containing \(A\), which contains at least one Riemann integrable function \(u\) not in \(A\), but in \(H^\infty\). Suppose that \(uH^\infty\subset B\). Then either \(B\subset H^\infty\) or \(C({\mathbb T})\subset B\). NEWLINENEWLINENEWLINEThe last section of the article contains several examples and counterexample, which illustrate the obtained results.