Let \(f\in L^1((0, 2\pi))\).
- Show that for an \(\epsilon>0\) one can write \(f = g+h\) where \(g\in L^2((0, 2\pi))\) and \(\norm{H}_1 < \epsilon\).
In the statement, \(\norm{H}_1\) is read as \(\norm{h}_1\).
Solution
1. It suffices to treat real \(f \ge 0\).
Write \(f = u^+ - u^- + i(v^+ - v^-)\) with \(u = \operatorname{Re} f\), \(v = \operatorname{Im} f\). If each of the four nonnegative parts is \(g_j + h_j\) with \(g_j \in L^2\) and \(\|h_j\|_1 < \eps/4\), the corresponding combinations give \(f = g + h\) with \(g \in L^2\) and \(\|h\|_1 < \eps\).
2. For \(f \ge 0\) and \(M > 0\), put \(g_M = \min(f, M)\) and \(h_M = f - g_M = (f - M)^+\). Then \(g_M \in L^2((0,2\pi))\).
\(0 \le g_M \le M\) and \(g_M \le f\), so \(\int g_M^2 \le M\int g_M \le M\|f\|_1\).
3. \(\|h_M\|_1 \to 0\) as \(M \to \infty\).
\(h_M \to 0\) at every point where \(f < \infty\), hence a.e., and \(0 \le h_M \le f \in L^1\), so dominated convergence applies.
4. Q.E.D.
By steps <1>1–<1>3, choosing \(M\) with \(\|h_M\|_1 < \eps\) gives \(f = g_M + h_M\) with \(g_M \in L^2\).