[proofplan]
We separate the [bilinear form](/page/Bilinear%20Form) into its principal elliptic part and its lower-order part. Uniform ellipticity gives a positive lower bound for the principal part in terms of $\|\nabla u\|_{L^2(U)}^2$, while the hypothesis bounds the lower-order contribution from below by $-\beta\|\nabla u\|_{L^2(U)}^2$. Since $\beta < \theta$, the sum still controls the gradient norm, and Poincare's inequality converts gradient control into full $H^1_0(U)$ coercivity.
[/proofplan]
[step:Bound the principal part from below by uniform ellipticity]
Fix $u \in H^1_0(U)$. The scalar field is real throughout: $H^1_0(U)$ denotes the real [Sobolev space](/page/Sobolev%20Space), and $a_{ij}, b_i, c$ are real-valued coefficient functions. For $v \in H^1_0(U)$, define the bilinear form $B: H^1_0(U) \times H^1_0(U) \to \mathbb{R}$ by
\begin{align*}
B[u,v] = \int_U \sum_{i,j=1}^n a_{ij}\,\partial_{x_j}u\,\partial_{x_i}v\,d\mathcal{L}^n(x) + \int_U \sum_{i=1}^n b_i\,\partial_{x_i}u\,v\,d\mathcal{L}^n(x) + \int_U c\,u\,v\,d\mathcal{L}^n(x).
\end{align*}
Each integral is finite by $L^\infty$ boundedness of the coefficients and the [Cauchy-Schwarz inequality](/theorems/432) applied to the relevant $L^2(U)$ factors. Since $a_{ij} \in L^\infty(U)$ and $\partial_{x_i}u, \partial_{x_j}u \in L^2(U)$, the Cauchy-Schwarz inequality gives $a_{ij}\,\partial_{x_j}u\,\partial_{x_i}u \in L^1(U)$ for each $1 \leq i,j \leq n$. Hence the finite sum below is integrable. Define the principal contribution $B_0[u,u] \in \mathbb{R}$ by
\begin{align*}
B_0[u,u] = \int_U \sum_{i,j=1}^n a_{ij}\,\partial_{x_j}u\,\partial_{x_i}u\,d\mathcal{L}^n(x).
\end{align*}
For $\mathcal{L}^n$-a.e. $x \in U$, the vector $\nabla u(x) \in \mathbb{R}^n$ is defined, so uniform ellipticity applied with $\xi = \nabla u(x)$ gives
\begin{align*}
\sum_{i,j=1}^n a_{ij}(x)\,\partial_{x_j}u(x)\,\partial_{x_i}u(x) \geq \theta |\nabla u(x)|^2.
\end{align*}
Integrating this pointwise inequality over $U$ with respect to $\mathcal{L}^n$ yields
\begin{align*}
B_0[u,u] \geq \theta \int_U |\nabla u|^2\,d\mathcal{L}^n(x) = \theta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
[/step]
[step:Add the lower-order estimate to obtain gradient coercivity]
Since $b_i \in L^\infty(U)$ and $u, \partial_{x_i}u \in L^2(U)$, the Cauchy-Schwarz inequality gives $b_i\,\partial_{x_i}u\,u \in L^1(U)$ for each $1 \leq i \leq n$. Since $c \in L^\infty(U)$ and $u \in L^2(U)$, we also have $c\,u^2 \in L^1(U)$. Define the lower-order contribution $B_1[u,u] \in \mathbb{R}$ by
\begin{align*}
B_1[u,u] = \int_U \sum_{i=1}^n b_i\,\partial_{x_i}u\,u\,d\mathcal{L}^n(x) + \int_U c\,u^2\,d\mathcal{L}^n(x).
\end{align*}
By the assumed lower-order estimate,
\begin{align*}
B_1[u,u] \geq -\beta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Since $B[u,u] = B_0[u,u] + B_1[u,u]$, combining the preceding inequality with the principal-part estimate gives
\begin{align*}
B[u,u] \geq \theta \|\nabla u\|_{L^2(U)}^2 - \beta \|\nabla u\|_{L^2(U)}^2 = (\theta-\beta)\|\nabla u\|_{L^2(U)}^2.
\end{align*}
Because $\beta < \theta$, the constant $\theta-\beta$ is strictly positive.
[guided]
We now combine the two estimates in the only place where the smallness condition $\beta < \theta$ is used, but first we verify the pieces being added. Since $a_{ij}, b_i, c \in L^\infty(U;\mathbb{R})$ and $u, \partial_{x_i}u \in L^2(U)$ for each $1 \leq i \leq n$, the Cauchy-Schwarz inequality shows that all products appearing below belong to $L^1(U)$. Thus the quantities
\begin{align*}
B_0[u,u] = \int_U \sum_{i,j=1}^n a_{ij}\,\partial_{x_j}u\,\partial_{x_i}u\,d\mathcal{L}^n(x)
\end{align*}
and
\begin{align*}
B_1[u,u] = \int_U \sum_{i=1}^n b_i\,\partial_{x_i}u\,u\,d\mathcal{L}^n(x) + \int_U c\,u^2\,d\mathcal{L}^n(x)
\end{align*}
are well-defined [real numbers](/page/Real%20Numbers), and the definition of $B$ gives
\begin{align*}
B[u,u] = B_0[u,u] + B_1[u,u].
\end{align*}
For the principal part, uniform ellipticity applies pointwise for $\mathcal{L}^n$-a.e. $x \in U$ to the real vector $\xi = \nabla u(x) \in \mathbb{R}^n$. Hence
\begin{align*}
\sum_{i,j=1}^n a_{ij}(x)\,\partial_{x_j}u(x)\,\partial_{x_i}u(x) \geq \theta |\nabla u(x)|^2
\end{align*}
for $\mathcal{L}^n$-a.e. $x \in U$. Integrating this pointwise inequality with respect to $\mathcal{L}^n$ gives
\begin{align*}
B_0[u,u] \geq \theta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
For the lower-order part, the theorem assumes exactly that
\begin{align*}
B_1[u,u] \geq -\beta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Adding these two lower bounds gives
\begin{align*}
B[u,u] = B_0[u,u] + B_1[u,u] \geq \theta \|\nabla u\|_{L^2(U)}^2 - \beta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Therefore
\begin{align*}
B[u,u] \geq (\theta-\beta)\|\nabla u\|_{L^2(U)}^2.
\end{align*}
This is the central estimate: the lower-order terms may be negative, but their negative part is dominated by less than the ellipticity strength of the principal part. The strict inequality $\beta < \theta$ ensures that $\theta-\beta > 0$, so the gradient still has a positive coefficient.
[/guided]
[/step]
[step:Use Poincare's inequality to convert gradient control into $H^1_0(U)$ coercivity]
We equip $H^1_0(U)$ with the norm inherited from $H^1(U)$, namely
\begin{align*}
\|u\|_{H^1(U)}^2 = \|u\|_{L^2(U)}^2 + \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Let $C_P>0$ denote a Poincare constant for $U$, so Poincare's inequality gives
\begin{align*}
\|u\|_{L^2(U)} \leq C_P \|\nabla u\|_{L^2(U)}.
\end{align*}
Squaring both sides gives
\begin{align*}
\|u\|_{L^2(U)}^2 \leq C_P^2 \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Hence
\begin{align*}
\|u\|_{H^1(U)}^2 \leq (1+C_P^2)\|\nabla u\|_{L^2(U)}^2.
\end{align*}
Equivalently,
\begin{align*}
\|\nabla u\|_{L^2(U)}^2 \geq \frac{1}{1+C_P^2}\|u\|_{H^1(U)}^2.
\end{align*}
Combining this with the gradient coercivity estimate gives
\begin{align*}
B[u,u] \geq \frac{\theta-\beta}{1+C_P^2}\|u\|_{H^1(U)}^2.
\end{align*}
Define $\alpha \in \mathbb{R}$ by
\begin{align*}
\alpha = \frac{\theta-\beta}{1+C_P^2}.
\end{align*}
Since $\theta-\beta > 0$ and $1+C_P^2 > 0$, we have $\alpha > 0$. Therefore, for every $u \in H^1_0(U)$,
\begin{align*}
B[u,u] \geq \alpha\|u\|_{H^1(U)}^2.
\end{align*}
Thus $B$ is coercive on $H^1_0(U)$ with respect to the inherited $H^1(U)$ norm.
[guided]
The gradient estimate from the previous step is not yet the stated coercivity estimate, because coercivity on $H^1_0(U)$ is measured using the full inherited $H^1(U)$ norm. We therefore use Poincare's inequality to compare the missing $L^2(U)$ term with the gradient term. Equip $H^1_0(U)$ with
\begin{align*}
\|u\|_{H^1(U)}^2 = \|u\|_{L^2(U)}^2 + \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Let $C_P>0$ be a Poincare constant for $U$. Since $u \in H^1_0(U)$ and Poincare's inequality holds on $U$, we have
\begin{align*}
\|u\|_{L^2(U)} \leq C_P \|\nabla u\|_{L^2(U)}.
\end{align*}
Squaring this estimate gives
\begin{align*}
\|u\|_{L^2(U)}^2 \leq C_P^2 \|\nabla u\|_{L^2(U)}^2.
\end{align*}
Substituting this into the definition of the $H^1(U)$ norm yields
\begin{align*}
\|u\|_{H^1(U)}^2 \leq (1+C_P^2)\|\nabla u\|_{L^2(U)}^2.
\end{align*}
Equivalently,
\begin{align*}
\|\nabla u\|_{L^2(U)}^2 \geq \frac{1}{1+C_P^2}\|u\|_{H^1(U)}^2.
\end{align*}
Combining this with
\begin{align*}
B[u,u] \geq (\theta-\beta)\|\nabla u\|_{L^2(U)}^2
\end{align*}
gives
\begin{align*}
B[u,u] \geq \frac{\theta-\beta}{1+C_P^2}\|u\|_{H^1(U)}^2.
\end{align*}
Define $\alpha \in \mathbb{R}$ by
\begin{align*}
\alpha = \frac{\theta-\beta}{1+C_P^2}.
\end{align*}
Because $\beta<\theta$ and $C_P>0$, we have $\alpha>0$. Hence for every $u\in H^1_0(U)$,
\begin{align*}
B[u,u] \geq \alpha\|u\|_{H^1(U)}^2,
\end{align*}
which is precisely coercivity of $B$ on $H^1_0(U)$ with respect to the inherited $H^1(U)$ norm.
[/guided]
[/step]
