Note: (a) is a repeat.
Let \(\Lambda\in \dualof{L^2(X)}\).
Show that \(M\definedas \theset{f\in L^2(X) \suchthat \Lambda(f) = 0} \subseteq L^2(X)\) is a closed subspace, and \(L^2(X) = M \oplus M\perp\).
Prove that there exists a unique \(g\in L^2(X)\) such that \(\Lambda(f) = \int_X g \bar f\).
Erratum: since \(\Lambda\) is linear and \(f \mapsto \int_X g\bar f\) is conjugate-linear, the formula in the statement holds only for \(\Lambda = 0\) over \(\CC\). The representation is \(\Lambda(f) = \int_X f\bar g\); equivalently, \(\int_X g\bar f\) represents the conjugate-linear functional \(\overline{\Lambda}\). Over \(\RR\) the two formulas agree.
Solution
Write \(\langle f, h\rangle = \int_X f\bar h\), linear in \(f\). For a linear \(\Lambda\) the representing formula is \(\Lambda(f) = \langle f, g\rangle = \int_X f\bar g\); see the remark below.
1. \(M = \ker \Lambda\) is a closed subspace, and \(L^2(X) = M \oplus M^\perp\).
\(M\) is the kernel of the bounded linear functional \(\Lambda\), so it is a subspace, and it is closed because \(\Lambda\) is continuous. In a Hilbert space every closed subspace \(M\) satisfies \(H = M \oplus M^\perp\) by the projection theorem.
2. If \(\Lambda \neq 0\), then \(M^\perp = \operatorname{span}\theset{g_0}\) for some \(g_0\) with \(\|g_0\| = 1\).
\(M \neq L^2(X)\), so \(M^\perp \neq \theset0\) by step 1; choose \(g_0 \in M^\perp\) with \(\|g_0\| = 1\), so \(\Lambda(g_0) \neq 0\). For \(h \in M^\perp\), the vector \(\Lambda(h)g_0 - \Lambda(g_0)h\) lies in \(M^\perp\) and in \(M\), hence is \(0\), so \(h\) is a multiple of \(g_0\).
3. There is \(g \in L^2(X)\) with \(\Lambda(f) = \langle f, g\rangle\) for all \(f\).
If \(\Lambda = 0\) take \(g = 0\). Otherwise, by steps 1 and 2, \(f = m + \langle f, g_0\rangle g_0\) with \(m \in M\), so \(\Lambda(f) = \langle f, g_0\rangle\Lambda(g_0) = \langle f, \overline{\Lambda(g_0)}\,g_0\rangle\). Take \(g = \overline{\Lambda(g_0)}\,g_0\).
4. \(g\) is unique in \(L^2(X)\).
If \(\langle f, g\rangle = \langle f, g'\rangle\) for all \(f\), take \(f = g - g'\) to get \(\|g - g'\|^2 = 0\).