[proofplan]
The theorem statement defines the Dirichlet resolvent $Kf$ as the unique weak Dirichlet solution of $-\Delta u=f$ in $H^1_0(U)$, with all integrals taken with respect to the $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^n$. We first use this weak formulation, the Cauchy-Schwarz inequality, and the Poincare inequality on the bounded [open set](/page/Open%20Set) $U$ to obtain an a priori bound of $Kf$ in $H^1_0(U)$ by $\|f\|_{L^2(U)}$. This shows that $K$ sends bounded subsets of $L^2(U)$ into bounded subsets of $H^1_0(U)$, and compactness follows from the Rellich-Kondrachov compact embedding, giving compactness as an operator on $L^2(U)$. We also prove linearity of $K$ from linearity of the weak formulation and uniqueness of weak Dirichlet solutions. Self-adjointness is then obtained by testing the weak equations for $Kf$ and $Kg$ against each other, and positivity follows by testing the weak equation for $Kf$ against $Kf$ itself.
[/proofplan]
[step:Estimate the $H^1_0(U)$ norm of the weak solution]
Let $f \in L^2(U)$ be fixed, and define $u := Kf \in H^1_0(U)$, where $Kf$ is the weak Dirichlet solution specified in the theorem statement. The weak formulation gives, for every $v \in H^1_0(U)$,
\begin{align*}
\int_U \nabla u(x) \cdot \nabla v(x)\, d\mathcal{L}^n(x) = \int_U f(x)v(x)\, d\mathcal{L}^n(x).
\end{align*}
Taking $v=u$ gives
\begin{align*}
\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \int_U f(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
By the [Cauchy-Schwarz inequality](/theorems/432) in $L^2(U)$,
\begin{align*}
\left|\int_U f(x)u(x)\, d\mathcal{L}^n(x)\right| \leq \|f\|_{L^2(U)}\|u\|_{L^2(U)}.
\end{align*}
Since $U \subset \mathbb{R}^n$ is bounded and open and $u \in H^1_0(U)$, the hypotheses of the [Poincare inequality with zero trace](/theorems/76) on $H^1_0(U)$ are satisfied. Hence there exists a constant $C_P(U)>0$ such that
\begin{align*}
\|u\|_{L^2(U)} \leq C_P(U)\|\nabla u\|_{L^2(U)}.
\end{align*}
Therefore
\begin{align*}
\|\nabla u\|_{L^2(U)}^2 \leq C_P(U)\|f\|_{L^2(U)}\|\nabla u\|_{L^2(U)}.
\end{align*}
If $\|\nabla u\|_{L^2(U)}=0$, the desired estimate is immediate. Otherwise, dividing by $\|\nabla u\|_{L^2(U)}$ yields
\begin{align*}
\|\nabla u\|_{L^2(U)} \leq C_P(U)\|f\|_{L^2(U)}.
\end{align*}
Combining this with Poincare once more gives
\begin{align*}
\|u\|_{H^1_0(U)} \leq C_H(U)\|f\|_{L^2(U)}
\end{align*}
for a constant $C_H(U)>0$ depending only on the Poincare constant of $U$ and on the chosen equivalent norm on $H^1_0(U)$.
[guided]
Let $f \in L^2(U)$ be fixed, and define $u := Kf \in H^1_0(U)$, where $Kf$ is the unique weak Dirichlet solution of $-\Delta u=f$ from the theorem statement. The point of this step is to show that the solution cannot be large in $H^1_0(U)$ unless the data $f$ is large in $L^2(U)$. This is the estimate that will later feed into compactness.
The weak formulation says that, for every [test function](/page/Test%20Function) $v \in H^1_0(U)$,
\begin{align*}
\int_U \nabla u(x) \cdot \nabla v(x)\, d\mathcal{L}^n(x) = \int_U f(x)v(x)\, d\mathcal{L}^n(x).
\end{align*}
We choose $v=u$, which is admissible because $u=Kf$ belongs to $H^1_0(U)$. This gives the energy identity
\begin{align*}
\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \int_U f(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
We now estimate the right-hand side. Since $f,u \in L^2(U)$, the [Cauchy-Schwarz inequality](/theorems/432) in the [Hilbert space](/page/Hilbert%20Space) $L^2(U)$ applies and gives
\begin{align*}
\left|\int_U f(x)u(x)\, d\mathcal{L}^n(x)\right| \leq \|f\|_{L^2(U)}\|u\|_{L^2(U)}.
\end{align*}
The Poincare inequality is where the zero boundary condition is used. The [Poincare inequality with zero trace](/theorems/76) on $H^1_0(U)$ requires $U \subset \mathbb{R}^n$ to be a bounded open set and the function to lie in $H^1_0(U)$. These hypotheses hold because the theorem assumes $U$ is bounded and open, and because $u=Kf$ belongs to $H^1_0(U)$ by the definition of the Dirichlet resolvent. Therefore there exists a constant $C_P(U)>0$ depending only on $U$ such that
\begin{align*}
\|u\|_{L^2(U)} \leq C_P(U)\|\nabla u\|_{L^2(U)}.
\end{align*}
Substituting this into the Cauchy-Schwarz estimate gives
\begin{align*}
\|\nabla u\|_{L^2(U)}^2 \leq C_P(U)\|f\|_{L^2(U)}\|\nabla u\|_{L^2(U)}.
\end{align*}
If $\|\nabla u\|_{L^2(U)}=0$, then the gradient part is already bounded by $0$, and Poincare also gives $\|u\|_{L^2(U)}=0$. If $\|\nabla u\|_{L^2(U)}>0$, we divide the preceding inequality by $\|\nabla u\|_{L^2(U)}$ and obtain
\begin{align*}
\|\nabla u\|_{L^2(U)} \leq C_P(U)\|f\|_{L^2(U)}.
\end{align*}
Using Poincare once more gives
\begin{align*}
\|u\|_{L^2(U)} \leq C_P(U)^2\|f\|_{L^2(U)}.
\end{align*}
Thus the full $H^1_0(U)$ size of $u$ is controlled by the $L^2(U)$ size of $f$. Equivalently, for a constant $C_H(U)>0$ depending only on the Poincare constant of $U$,
\begin{align*}
\|Kf\|_{H^1_0(U)} \leq C_H(U)\|f\|_{L^2(U)}.
\end{align*}
[/guided]
[/step]
[step:Use uniqueness of weak solutions to prove linearity of $K$]
Let $f,g \in L^2(U)$ and let $a,b \in \mathbb{R}$. Define $u:=Kf \in H^1_0(U)$ and $w:=Kg \in H^1_0(U)$. Since $H^1_0(U)$ is a real [vector space](/page/Vector%20Space), the function $z:=au+bw$ belongs to $H^1_0(U)$. For every $v \in H^1_0(U)$, linearity of the integral and of the Euclidean dot product gives
\begin{align*}
\int_U \nabla z(x) \cdot \nabla v(x)\, d\mathcal{L}^n(x) = \int_U (af(x)+bg(x))v(x)\, d\mathcal{L}^n(x).
\end{align*}
Thus $z$ is a weak Dirichlet solution of $-\Delta z=af+bg$ in $H^1_0(U)$. By uniqueness in the definition of the Dirichlet resolvent, $z=K(af+bg)$. Therefore $K(af+bg)=aKf+bKg$, so $K:L^2(U;\mathbb{R})\to L^2(U;\mathbb{R})$ is linear.
[/step]
[step:Use Rellich-Kondrachov to prove compactness of $K$]
Let $(f_m)_{m=1}^{\infty}$ be a bounded sequence in $L^2(U)$, and define $u_m := Kf_m \in H^1_0(U)$ for each $m \in \mathbb{N}$. By the estimate from the previous step, the sequence $(u_m)_{m=1}^{\infty}$ is bounded in $H^1_0(U)$.
Since $U \subset \mathbb{R}^n$ is bounded and open, the hypotheses of the [Rellich-Kondrachov Compactness Theorem](/theorems/64) apply to the inclusion of $H^1_0(U)$ into $L^2(U)$. Therefore the inclusion map $j: H^1_0(U) \to L^2(U)$, defined by $j(u)=u$ for each $u \in H^1_0(U)$, is compact. Therefore the bounded sequence $(u_m)_{m=1}^{\infty}$ has a subsequence $(u_{m_k})_{k=1}^{\infty}$ that converges strongly in $L^2(U)$. Since $u_{m_k}=Kf_{m_k}$, every bounded sequence in the image of $K$ has an $L^2(U)$-convergent subsequence. Hence $K:L^2(U)\to L^2(U)$ is a [compact operator](/page/Compact%20Operator).
[guided]
Let $(f_m)_{m=1}^{\infty}$ be a bounded sequence in $L^2(U)$, and define $u_m:=Kf_m \in H^1_0(U)$ for each $m \in \mathbb{N}$. To prove compactness of $K:L^2(U)\to L^2(U)$, it is enough to show that every bounded input sequence has a subsequence whose image under $K$ converges strongly in $L^2(U)$.
The estimate proved in the first step gives a constant $C_H(U)>0$ such that
\begin{align*}
\|u_m\|_{H^1_0(U)} \leq C_H(U)\|f_m\|_{L^2(U)}
\end{align*}
for every $m \in \mathbb{N}$. Since $(f_m)_{m=1}^{\infty}$ is bounded in $L^2(U)$, the sequence $(u_m)_{m=1}^{\infty}$ is bounded in $H^1_0(U)$.
We now apply the [Rellich-Kondrachov Compactness Theorem](/theorems/64). Its relevant hypotheses here are that $U \subset \mathbb{R}^n$ is bounded and open, and that we are considering the Sobolev inclusion from $H^1_0(U)$ into $L^2(U)$. These hypotheses hold by the theorem statement and by the definition of $H^1_0(U)$. Therefore the inclusion map
\begin{align*}
j:H^1_0(U) \to L^2(U)
\end{align*}
defined by $j(u)=u$ for each $u \in H^1_0(U)$ is compact.
Compactness of the inclusion means that every bounded sequence in $H^1_0(U)$ has a subsequence converging strongly in $L^2(U)$. Applying this to $(u_m)_{m=1}^{\infty}$, there exists a strictly increasing sequence of indices $(m_k)_{k=1}^{\infty}$ such that $(u_{m_k})_{k=1}^{\infty}$ converges strongly in $L^2(U)$. Since $u_{m_k}=Kf_{m_k}$, the sequence $(Kf_{m_k})_{k=1}^{\infty}$ converges strongly in $L^2(U)$. Thus $K$ maps bounded sequences in $L^2(U)$ to sequences with strongly convergent subsequences in $L^2(U)$, which is exactly the sequential characterization of a [compact operator](/page/Compact%20Operator) between metric normed spaces.
[/guided]
[/step]
[step:Test the two weak equations against each other to prove self-adjointness]
Let $f,g \in L^2(U)$, and define $u:=Kf \in H^1_0(U)$ and $w:=Kg \in H^1_0(U)$. Applying the weak formulation for $u$ with test function $v=w$ gives
\begin{align*}
\int_U \nabla u(x)\cdot \nabla w(x)\, d\mathcal{L}^n(x) = \int_U f(x)w(x)\, d\mathcal{L}^n(x).
\end{align*}
Applying the weak formulation for $w$ with test function $v=u$ gives
\begin{align*}
\int_U \nabla w(x)\cdot \nabla u(x)\, d\mathcal{L}^n(x) = \int_U g(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
The Euclidean dot product is symmetric, so the two left-hand sides are equal. Therefore
\begin{align*}
\int_U f(x)w(x)\, d\mathcal{L}^n(x) = \int_U u(x)g(x)\, d\mathcal{L}^n(x).
\end{align*}
Since $u=Kf$ and $w=Kg$, this is exactly the symmetry identity for the real $L^2(U)$ [inner product](/page/Inner%20Product):
\begin{align*}
(f,Kg)_{L^2(U)} = (Kf,g)_{L^2(U)}.
\end{align*}
Because the preceding step proves that $K$ is linear on the real Hilbert space $L^2(U;\mathbb{R})$, this identity shows that $K$ is a self-adjoint operator.
[guided]
Let $f,g \in L^2(U)$, and define $u:=Kf \in H^1_0(U)$ and $w:=Kg \in H^1_0(U)$. The goal is to prove the symmetry identity for the real $L^2(U)$ inner product:
\begin{align*}
(f,Kg)_{L^2(U)} = (Kf,g)_{L^2(U)}.
\end{align*}
Since linearity of $K$ was proved above, this identity is the defining condition for $K$ to be a self-adjoint operator on the real Hilbert space $L^2(U;\mathbb{R})$.
The weak formulation for $u=Kf$ is valid against every test function in $H^1_0(U)$. Since $w=Kg$ belongs to $H^1_0(U)$, we may choose $v=w$ and obtain
\begin{align*}
\int_U \nabla u(x)\cdot \nabla w(x)\, d\mathcal{L}^n(x) = \int_U f(x)w(x)\, d\mathcal{L}^n(x).
\end{align*}
Likewise, the weak formulation for $w=Kg$ is valid against every test function in $H^1_0(U)$. Since $u=Kf$ belongs to $H^1_0(U)$, choosing $v=u$ gives
\begin{align*}
\int_U \nabla w(x)\cdot \nabla u(x)\, d\mathcal{L}^n(x) = \int_U g(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
The Euclidean dot product is symmetric pointwise, so $\nabla u(x)\cdot \nabla w(x)=\nabla w(x)\cdot \nabla u(x)$ for $\mathcal{L}^n$-almost every $x \in U$. Hence the two left-hand sides are equal, and therefore
\begin{align*}
\int_U f(x)w(x)\, d\mathcal{L}^n(x) = \int_U u(x)g(x)\, d\mathcal{L}^n(x).
\end{align*}
Substituting $u=Kf$ and $w=Kg$ converts this equality into
\begin{align*}
(f,Kg)_{L^2(U)} = (Kf,g)_{L^2(U)}.
\end{align*}
Thus $K$ is self-adjoint on $L^2(U;\mathbb{R})$.
[/guided]
[/step]
[step:Test the weak equation against $Kf$ to prove positivity]
Let $f \in L^2(U)$, and define $u:=Kf \in H^1_0(U)$. Taking $v=u$ in the weak formulation gives
\begin{align*}
\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \int_U f(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
Since $u=Kf$ and the real $L^2(U)$ inner product is symmetric,
\begin{align*}
(Kf,f)_{L^2(U)} = \int_U u(x)f(x)\, d\mathcal{L}^n(x) = \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x).
\end{align*}
The integrand $|\nabla u(x)|^2$ is non-negative for $\mathcal{L}^n$-almost every $x \in U$, hence
\begin{align*}
(Kf,f)_{L^2(U)} \geq 0.
\end{align*}
Therefore $K$ is a positive operator on the real Hilbert space $L^2(U;\mathbb{R})$. Together with linearity, compactness, and self-adjointness proved above, this completes the proof.
[guided]
Let $f \in L^2(U)$, and define $u:=Kf \in H^1_0(U)$. To prove that $K$ is positive, we must show that
\begin{align*}
(Kf,f)_{L^2(U)} \geq 0
\end{align*}
for every $f \in L^2(U)$.
The weak formulation for $u=Kf$ is valid for every $v \in H^1_0(U)$. Since $u \in H^1_0(U)$, we may test with $v=u$. This gives
\begin{align*}
\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \int_U f(x)u(x)\, d\mathcal{L}^n(x).
\end{align*}
The right-hand side is the real $L^2(U)$ inner product of $f$ with $u=Kf$. Since the real $L^2(U)$ inner product is symmetric, we have
\begin{align*}
(Kf,f)_{L^2(U)} = \int_U u(x)f(x)\, d\mathcal{L}^n(x) = \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x).
\end{align*}
For $\mathcal{L}^n$-almost every $x \in U$, the quantity $|\nabla u(x)|^2$ is non-negative. Therefore its [Lebesgue integral](/page/Lebesgue%20Integral) is non-negative:
\begin{align*}
\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) \geq 0.
\end{align*}
Combining the preceding identities gives
\begin{align*}
(Kf,f)_{L^2(U)} \geq 0.
\end{align*}
Since this holds for every $f \in L^2(U)$, $K$ is a positive operator on $L^2(U;\mathbb{R})$.
[/guided]
[/step]
