[proofplan]
The coercive estimate turns $Q_q$ itself into a Hilbert norm on the form domain $\mathcal D_q$, equivalent to the usual closed-form graph norm. For each datum $f \in H_q$, the functional $v \mapsto (f,v)_{H_q}$ is bounded with respect to this coercive norm, so Lax-Milgram produces a unique $u \in \mathcal D_q$ solving the variational equation $Q_q(u,v)=(f,v)_{H_q}$. The definition of the operator associated to the closed form then identifies this variational solution as an element of $\operatorname{Dom}(\Box_q)$ satisfying $\Box_q u=f$, and the original coercive estimate gives the stated operator norm bound.
[/proofplan]
[step:Turn the form into an equivalent Hilbert norm on $\mathcal D_q$]
Define
\begin{align*}
\|u\|_{Q_q} := Q_q(u,u)^{1/2}
\end{align*}
for $u \in \mathcal D_q$. The assumed estimate implies that $\|u\|_{Q_q}=0$ forces $\|u\|_{H_q}=0$, hence $u=0$ in $H_q$. Thus $\|\cdot\|_{Q_q}$ is a norm.
Since $Q_q$ is closed, $\mathcal D_q$ is complete for the graph norm
\begin{align*}
\|u\|_{\mathrm{gr}} := \left(\|u\|_{H_q}^2+Q_q(u,u)\right)^{1/2}.
\end{align*}
The coercive estimate gives
\begin{align*}
\|u\|_{\mathrm{gr}}^2
=
\|u\|_{H_q}^2+Q_q(u,u)
\le
(C+1)Q_q(u,u)
=
(C+1)\|u\|_{Q_q}^2.
\end{align*}
The reverse inequality $\|u\|_{Q_q}\le \|u\|_{\mathrm{gr}}$ follows from the definition of $\|\cdot\|_{\mathrm{gr}}$. Therefore $\|\cdot\|_{Q_q}$ and $\|\cdot\|_{\mathrm{gr}}$ are equivalent norms on $\mathcal D_q$. Since $\mathcal D_q$ is complete for $\|\cdot\|_{\mathrm{gr}}$, it is complete for $\|\cdot\|_{Q_q}$.
[guided]
We want to use $Q_q$ as the Hilbert-space inner product on the form domain, so first we must check that the form seminorm is actually a complete norm. Define
\begin{align*}
\|u\|_{Q_q} := Q_q(u,u)^{1/2}
\end{align*}
for $u \in \mathcal D_q$. If $\|u\|_{Q_q}=0$, then $Q_q(u,u)=0$. The assumed coercive estimate gives
\begin{align*}
\|u\|_{H_q}^2 \le C Q_q(u,u)=0,
\end{align*}
so $u=0$ as an element of $H_q$. Hence $\|\cdot\|_{Q_q}$ is a norm on $\mathcal D_q$.
The closedness of the form means that $\mathcal D_q$ is complete for the graph norm
\begin{align*}
\|u\|_{\mathrm{gr}} := \left(\|u\|_{H_q}^2+Q_q(u,u)\right)^{1/2}.
\end{align*}
The coercive estimate controls the $H_q$ part of this graph norm by the form part:
\begin{align*}
\|u\|_{H_q}^2 \le C Q_q(u,u).
\end{align*}
Therefore, for every $u \in \mathcal D_q$,
\begin{align*}
\|u\|_{\mathrm{gr}}^2
=
\|u\|_{H_q}^2+Q_q(u,u)
\le
(C+1)Q_q(u,u)
=
(C+1)\|u\|_{Q_q}^2.
\end{align*}
This proves that the graph norm is bounded above by $\sqrt{C+1}\,\|\cdot\|_{Q_q}$.
Conversely, the definition of the graph norm gives
\begin{align*}
\|u\|_{Q_q}^2
=
Q_q(u,u)
\le
\|u\|_{H_q}^2+Q_q(u,u)
=
\|u\|_{\mathrm{gr}}^2.
\end{align*}
Thus $\|u\|_{Q_q}\le \|u\|_{\mathrm{gr}}$ for every $u \in \mathcal D_q$. The two inequalities show that $\|\cdot\|_{Q_q}$ and $\|\cdot\|_{\mathrm{gr}}$ are equivalent norms on $\mathcal D_q$.
Now take a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal D_q$ that is Cauchy with respect to $\|\cdot\|_{Q_q}$. The estimate
\begin{align*}
\|u_k-u_m\|_{\mathrm{gr}}
\le
\sqrt{C+1}\,\|u_k-u_m\|_{Q_q}
\end{align*}
shows that $(u_k)_{k=1}^{\infty}$ is Cauchy with respect to $\|\cdot\|_{\mathrm{gr}}$. Since $\mathcal D_q$ is complete for $\|\cdot\|_{\mathrm{gr}}$, there exists $u \in \mathcal D_q$ such that
\begin{align*}
\|u_k-u\|_{\mathrm{gr}} \to 0.
\end{align*}
Using the reverse estimate $\|w\|_{Q_q}\le \|w\|_{\mathrm{gr}}$ with $w=u_k-u$, we obtain
\begin{align*}
\|u_k-u\|_{Q_q}\le \|u_k-u\|_{\mathrm{gr}} \to 0.
\end{align*}
Therefore $\mathcal D_q$ is complete for $\|\cdot\|_{Q_q}$. This is the point at which the two hypotheses combine: closedness gives completeness in the graph norm, and coercivity lets the form norm alone carry the same topology.
[/guided]
[/step]
[step:Represent the datum by a variational solution]
Fix $f \in H_q$. Define the conjugate-linear functional
\begin{align*}
\Lambda_f: \mathcal D_q &\to \mathbb C, \\
v &\mapsto (f,v)_{H_q}.
\end{align*}
For every $v \in \mathcal D_q$, the [Cauchy-Schwarz inequality](/theorems/432) in $H_q$ and the coercive estimate give
\begin{align*}
|\Lambda_f(v)|
=
|(f,v)_{H_q}|
\le
\|f\|_{H_q}\|v\|_{H_q}
\le
C^{1/2}\|f\|_{H_q}\|v\|_{Q_q}.
\end{align*}
Hence $\Lambda_f$ is bounded on the [Hilbert space](/page/Hilbert%20Space) $(\mathcal D_q,\|\cdot\|_{Q_q})$.
By the [Lax-Milgram theorem](/theorems/91) applied to the coercive Hermitian form $Q_q$ on $\mathcal D_q$ and the bounded functional $\Lambda_f$ (citing a result not yet in the wiki: [Lax-Milgram Theorem](/theorems/91)), there exists a unique $u_f \in \mathcal D_q$ such that
\begin{align*}
Q_q(u_f,v)=(f,v)_{H_q}
\end{align*}
for every $v \in \mathcal D_q$.
[guided]
Fix a datum $f \in H_q$. We want to solve $\Box_q u=f$, but $\Box_q$ is defined through the form $Q_q$, so the natural first target is the weak equation
\begin{align*}
Q_q(u,v)=(f,v)_{H_q}
\end{align*}
for every test element $v \in \mathcal D_q$.
Define
\begin{align*}
\Lambda_f: \mathcal D_q &\to \mathbb C, \\
v &\mapsto (f,v)_{H_q}.
\end{align*}
The [Cauchy-Schwarz inequality](/theorems/432) in the [Hilbert space](/page/Hilbert%20Space) $H_q$ gives
\begin{align*}
|\Lambda_f(v)|
=
|(f,v)_{H_q}|
\le
\|f\|_{H_q}\|v\|_{H_q}.
\end{align*}
The coercive estimate then converts the $H_q$ norm of $v$ into the form norm:
\begin{align*}
\|v\|_{H_q}
\le
C^{1/2}Q_q(v,v)^{1/2}
=
C^{1/2}\|v\|_{Q_q}.
\end{align*}
Combining the two inequalities,
\begin{align*}
|\Lambda_f(v)|
\le
C^{1/2}\|f\|_{H_q}\|v\|_{Q_q}.
\end{align*}
Thus $\Lambda_f$ is bounded on the [Hilbert space](/page/Hilbert%20Space) $(\mathcal D_q,\|\cdot\|_{Q_q})$.
We now apply the [Lax-Milgram theorem](/theorems/91) to the [Hilbert space](/page/Hilbert%20Space) $\mathcal D_q$ with inner product $Q_q$ and to the bounded functional $\Lambda_f$ (citing a result not yet in the wiki: [Lax-Milgram Theorem](/theorems/91)). The hypotheses have been verified: $\mathcal D_q$ is complete for $\|\cdot\|_{Q_q}$, the form $Q_q$ is the Hilbert inner product on this space, and $\Lambda_f$ is bounded. Therefore there exists a unique $u_f \in \mathcal D_q$ satisfying
\begin{align*}
Q_q(u_f,v)=(f,v)_{H_q}
\end{align*}
for all $v \in \mathcal D_q$.
[/guided]
[/step]
[step:Identify the variational solution with the operator solution]
By the definition of the operator associated to $Q_q$, an element $u \in \mathcal D_q$ belongs to $\operatorname{Dom}(\Box_q)$ exactly when there exists $g \in H_q$ such that
\begin{align*}
Q_q(u,v)=(g,v)_{H_q}
\end{align*}
for every $v \in \mathcal D_q$, and then $\Box_q u=g$. For $u=u_f$, the previous step gives this identity with $g=f$. Hence
\begin{align*}
u_f \in \operatorname{Dom}(\Box_q),
\qquad
\Box_q u_f=f.
\end{align*}
Define
\begin{align*}
N_q: H_q &\to \operatorname{Dom}(\Box_q), \\
f &\mapsto u_f.
\end{align*}
This map is well-defined by existence and uniqueness of $u_f$.
[guided]
The weak solution becomes a strong operator solution because $\Box_q$ was defined as the operator represented by the closed form $Q_q$. The definition says: if $u \in \mathcal D_q$ and there is some $g \in H_q$ such that
\begin{align*}
Q_q(u,v)=(g,v)_{H_q}
\end{align*}
for every $v \in \mathcal D_q$, then $u$ lies in $\operatorname{Dom}(\Box_q)$ and $\Box_q u=g$.
For the element $u_f$ constructed by Lax-Milgram, we have exactly this identity with $g=f$:
\begin{align*}
Q_q(u_f,v)=(f,v)_{H_q}
\end{align*}
for every $v \in \mathcal D_q$. Therefore
\begin{align*}
u_f \in \operatorname{Dom}(\Box_q),
\qquad
\Box_q u_f=f.
\end{align*}
We may consequently define
\begin{align*}
N_q: H_q &\to \operatorname{Dom}(\Box_q), \\
f &\mapsto u_f.
\end{align*}
The uniqueness part of Lax-Milgram ensures that this definition assigns only one element of $\operatorname{Dom}(\Box_q)$ to each $f \in H_q$.
[/guided]
[/step]
[step:Prove injectivity and conclude that $N_q$ is the inverse]
The previous step proves that $\Box_q N_q f=f$ for every $f \in H_q$, so $\Box_q$ is surjective from $\operatorname{Dom}(\Box_q)$ onto $H_q$.
If $u \in \operatorname{Dom}(\Box_q)$ and $\Box_q u=0$, then by the associated-form identity,
\begin{align*}
Q_q(u,v)=(\Box_q u,v)_{H_q}=0
\end{align*}
for every $v \in \mathcal D_q$. Taking $v=u$ gives
\begin{align*}
Q_q(u,u)=0.
\end{align*}
The coercive estimate gives
\begin{align*}
\|u\|_{H_q}^2 \le C Q_q(u,u)=0,
\end{align*}
so $u=0$ in $H_q$. Hence $\Box_q$ is injective. Therefore $\Box_q:\operatorname{Dom}(\Box_q)\to H_q$ is bijective, and $N_q$ is its inverse.
[/step]
[step:Derive the boundedness estimate]
Fix $f \in H_q$ and set $u=N_q f$. If $u=0$, then the desired estimate holds. If $u\ne 0$, use the variational identity with $v=u$:
\begin{align*}
Q_q(u,u)=(f,u)_{H_q}.
\end{align*}
Taking absolute values and applying Cauchy-Schwarz in $H_q$,
\begin{align*}
Q_q(u,u)
\le
\|f\|_{H_q}\|u\|_{H_q}.
\end{align*}
Combining this with the coercive estimate,
\begin{align*}
\|u\|_{H_q}^2
\le
C Q_q(u,u)
\le
C\|f\|_{H_q}\|u\|_{H_q}.
\end{align*}
Since $\|u\|_{H_q}>0$, division by $\|u\|_{H_q}$ gives
\begin{align*}
\|N_q f\|_{H_q}
=
\|u\|_{H_q}
\le
C\|f\|_{H_q}.
\end{align*}
Together with $\Box_qN_qf=f$, this is the asserted bounded inverse estimate.
[/step]
