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A compact embedding of Banach spaces $X,Y$ is an injective continuous linear map $i:X\to Y$ that sends every bounded set to a relatively compact subset. Equivalently, every bounded sequence in $X$ has a subsequence converging in $Y$. One identifies $X$ with its image when $i$ is the natural inclusion. The Rellich-Kondrachov theorem supplies many compact embeddings between Sobolev spaces and Lp spaces on bounded domains.

For a symmetric uniformly elliptic divergence-form operator with zero conormal flux and zero reaction, Lax-Milgram theorem constructs the shifted inverse $K:L^2\to H^1$. The compact Rellich-Kondrachov theorem embedding makes it compact on $L^2$, and symmetry makes it self-adjoint. If the homogeneous kernel consists of constants, then $\ker (I-K)$ is the constant line and $K1=1$. The Fredholm alternative makes solvability equivalent to $Kf$ being orthogonal to constants, or $⟨Kf,1⟩=⟨f,K1⟩=\int f=0$. The weak solution is unique up to an additive constant. For a disconnected domain the identical argument gives one compatibility condition per connected component.

We apply the closed-bounded-domain version of the Schauder fixed-point theorem in the Banach space $L^2(\Omega )$. Write $m=|\Omega |$ and use $G,T,C_P$ from part (b)(ii). ChooseThe set $C$ is nonempty, bounded, closed and a convex set. For closedness of its positivity condition, if $w_j\ge 0$ and $w_j\to w$ in $L^2$, then pointwise $w^{-}\le |w-w_j|$, whence $\Vert w^{-}{\Vert }_2\to 0$.

The weak maximum principle already gives $Tw\ge 0$. Our estimates give, for $w\in C$,Thus $T(C)\subset C$.

To prove continuity, use the pointwise square root estimate $|\sqrt{s}-\sqrt{t}{|}^2\le |s-t|$ for $s,t\ge 0$. The Cauchy-Schwarz inequality then givesHence $T$ is a continuous map on $C$, even though the square root need not have a bounded derivative near zero.

Finally,The Rellich-Kondrachov theorem says that $H_0^1(\Omega )$ embeds compactly into $L^2(\Omega )$ on a bounded smooth domain. For $n>2$, $2<2n/(n-2)$, and for $n=2$ every finite target exponent is available, so the required compact embedding holds in every dimension in the question. Thus $T(C)$ is relatively compact in $L^2(\Omega )$.

All hypotheses of the Schauder fixed-point theorem are met. It gives $u=Tu\in C$, and part (b)(ii) converts this fixed point into the required weak solution:Only existence is needed. The construction does not assume that $\Omega$ is small or that $T$ is a contraction mapping.

The Lax-Milgram theorem states that a bounded bilinear form $a$ on a real Hilbert space $V$, satisfying $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, has for each $F\in V^{\ast }$ a unique $u\in V$ with $a(u,v)=F(v)$ for all $v$. Moreover $\Vert u{\Vert }_V\le \Vert F\Vert /\alpha$.

For a compact operator $K$ on a Hilbert space, the Fredholm alternative says that $I-\lambda K$ has closed range, finite-dimensional kernel and cokernel of equal dimension, andThus a zero kernel gives unique solvability for every right-hand side; a nonzero kernel gives solvability exactly under the adjoint orthogonality conditions, and then all solutions differ by a kernel vector.

Take $V=H_0^1(\Omega )$. Gårding's inequality supplies constants $a>0$, $b\ge 0$ with $B[v,v]\ge a\Vert v{\Vert }_{H^1}^2-b\Vert v{\Vert }_2^2$. Therefore for $\lambda \le -b$ the shifted form $B_{\lambda }[u,v]=B[u,v]-\lambda (u,v{)}_{L^2}$ is coercive. It is bounded for every fixed $\lambda$, and $v\mapsto (f,v)$ is bounded on $V$. Lax-Milgram gives unique weak solvability for $\lambda \le -\gamma$, with $\gamma =b$. The inequality itself follows, for example, by absorbing the drift cross-term $\Vert b{\Vert }_{\infty }\Vert \nabla v{\Vert }_2\Vert v{\Vert }_2$ using Young's inequality and allowing a lower $L^2$ correction for $c$.

For the compact inverse and adjoint of a coercive elliptic operator, under the additional zero-shift coercivity assumption, define $Gg\in V$ by $B[Gg,v]=(g,v)$ for all $v\in V$, and let $K$ be this solution viewed in $L^2$. The solution operator is linear and bounded into $V$, and the Rellich-Kondrachov compact embedding $V\hookrightarrow L^2$ makes $K$ compact. Rewriting the weak equation as $B[u,v]=(f+\lambda u,v)$ givesConversely an $L^2$ solution of this operator equation lies in $V$, because it equals $K(f+\lambda u)$, and satisfies the weak differential equation.

For constant drift coefficients and zero Dirichlet traces, integration by parts gives $\int b_i{D}_{i}vu=-\int b_{i}v{D}_{i}u$. Symmetry of the principal coefficients therefore makes $B^{\ast }[u,v]=B[v,u]$ the unshifted weak form of the displayed formal adjoint. The shifted adjoint equation requiresThe spectral shift in an adjoint weak formulation is essential: the printed sentence omitting it is false for nonzero $\lambda$. For instance take a smooth ball, $L=-\Delta +1$, any nonzero $u\in C_c^{\infty }(\Omega )$ and $f=-\Delta u+u$. The printed unshifted identity holds, but $(L^{\ast }-\lambda )u=f-\lambda u\ne f$ whenever $\lambda \ne 0$. This counterexample also meets zero-shift coercivity and constant-drift assumptions. The corrected identity above is used for the remainder of the argument.

Let $G_{\ast }g\in V$ solve $B^{\ast }[G_{\ast }g,v]=(g,v)$, which exists by Lax-Milgram because the diagonal coercivity is unchanged. For $p,q\in L^2$,Thus $G_{\ast }$, viewed as an $L^2$ operator, is exactly $K^{\ast }$. Applying this inverse to the corrected shifted adjoint equation givesand the converse follows from its range lying in $V$, just as before.

Finally let the original homogeneous equation have a nonzero solution. Its parameter is nonzero, since the unshifted problem is coercive. Fredholm gives solvability precisely when $(Kf,w)=0$ for every $w\in \ker (I-\lambda K^{\ast })$. Such a $w$ satisfies $K^{\ast }w=w/\lambda$, so $(Kf,w)=(f,w)/\lambda$. Hence the necessary and sufficient condition isWhen this holds, the solutions are an affine translate of the original finite-dimensional homogeneous solution space. The compatibility space is the adjoint kernel; a drift term does not generally permit replacing it by the original kernel.

The spectral theorem for compact self-adjoint operators used here concerns a compact self-adjoint operator: its nonzero eigenvalues are real, have finite multiplicity and can accumulate only at zero; the orthogonal complement of its kernel is the closed span of its orthonormal eigenvectors. Self-adjointness is essential for this eigenbasis statement.

Let $Kf$ be the unique zero-parameter weak solution. Uniqueness gives linearity. Testing with $Kf$ and using coercivity gives $\Vert Kf{\Vert }_{H^1}\le {\nu }^{-1}\Vert f{\Vert }_2$. The Rellich-Kondrachov theorem makes the embedding $H_0^1(\Omega )\hookrightarrow L^2(\Omega )$ compact on this bounded $C^1$ domain, so $K$ is compact on $L^2$. Symmetry givesIt is self-adjoint and positive. It is injective: if $Kf=0$, the weak equation pairs $f$ to zero with all $H_0^1$ tests, dense in $L^2$, so $f=0$. Moreover $⟨Kf,f⟩=B[Kf,Kf]>0$ for $f\ne 0$.

The parameter equation is equivalent to $u=K(f+\lambda u)$, that isA solution of this operator equation in $L^2$ lies in $H_0^1$, because its right-side representation is $K(f+\lambda u)$, and the equivalence follows in both directions. For $\lambda \le 0$, the form $B[u,v]-\lambda ⟨u,v⟩$ remains coercive; the preceding minimization argument gives a unique weak solution.

The spectral theorem now supplies a complete orthonormal basis $({\varphi }_j)$ of $L^2$ with $K{\varphi }_j={\kappa }_j{\varphi }_j$, ${\kappa }_j>0$, and ${\kappa }_j\to 0$. The elliptic eigenvalues are ${\lambda }_j=1/{\kappa }_j>0$, with finite multiplicity and tending to infinity, and $L{\varphi }_j={\lambda }_j{\varphi }_j$ weakly. If $\lambda$ is not one of them, thenThis converges even in $H_0^1$: its squared energy norm is ${\sum }_j{\lambda }_j|⟨f,{\varphi }_j⟩{|}^2/|{\lambda }_j-\lambda {|}^2$, finite because the multiplier is bounded over $j$. Partial sums solve the weak equation with the corresponding $L^2$ projections of $f$. Passing to the limit proves the equation for every test; the same coefficients force uniqueness. Thus the expansion represents the unique weak solution away from the elliptic eigenvalues.

The weak formulation isThe boundary condition is encoded by $u\in H_0^1$. The Sobolev embedding in dimension three gives $H_0^1\hookrightarrow L^6$, hence also $L^4,L^3,L^2$ on the bounded domain. All energy terms and all the displayed pairings are therefore finite.

Pointwise, $t^4/4-t^3/3\ge t^4/8-C$ for a constant $C$, since the difference is a quartic with positive leading coefficient. The Poincaré inequality and Young inequality give $\int fv\ge -\frac{1}{4}\Vert Dv{\Vert }_2^2-C_f$. ConsequentlyThus the infimum is finite and a minimizing sequence is bounded in $H_0^1$. Extract a weakly convergent subsequence there. The Rellich-Kondrachov theorem states that on a bounded $C^1$ domain in dimension three the embedding $H^1\hookrightarrow L^q$ is compact for $q<6$. A subsequence therefore converges strongly in $L^2,L^3,L^4$ to the same limit $u$. The potential terms and forcing converge, while the gradient norm is weakly lower semicontinuous. The weak limit retains zero trace. Hence $I(u)\le \mathrm{liminf}I(v_j)$ and $u$ attains the absolute minimum.

For any $\varphi \in H_0^1$, the function $t\mapsto I(u+t\varphi )$ has a minimum at zero. Differentiating its polynomial terms gives $\int [Du\cdot D\varphi +(u^3-u^2+f)\varphi ]=0$. These differentiations are justified by the Sobolev integrability already noted. This is the weak formulation, proving every energy minimizer is a weak solution.

This is the zero-boundary special case of the Rellich-Kondrachov theorem. For every bounded open $\Omega \subset {\mathbb{R}}^d$, the inclusion $H_0^1(\Omega )\hookrightarrow L^2(\Omega )$ is compact.