问题: $(\Omega, \mathscr{B}, \mu)$ 是测度空间，$X=L^2(\Omega, \mu), X^*=X$ ，函数 $K: \Omega \times \Omega \rightarrow$ $\mathbb{K},(x, y) \mapsto K(x, y)$ 且 $$ \int_{\Omega \times \Omega}|K(x, y)|^2 \mathrm{~d} \mu_x \mathrm{~d} \mu_y:=M<+\infty $$ 对 $\forall u \in L^2$ ，定义 $$ T u(x):=\iint_{\Omega} K(x, y) u(y) \mathrm{d} \mu_y $$ 则（1）$T \in \mathscr{L}\left(L^2, L^2\right)$
（2）对 $\forall u \in\left(L^2\right)^*=L^2,\left(T^* v\right)(y)=\int_{\Omega} K(x, y) v(x) \mathrm{d} \mu_x$

---

---

解答: ( ID: 管理员[Alina Lagrange] math@lamu.run )

证明．（1）对 $\forall u \in L^2$ 有 $$ \begin{aligned} \|T u\|_{L^2}^2 & =\int_{\Omega}|T u|^2 \mathrm{~d} \mu_x \\ & =\int_{\Omega}\left|\int_{\Omega} K(x, y) u(y) \mathrm{d} \mu_y\right| \mathrm{d} \mu_x \\ & \leq \int_{\Omega}\left(\int_{\Omega}|K(x, y) u(y)| \mathrm{d} \mu_y\right)^2 \mathrm{~d} \mu_x \\ & \leq \int_{\Omega}\left(\int_{\Omega}|K(x, y)|^2 \mathrm{~d} \mu_y \int_{\Omega}|u(y)|^2 \mathrm{~d} \mu_y\right) \mathrm{d} \mu_x \quad \text { Holder } \\ & \leq M \cdot\|u\|_{L^2}^2 \end{aligned} $$ 因此 $\|T\| \leq \sqrt{M}$ ． （2）对 $\forall v(x) \in\left(L^2\right)^*=L^2, \forall u(y) \in L^2$ ． $$ \begin{aligned} \left(T^* v\right)(u) & =v(T u) \\ & =\int_{\Omega}(T u)(x) \cdot v(x) \mathrm{d} \mu_x \\ & =\int_{\Omega}\left(\int_{\Omega} K(x, y) u(y) \mathrm{d} \mu_y\right) v(x) \mathrm{d} \mu_x \\ & \xlongequal{\text { Fubini }} \int_{\Omega} u(y)\left(\int_{\Omega} K(x, y) v(x) \mathrm{d} \mu_x\right) \mathrm{d} \mu_y \\ & =\left(\int_{\Omega} K(x, y) v(x) \mathrm{d} \mu_x\right)(u) \end{aligned} $$