[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*}
[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]
[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]
