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Take $V_h={\mathcal{U}}_h\times {\mathcal{W}}_h\subset V$. The Galerkin method asks for $U_h=(u_h,w_h)\in V_h$ such thatSince a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. The bounded bilinear form and coercive bilinear form estimates in part (a) restrict to $V_h$ with the same constants. The Lax-Milgram theorem therefore gives exactly one $U_h$ for every $h$.

Put $E=U-U_h=(e_u,e_w)$. Subtracting the continuous and discrete weak formulations gives Galerkin orthogonality:For any $Y_h=(v_h,z_h)\in V_h$, the difference $Y_h-U_h$ also belongs to $V_h$, soThe last step uses the constant-one Cauchy-Schwarz inequality estimate from part (a). Bounding the four terms separately by a sum would obscure the precise constant requested here.

The two minimization problems over ${\mathcal{U}}_h$ and ${\mathcal{W}}_h$ are independent. Each has a best approximation because its vector subspace is finite-dimensional. Consequently, ifthen the preceding inequality gives exactlyBoth errors belong to the zero-boundary Sobolev space. Applying the Poincaré inequality componentwise now yieldsWhen $U=U_h$ the desired estimate is automatic; otherwise divide by the nonzero norm. ThusThe constant $C=1+K$ is independent of $h$. This is the Céa lemma estimate for this particular skew-coupled elliptic system.

For a conforming triangular finite element mesh, takeThis is the quadratic conforming finite element space. Across a common edge of a polygon, the restrictions from neighboring triangles are univariate quadratic polynomials. Their agreement at the two endpoints and the midpoint makes their difference a degree-at-most-two polynomial with three distinct roots of a polynomial, so it vanishes on the entire common edge of a polygon. Thus matching the local finite element degrees of freedom produces the required continuity. Similarly, setting the three nodal values on every boundary edge carrying the Dirichlet boundary condition to zero makes its entire restriction vanish.

Continuity across element interfaces means the piecewise gradients give the global weak gradient, with no jump contribution, so $V_h\subset V$. There are finitely many local finite element degrees of freedom, hence $V_h$ is finite-dimensional. The Galerkin method seeksThe bounded bilinear form and coercive bilinear form estimates from part (b) remain valid on $V_h$. Since a finite-dimensional subspace is closed, the Lax-Milgram theorem gives exactly one $u_h$. Alternatively, the stiffness matrix plus the mass matrix is a positive-definite matrix in any basis of $V_h$. The same existence and uniqueness conclusion holds for any finite-dimensional vector subspace of this full space.

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets; equivalently, every countable intersection of open dense sets is dense.

Let $E_n=\mathrm{span}\{e_1,\dots ,e_n\}$ in the space of finitely supported sequences. Each $E_n$ is a closed linear subspace for every norm, since a finite-dimensional subspace is closed. Henceis a closed linear subspace of the Banach space $X$. Every finitely supported sequence lies in some $E_n$, so $X={\bigcup }_{n\ge 1}{F}_n$. By the Baire category theorem, some $F_N$ has nonempty interior.

A linear subspace with nonempty interior is the whole space: translate a ball in it to zero and scale any vector into that ball. Thus $F_N=X$, giving $T(X)\subseteq E_N$ andThe proof uses no particular choice of norm on the space of finitely supported sequences.

By the mean value theorem, a continuously differentiable function with zero derivative on $[-1,1]$ is constant. Thus the kernel of the differentiation operator isIt is a closed set by the finite-dimensional subspace is closed theorem, applied in the normed space $(C^1([-1,1]),\Vert \cdot {\Vert }_{\infty })$. It is proper because the function $x\mapsto x$ has nonzero derivative.

For $n\ge 1$, take $u_n(x)=\sin (nx)$. Its supremum norm is at most one, whereas $u_n^{\prime }(0)=n$ and $|u_n^{\prime }(x)|\le n$. Hence

Let $V$ be a real Hilbert space, $a:V\times V\to \mathbb{R}$ a bounded bilinear form, and $\ell \in V^{\prime }$ a bounded linear functional. Assume constants $C<\infty$, $\alpha >0$ satisfyThe second hypothesis says that $a$ is a coercive bilinear form. The Lax-Milgram theorem then supplies a unique weak solution $u\in V$ with $a(u,v)=\ell (v)$ for every $v\in V$, and $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$.

For a finite-dimensional trial linear subspace $V_h\subset V$, the same boundedness and coercivity constants apply to the restricted form. Since a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. Thus there is a unique Galerkin method solution $u_h\in V_h$ satisfying $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. The Galerkin orthogonality $a(u-u_h,v_h)=0$ gives the Céa lemma:In particular, approximation by the trial spaces implies numerical convergence. For a complex Hilbert space, replace bilinearity by a sesquilinear form, take $\ell$ antilinear in the test variable, and require $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_V^2$.

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.

Suppose for contradiction that the vector space of univariate polynomials $P$ were complete in the given norm. For $n\ge 0$, letEach $P_n$ is finite-dimensional and therefore closed by the fact that a finite-dimensional subspace is closed. It is a proper linear subspace of $P$, so it has empty interior: if a linear subspace contained any open ball, translation and scalar multiplication would make it contain every vector. Hence every $P_n$ is a nowhere dense set.

But every polynomial has finite degree, soThis writes the supposed complete metric space $P$ as a countable union of nowhere-dense closed subsets, contradicting the Baire category theorem. Thus no norm can make $P$ complete. Equivalently, this is the countable-basis obstruction that a Banach space has uncountable Hamel dimension.