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Let $\left( X,\Omega ,\mu \right) $ be a measure space and $k\in L^{2}\left( X\times X,\Omega \times \Omega ,\mu \times \mu \right) $, define \[ \left( Kf\right) \left( x\right) =\int k\left( x,y\right) f\left( y\right) d\mu \left( y\right) \text{.} \]

Would you help me to prove that $K$ is a compact operator.

Let $\left\{ e_{i}:i\in I\right\} $ is a basis for $L^{2}\left( X,\Omega ,\mu \right) \,\ $and $\phi _{ij}\left( x,y\right) =e_{j}\left( x\right) \overline{e_{i}\left( y\right) }$ for $i,j$ in $I$ and $x,y$ in $X$, would you help me to show that $\left\{ \phi _{ij}:i,j\in I\right\} $ is an orthonormal set in $L^{2}\left( X\times X,\Omega \times \Omega ,\mu \times \mu \right) $ and $\left\langle k,\phi _{ij}\right\rangle =\left\langle Ke_{j},e_{i}\right\rangle $.

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Related:… – Jonas Meyer Oct 23 '12 at 6:05
up vote 4 down vote accepted

As simple functions with "rectangles" as "steps" are dense in $L^2(\Omega^2)$, there is a sequence $k_n \in L^2(\Omega^2)$ of functions of the form \[ k_n(x,y) = \sum_{i=1}^{N_n} a_{i}\chi_{A^n_i}(x)\chi_{B^n_i}(y) \] where $A^n_i$, $B^n_j \subseteq \Omega$ are of finite measure, such that $k_n \to k$ in $L^2(\Omega^2)$.

Define $K_n$ as $K$, but with $k$ replaced by $k_n$. Then we have for $f \in L^2(\Omega)$ and $x \in \Omega$ \begin{align*} (K_nf)(x) &= \int_\Omega k_n(x,y)f(y)\,d\mu(y)\\ &= \sum_{i=1}^{N_n}a_i\chi_{A^n_i}(x)\int_\Omega \chi_{B^n_i}\, d\mu\\ &= \sum_{i=1}^{N_n}a_i \mu(B^n_i) \cdot \chi_{A^n_i}(x) \end{align*} That is $K_n f \in \operatorname{span}\{\chi_{A^n_i} \mid 1 \le i \le N_n\}$. As $f$ was arbitrary, $\dim\operatorname{ran} K_n \le N_n$, so $K_n$ is compact. Also we have \begin{align*} \|K_n f - K f\| &\le \left( \int_\Omega\left|\int_\Omega (k_n - k)(x,y)f(y)\,dy\right|^2\, dx\right)^{1/2}\\ &\le \left(\int_\Omega \int_\Omega |(k_n-k)(x,y)|^2\, dy\cdot \|f\|^2\, dx\right)^{1/2}\\ &\le \|k_n - k\|_{L^2(\Omega)^2}\|f\|_{L^2(\Omega)} \end{align*} So $\|K_n - K\| \le \|k_n - k\| \to 0$ and $K$ is compact as a limit of compact operators.

First we show that $(\phi_{ij})$ is orthonormal. We have for $i,j,k,l \in I$ that $\def\skp#1{\left\langle#1\right\rangle}$ \begin{align*} \skp{\phi_{ij}, \phi_{kl}} &= \int_{\Omega^2} e_j(x)\overline{e_i(y)}\overline{e_l(x)\overline{e_k(y)}} \, d\mu^2(x,y)\\ &= \int_\Omega e_j\overline{e_l}\, d\mu \cdot \int_\Omega \overline{e_i}e_k\, d\mu\\ &= \skp{e_j,e_l}\cdot \skp{e_i, e_k}\\ &= \delta_{jl}\delta_{ik} \end{align*} Moreover \begin{align*} \skp{k, \phi_{ij}} &= \int_{\Omega^2} k(x,y)\overline{e_j(x)\overline{e_i}(y)}\, d\mu^2(x,y)\\ &= \int_\Omega \left(\int_\Omega k(x,y)e_i(y)\,d\mu(y)\right) \overline{e_j(x)}\, dx\\ &= \int_\Omega Ke_i(x)\overline{e_j(x)}\, d\mu\\ &=\skp{Ke_i, e_j} \end{align*}

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