Hilbert-Schmidt Theorem, final statement: Kernel is contained in an orthogonal complement

Question

I am reading the Royden, Fitzpatrick's Real Analysis, Fourth Edition, p.327~p.328, proof of the Hilbert-Schmidt Theorem and stuck at final statement :

There, $Q_T$ is more precisely $Q_K$, which is defined as the quadratic form ( royden's book p.321 ) $Q_K : H \to \mathbb{R}$ ; $Q_K(u) := \langle K(u) , u \rangle $ for all $u\in H$. The argument in the final paragraph of the proof means that $Q|_{K|_{H_0^{\perp}}} =0$, which is equivalent to that $K|_{H_0^{\perp}} =0$ ( By the polarization identity ; see royden's book, remark in p.321 ), which is equivalent to that $K=0$ on $H_0^{\perp}$, which means that $H_0^{\perp} \subseteq \ker K$, Now I don't know why $\ker K \subseteq H_0^{\perp}$, as implicitly claimed in the underlined statement.

Answer 1

There are a few ways to see this. Here is perhaps the most straightforward. Let $h\in {\rm ker} \, K$. Let $k\in \mathbb{N}$. Since $K$ is symmetric and $\lambda_{k} \in \mathbb{R}$ we have \begin{equation} 0 = \langle K(h), \psi_{k} \rangle = \langle h, K(\psi_{k}) \rangle = \lambda_{k} \langle h, \psi_{k} \rangle . \end{equation} As $\lambda_{k} \neq 0$ it follows that $\langle h, \psi_{k} \rangle = 0$. This shows that $h\in \{\psi_{k} : k\in\mathbb{N} \}^{\perp}$. By this result we have $\{\psi_{k} : k\in\mathbb{N} \}^{\perp} = (\overline{{\rm span} \{\psi_{k} : k\in\mathbb{N} \}})^{\perp} = H_{0}^{\perp}$. Hence ${\rm ker} \, K \subseteq H_{0}^{\perp}$.

Another way is to note that since the sequence $(\lambda_{k})_{k\in\mathbb{N}}$ is bounded and $K(\psi_{k}) = \lambda_{k} \psi_{k}$ for all $k\in\mathbb{N}$, it follows from the Bessel inequality and taking limits that \begin{equation} K(h) = \sum_{k=1}^{\infty} \lambda_{k} \langle h, \psi_{k} \rangle \psi_{k} \tag{1} \end{equation} for all $h\in \overline{{\rm span} \{\psi_{k} : k\in\mathbb{N} \}} = H_{0}$. Let $h\in {\rm ker} \, K \cap H_{0}$. Then by $(1)$ we have \begin{equation} \sum_{k=1}^{\infty} \lambda_{k} \langle h, \psi_{k} \rangle \psi_{k} = 0 . \tag{2} \end{equation} Because $\{\psi_{k} : k\in\mathbb{N} \}$ is an orthonormal set it follows from $(2)$ that $\lambda_{k} \langle h, \psi_{k} \rangle = 0$ for all $k\in \mathbb{N}$. For each $k\in\mathbb{N}$ we have $\lambda_{k} \neq 0$ and hence $\langle h, \psi_{k} \rangle = 0$. Then as $\{\psi_{k} : k\in\mathbb{N} \}$ is an orthonormal basis for $H_{0}$ it follows that $h = 0$. This shows that ${\rm ker} \, K \cap H_{0} = \{0\}$.

Now take $h\in {\rm ker} \, K$. As $H_{0}$ is a closed subspace of $H$ we can write $h = h_{0} + h_{1}$ with $h_{0} \in H_{0}$ and $h_{1} \in H_{0}^{\perp}$. Then $K(h_{0}) + K(h_{1}) = K(h) = 0$. Because $K(H_{0}) \subseteq H_{0}$ and $K(H_{0}^{\perp}) \subseteq H_{0}^{\perp}$ we have $K(h_{0}) = 0$ (and $K(h_{1}) = 0$). Then from ${\rm ker} \, K \cap H_{0} = \{0\}$ it follows that $h_{0} = 0$. Hence $h = h_{1} \in H_{0}^{\perp}$ and we have ${\rm ker} \, K \subseteq H_{0}^{\perp}$.