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Question

Math 6210 • Real Analysis

1992 • Question 1

Let \(f\in L^\infty (\mathbb R)\) define a bounded linear operator \(M_f\) from \(L^2(\mathbb R)\) to \(L^2(\mathbb R)\) by \(M_f(g)=fg\).

(a) Prove that \(\lVert M_f\rVert =\lVert f\rVert _\infty \).

(b) Prove that if \(M_f\) is one to one and onto then there exists \(\epsilon >0\) and a set \(E\) of measure zero such that \(|f(x)|\ge \epsilon \) for \(x\in \mathbb R-E\).

Raw LaTeX

Let \(f\in L^\infty(\mathbb R)\) define a bounded linear operator
\(M_f\) from \(L^2(\mathbb R)\) to \(L^2(\mathbb R)\) by \(M_f(g)=fg\).

(a) Prove that \(\lVert M_f\rVert=\lVert f\rVert_\infty\).

(b) Prove that if \(M_f\) is one to one and onto then there exists \(\epsilon>0\)
and a set \(E\) of measure zero such that \(|f(x)|\ge\epsilon\) for
\(x\in\mathbb R-E\).

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