[proofplan] We use the Rayleigh quotient definition of the first Dirichlet eigenvalue to control the $L^2(U)$ norm of any $u \in H^1_0(U)$, the zero-trace [Sobolev space](/page/Sobolev%20Space), by its gradient norm. This turns the shifted quadratic form \begin{align*} \int_U |\nabla u|^2 \, d\mathcal{L}^n - \lambda \int_U |u|^2 \, d\mathcal{L}^n \end{align*} into a positive multiple of \begin{align*} \int_U |\nabla u|^2 \, d\mathcal{L}^n, \end{align*} with separate bookkeeping when $\lambda < 0$. The same Rayleigh quotient inequality also compares the full $H^1_0(U)$ norm with the gradient norm, yielding the desired coercivity constant. [/proofplan]

[step:Use the Rayleigh quotient to control the $L^2(U)$ norm] Define the gradient energy functional $G: H^1_0(U) \to [0,\infty)$ by, for each $u \in H^1_0(U)$, \begin{align*} G[u] = \int_U |\nabla u|^2 \, d\mathcal{L}^n. \end{align*} Define the squared $L^2(U)$ functional $L: H^1_0(U) \to [0,\infty)$ by, for each $u \in H^1_0(U)$, \begin{align*} L[u] = \int_U |u|^2 \, d\mathcal{L}^n. \end{align*} If $u = 0$ in $H^1_0(U)$, then $G[u] = L[u] = 0$. If $u \neq 0$, then $L[u] > 0$ and the definition of $\lambda_1(U)$ in the theorem statement as the infimum of the Rayleigh quotient over $H^1_0(U) \setminus \{0\}$ gives \begin{align*} \lambda_1(U) \leq \frac{G[u]}{L[u]}. \end{align*} Because $U$ is bounded and open, the [Poincare Inequality](/theorems/76) applies on $H^1_0(U)$ and gives a constant $C_U > 0$ such that \begin{align*} L[v] \leq C_U^2 G[v] \end{align*} for every $v \in H^1_0(U)$. Taking the infimum over nonzero $v \in H^1_0(U)$ in the Rayleigh quotient therefore gives $\lambda_1(U) \geq C_U^{-2} > 0$. Hence \begin{align*} L[u] \leq \frac{1}{\lambda_1(U)} G[u]. \end{align*} Thus this inequality holds for every $u \in H^1_0(U)$. [/step]

[step:Bound the shifted energy from below by the gradient energy]Define the shifted Dirichlet energy functional $E_\lambda: H^1_0(U) \to \mathbb{R}$ by \begin{align*} E_\lambda(u) = \int_U |\nabla u|^2 \, d\mathcal{L}^n - \lambda \int_U |u|^2 \, d\mathcal{L}^n. \end{align*} Set \begin{align*} c_\lambda = 1 - \frac{\max\{\lambda,0\}}{\lambda_1(U)}. \end{align*} Because $\lambda < \lambda_1(U)$ and $\lambda_1(U) > 0$, we have $c_\lambda > 0$. If $\lambda \leq 0$, then $-\lambda L[u] \geq 0$, so \begin{align*} E_\lambda(u) = G[u] - \lambda L[u] \geq G[u] = c_\lambda G[u], \end{align*} because in this case $c_\lambda = 1$. If $0 < \lambda < \lambda_1(U)$, then the $L^2(U)$ estimate from the previous step gives \begin{align*} -\lambda L[u] \geq -\frac{\lambda}{\lambda_1(U)}G[u]. \end{align*} Adding $G[u]$ to both sides gives \begin{align*} E_\lambda(u) = G[u] - \lambda L[u] \geq \left(1 - \frac{\lambda}{\lambda_1(U)}\right)G[u] = c_\lambda G[u]. \end{align*} Therefore, for every $u \in H^1_0(U)$, \begin{align*} E_\lambda(u) \geq c_\lambda G[u]. \end{align*}[/step]

[guided]Define the gradient energy functional $G: H^1_0(U) \to [0,\infty)$ and the squared $L^2(U)$ functional $L: H^1_0(U) \to [0,\infty)$ by \begin{align*} G[u] = \int_U |\nabla u|^2 \, d\mathcal{L}^n \end{align*} and \begin{align*} L[u] = \int_U |u|^2 \, d\mathcal{L}^n \end{align*} for each $u \in H^1_0(U)$. Also define the shifted Dirichlet energy functional $E_\lambda: H^1_0(U) \to \mathbb{R}$ by \begin{align*} E_\lambda(u) = G[u] - \lambda L[u]. \end{align*} Thus \begin{align*} E_\lambda(u) = \int_U |\nabla u|^2 \, d\mathcal{L}^n - \lambda \int_U |u|^2 \, d\mathcal{L}^n. \end{align*} The only potentially dangerous term is the negative contribution coming from $-\lambda \int_U |u|^2 \, d\mathcal{L}^n$. If $\lambda \leq 0$, that term is not dangerous at all: since $L[u] \geq 0$, we have $-\lambda L[u] \geq 0$, and hence \begin{align*} E_\lambda(u) = G[u] - \lambda L[u] \geq G[u]. \end{align*} Before treating the case $0 < \lambda < \lambda_1(U)$, we verify within this guided argument that $\lambda_1(U) > 0$. Since $U$ is bounded and open, the [Poincare Inequality](/theorems/76) applies on $H^1_0(U)$ and gives a constant $C_U > 0$ such that \begin{align*} L[v] \leq C_U^2 G[v] \end{align*} for every $v \in H^1_0(U)$. For every nonzero $v \in H^1_0(U)$, this implies \begin{align*} \frac{G[v]}{L[v]} \geq C_U^{-2}. \end{align*} Taking the infimum over nonzero $v \in H^1_0(U)$ in the Rayleigh quotient definition of $\lambda_1(U)$ gives $\lambda_1(U) \geq C_U^{-2} > 0$. If $0 < \lambda < \lambda_1(U)$, the $L^2(U)$ term has to be controlled by the gradient term. The definition of $\lambda_1(U)$ in the theorem statement gives, for every nonzero $u \in H^1_0(U)$, \begin{align*} \lambda_1(U) \leq \frac{G[u]}{L[u]}. \end{align*} Equivalently, \begin{align*} L[u] \leq \frac{1}{\lambda_1(U)}G[u]. \end{align*} For $u = 0$ in $H^1_0(U)$, both sides are zero, so the same inequality holds for every $u \in H^1_0(U)$. Multiplying this inequality by the negative number $-\lambda$ reverses the inequality, giving \begin{align*} -\lambda L[u] \geq -\frac{\lambda}{\lambda_1(U)}G[u]. \end{align*} Adding $G[u]$ to both sides yields \begin{align*} E_\lambda(u) = G[u] - \lambda L[u] \geq \left(1 - \frac{\lambda}{\lambda_1(U)}\right)G[u]. \end{align*} These two cases are combined by defining \begin{align*} c_\lambda = 1 - \frac{\max\{\lambda,0\}}{\lambda_1(U)}. \end{align*} Because $\lambda < \lambda_1(U)$ and $\lambda_1(U) > 0$, this constant is positive. Thus, in all cases, \begin{align*} E_\lambda(u) \geq c_\lambda G[u]. \end{align*}[/guided]

[step:Compare the gradient energy with the $H^1_0(U)$ norm] Use the $H^1_0(U)$ norm convention specified in the theorem statement: \begin{align*} \|u\|_{H^1_0(U)}^2 = \int_U |u|^2 \, d\mathcal{L}^n + \int_U |\nabla u|^2 \, d\mathcal{L}^n. \end{align*} By the Rayleigh quotient estimate, \begin{align*} \|u\|_{H^1_0(U)}^2 = L[u] + G[u] \leq \left(1 + \frac{1}{\lambda_1(U)}\right)G[u]. \end{align*} Equivalently, \begin{align*} G[u] \geq \frac{\lambda_1(U)}{1+\lambda_1(U)}\|u\|_{H^1_0(U)}^2. \end{align*} [/step]

[step:Choose the coercivity constant] Define \begin{align*} \alpha = c_\lambda \frac{\lambda_1(U)}{1+\lambda_1(U)}. \end{align*} Since $c_\lambda > 0$ and $\lambda_1(U) > 0$, we have $\alpha > 0$. Combining the lower bound for $E_\lambda(u)$ with the comparison between $G[u]$ and $\|u\|_{H^1_0(U)}^2$, we obtain, for every $u \in H^1_0(U)$, \begin{align*} \int_U |\nabla u|^2 \, d\mathcal{L}^n - \lambda \int_U |u|^2 \, d\mathcal{L}^n = E_\lambda(u) \geq c_\lambda G[u] \geq \alpha \|u\|_{H^1_0(U)}^2. \end{align*} This proves the asserted coercivity estimate. [/step]