> [!definition] > > Let $H$ be a [[Separable Topological Space|separable]] [[Hilbert Space|Hilbert space]]. A linear operator $A \in L(H, H)$ if there exists a Hilbert space basis $\seq{u_j}$ such that > $ > \sum_{i \in \nat}\norm{Au_i}^2 < \infty > $ > For any Hilbert-Schmidt operator, define > $ > \angles{A, B} = \sum_{i = 1}^\infty \angles{Au_i, Bu_i} > $ > then the two above sums are independent of choice of basis, and absolutely convergent. > > *Proof*. Let $\seq{v_j}$ be another Hilbert space basis, then using Parseval's identity > $ > \begin{align*} > \sum_{i \in \nat}\norm{Au_i}^2 &= \sum_{i, j \in \nat}\abs{\angles{Au_i, v_j}} = \sum_{i, j \in \nat}\abs{\angles{u_i, A^*v_j}} \\ > &= \sum_{i \in \nat}\norm{A^*v_j}^2 > \end{align*} > $ > If $\seq{u_j} = \seq{v_j}$, then $\sum_{i \in \nat}\norm{Au_i}^2 = \sum_{i \in \nat}\norm{A^*u_i}^2$. Thus the equality holds for any choice of bases. By [[Hölder's Inequality|Hölder's inequality]], $\seq{\angles{Au_i, Bu_i}} \in L^1$, so the sum is absolutely convergent. > [!theorem] > > Let $A$ be a Hilbert-Schmidt operator, then $A$ is [[Compact Operator|compact]]. > [!theorem] > > Let $\Omega \subset \real^2$ be a bounded domain such that $L^2(\Omega)$ is separable, and $K: \real^2 \to \complex$ be a measurable function. The operator > $ > Kf(x) = \int K(x, y) f(y)dy > $ > is Hilbert-Schmidt if and only if $K \in L^2(\Omega)$. > > *Proof*. Let $\bracs{e_i \otimes e_j}_{i,j \in \nat}$ be a Hilbert basis, expressed as a product of bases along each axis. By Parseval's identity, > $ > \begin{align*} > \sum_{j \in \nat}\norm{Ke_j}^2 &= \sum_{i, j \in \nat}\abs{\angles{Ke_j, e_i}}^2 = \sum_{i, j \in \nat}\braks{\int K(x, y)(e_i \otimes e_j)dxdy}^2 \\ > &= \sum_{i, j \in \nat}\abs{\angles{K, e_i \otimes e_j}}^2 = \norm{K}_2^2 > \end{align*} > $