= Solution
{c}
Use the <Sobolev space with a partial Dirichlet condition>
$$
\boxed{V=\{v\in H^1(\Omega):Tv=0\text{ almost everywhere on }\Gamma_D\},}
$$
where $T$ is the <Sobolev trace operator>. The restricted <Sobolev trace operator> is continuous into $L^2(\Gamma_D)$; its <kernel> is consequently a <closed vector subspace> of $H^1(\Omega)$. Thus $V$, with the first-order <Sobolev norm>, is a <Hilbert space>. The endpoint convention for $\Gamma_D$ and $\Gamma_N$ does not affect this definition: finitely many boundary points have zero boundary length. It would be inappropriate to impose pointwise boundary values on arbitrary $H^1$ functions in two dimensions.
For a smooth solution, <Green's first identity> gives
$$
\int_\Omega(-\Delta u+u)v
=\int_\Omega(\nabla u\cdot\nabla v+uv)-\int_{\partial\Omega}\partial_nu\,Tv.
$$
The boundary term vanishes because $Tv=0$ on $\Gamma_D$ and the <normal derivative> is zero on $\Gamma_N$. This suggests the <weak formulation>
$$
\boxed{u\in V,\qquad a(u,v)=\ell(v)\quad(v\in V),\qquad
a(p,v)=(\nabla p,\nabla v)_0+(p,v)_0,\quad\ell(v)=(f,v)_0.}
$$
Here $a$ is exactly the first-order <Sobolev inner product>. In particular the <Cauchy-Schwarz inequality> gives
$$
|a(p,v)|\leq\|p\|_1\|v\|_1,\qquad a(v,v)=\|v\|_1^2,\qquad
|\ell(v)|\leq\|f\|_0\|v\|_1.
$$
Thus $a$ is a <bounded bilinear form> and a <coercive bilinear form>, both with constant one. The <Lax-Milgram theorem> proves existence and uniqueness, together with
$$
\boxed{\|u\|_1\leq\|f\|_0.}
$$
The positive reaction term already controls the entire <Sobolev norm>, so no <Poincaré inequality> is needed here. The zero <Neumann boundary condition> is the natural part of the <weak formulation>; it is not an additional restriction on the test space. For this <weak solution>, $-\Delta u+u=f$ holds as a <distributional identity>, and the zero <normal derivative> on $\Gamma_N$ is understood through the corresponding <Green's first identity>.
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