[proofplan]
Subtract the fixed trace representative $g$ and write each admissible function as $u_k=g+v_k$ with $v_k\in H^1_0(U)$. The assumed uniform Dirichlet-energy bound controls $\nabla u_k$, and hence controls $\nabla v_k=\nabla u_k-\nabla g$ uniformly in $L^2(U)$. The zero trace Poincare inequality on $H^1_0(U)$ then converts this gradient bound into a full $H^1(U)$ bound for $v_k$. Adding back the fixed function $g$ gives the desired uniform $H^1(U)$ bound for $u_k$.
[/proofplan]
[step:Decompose each admissible function into fixed trace data and a zero trace part]
For each $k\in\mathbb N$, the condition $u_k\in\mathcal A_g$ means precisely that $u_k-g\in H^1_0(U)$. Define
\begin{align*}
v_k:=u_k-g\in H^1_0(U).
\end{align*}
Then
\begin{align*}
u_k=g+v_k
\end{align*}
as elements of $H^1(U)$.
[/step]
[step:Use the Dirichlet energy bound to control the zero trace gradients]
Define the finite constant
\begin{align*}
M:=\sup_{k\in\mathbb N}\int_U |\nabla u_k|^2\,d\mathcal L^n.
\end{align*}
In this proof, $L^2(U)$ denotes the space of real-valued square-integrable functions on $U$ with respect to $\mathcal L^n$, and $L^2(U;\mathbb R^n)$ denotes the space of square-integrable maps from $U$ to $\mathbb R^n$ with respect to $\mathcal L^n$. Since $g\in H^1(U)$, define
\begin{align*}
G:=\|\nabla g\|_{L^2(U)}=\left(\int_U |\nabla g|^2\,d\mathcal L^n\right)^{1/2}<\infty.
\end{align*}
By linearity of [weak derivatives](/page/Weak%20Derivative),
\begin{align*}
\nabla v_k=\nabla u_k-\nabla g
\end{align*}
in $L^2(U;\mathbb R^n)$. Applying the triangle inequality in $L^2(U;\mathbb R^n)$ gives, for every $k\in\mathbb N$,
\begin{align*}
\|\nabla v_k\|_{L^2(U)}\le \|\nabla u_k\|_{L^2(U)}+\|\nabla g\|_{L^2(U)}\le M^{1/2}+G.
\end{align*}
Define
\begin{align*}
B:=M^{1/2}+G.
\end{align*}
Then
\begin{align*}
\sup_{k\in\mathbb N}\|\nabla v_k\|_{L^2(U)}\le B.
\end{align*}
[guided]
The reason for subtracting $g$ is that Poincare's inequality applies to functions with zero trace, not to arbitrary elements of the affine class. For each $k\in\mathbb N$, define
\begin{align*}
v_k:=u_k-g.
\end{align*}
Because $u_k\in\mathcal A_g=g+H^1_0(U)$, this definition gives $v_k\in H^1_0(U)$.
Now we estimate the gradient of $v_k$. Define
\begin{align*}
M:=\sup_{k\in\mathbb N}\int_U |\nabla u_k|^2\,d\mathcal L^n.
\end{align*}
The hypothesis says exactly that $M<\infty$. In this proof, $L^2(U)$ denotes the space of real-valued square-integrable functions on $U$ with respect to $\mathcal L^n$, and $L^2(U;\mathbb R^n)$ denotes the space of square-integrable maps from $U$ to $\mathbb R^n$ with respect to $\mathcal L^n$. Since $g\in H^1(U)$, its weak gradient is square-integrable, so the quantity
\begin{align*}
G:=\|\nabla g\|_{L^2(U)}=\left(\int_U |\nabla g|^2\,d\mathcal L^n\right)^{1/2}
\end{align*}
is finite.
[Weak differentiation](/page/Weak%20Derivative) is linear on $H^1(U)$, so
\begin{align*}
\nabla v_k=\nabla(u_k-g)=\nabla u_k-\nabla g
\end{align*}
in $L^2(U;\mathbb R^n)$. Taking the $L^2(U)$ norm and applying the triangle inequality gives
\begin{align*}
\|\nabla v_k\|_{L^2(U)}\le \|\nabla u_k\|_{L^2(U)}+\|\nabla g\|_{L^2(U)}.
\end{align*}
The first term is uniformly bounded because
\begin{align*}
\|\nabla u_k\|_{L^2(U)}=\left(\int_U |\nabla u_k|^2\,d\mathcal L^n\right)^{1/2}\le M^{1/2}.
\end{align*}
Therefore, with
\begin{align*}
B:=M^{1/2}+G,
\end{align*}
we have
\begin{align*}
\|\nabla v_k\|_{L^2(U)}\le B
\end{align*}
for every $k\in\mathbb N$. This is the uniform gradient bound needed for the zero trace sequence.
[/guided]
[/step]
[step:Apply the zero trace Poincare inequality to obtain a full $H^1$ bound for the zero trace parts]
Since $U$ is a bounded [open set](/page/Open%20Set), the zero trace Poincare inequality for Sobolev functions gives a constant $C_P=C_P(U)>0$ such that every $v\in H^1_0(U)$ satisfies
\begin{align*}
\|v\|_{L^2(U)}\le C_P\|\nabla v\|_{L^2(U)}.
\end{align*}
Applying this to $v_k$ gives
\begin{align*}
\|v_k\|_{L^2(U)}\le C_P B
\end{align*}
for every $k\in\mathbb N$. Using the standard $H^1(U)$ norm from the theorem statement,
\begin{align*}
\|v_k\|_{H^1(U)}^2=\|v_k\|_{L^2(U)}^2+\|\nabla v_k\|_{L^2(U)}^2\le C_P^2B^2+B^2=(C_P^2+1)B^2.
\end{align*}
Hence
\begin{align*}
\sup_{k\in\mathbb N}\|v_k\|_{H^1(U)}\le (C_P^2+1)^{1/2}B<\infty.
\end{align*}
[/step]
[step:Add back the fixed trace representative to bound the original sequence]
Since $g\in H^1(U)$ is fixed and $u_k=g+v_k$, the triangle inequality in $H^1(U)$ gives
\begin{align*}
\|u_k\|_{H^1(U)}\le \|g\|_{H^1(U)}+\|v_k\|_{H^1(U)}.
\end{align*}
Using the uniform bound for $(v_k)_{k=1}^{\infty}$ obtained above,
\begin{align*}
\|u_k\|_{H^1(U)}\le \|g\|_{H^1(U)}+(C_P^2+1)^{1/2}(M^{1/2}+G)
\end{align*}
for every $k\in\mathbb N$. The right-hand side is finite and independent of $k$. Consequently,
\begin{align*}
\sup_{k\in\mathbb N}\|u_k\|_{H^1(U)}<\infty.
\end{align*}
Thus $(u_k)_{k=1}^{\infty}$ is bounded in $H^1(U)$.
[/step]
