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Write $Z_t=z+B_t^1+i{B}_t^2$, where the coordinates are independent standard Brownian motions, and write $f=u+iv$. The Cauchy-Riemann equations imply $\Delta u=\Delta v=0$. Itô's formula therefore makescontinuous local martingales. Their quadratic variations and quadratic covariation areIndeed the two gradient norms are equal to $|f^{\prime }|$, and their dot product vanishes by the Cauchy-Riemann equations.

The clock is finite at every finite time, since a continuous Brownian path has compact range there and $f^{\prime }$ is continuous. To show its lifetime is infinite, choose a small disc on which $|f^{\prime }{|}^2\ge c>0$; such a disc exists because the entire function is nonconstant. A supporting recurrence result is that planar Brownian motion spends infinite total time in every nonempty open disc almost surely. One proof combines recurrent returns to a smaller disc with the strong Markov property: each visit has a uniformly positive chance of lasting a fixed positive time in the larger disc. Equivalently, use the divergence of a positive planar Brownian occupation integral. ThereforeFinally the common-clock time change of orthogonal local martingales states that two continuous local martingales starting at zero with common bracket $A$, zero cross bracket, and $A_{\infty }=\infty$ are the coordinates of a planar Brownian motion evaluated at $A_t$. Applying it gives a planar Brownian motion $\tilde{Z}$ starting at zero withThis proves conformal invariance of planar Brownian motion in the entire-function setting, including the divergence of the conformal Brownian clock for an entire function.

The odd-power family is also dense on $[1,2]$. It is $B=tA$, where $A$ is the even-power family in part (i). Multiplication by $t$ is an invertible bounded linear map on $C([1,2])$, with inverse division by $t$, since $t\ge 1$.

For an explicit approximation, take any $f\in C([1,2])$ and $ϵ>0$. The quotient $f(t)/t$ is continuous, so part (i) gives an even polynomial $a$ with $\Vert a-f/t{\Vert }_{\infty }<ϵ/2$. Then $ta$ belongs to the odd-power family andAlthough this family is not a subalgebra and does not contain the constant function exactly, it can approximate every continuous function, including constants. This proves even and odd polynomial density on a positive interval without incorrectly applying the unital algebra theorem directly to the odd family.

The even-power family is dense on $[1,2]$. It is the algebra $A=\{p(t^2):p\in \mathbb{C}[s]\}$. It contains constants, is closed under complex conjugation, and separates points because the real function $t\mapsto t^2$ is injective on this positive interval. The complex Stone-Weierstrass theorem therefore applies.

Equivalently, the squaring map is a homeomorphism from $[1,2]$ onto $[1,4]$. Given $f\in C([1,2])$, approximate the continuous function $s\mapsto f(\sqrt{s})$ uniformly on $[1,4]$ by complex polynomials, using the Weierstrass approximation theorem on the real and imaginary parts. Substitution $s=t^2$ produces exactly the required even-power approximants. The positive interval is what lets even powers distinguish all its points.

Take the Beth numberCantor's theorem makes this sequence strictly increasing. It is a countable cofinal sequence in $\kappa$, while no finite set is cofinal in the limit ordinal $\kappa$. Hence $\mathrm{cf}(\kappa )={\aleph }_0$. ConsequentlyThus Beth omega is a singular strong limit above the continuum. No Continuum hypothesis is needed for the example.

Put $\mu =\mathrm{cf}(\kappa )$ and choose cofinal ordinals ${\alpha }_i<\kappa$, $i<\mu$. With $a_i=|{\alpha }_i|$, the same union comparison and infinite cardinal arithmetic give ${\sum }_{i<\mu }{a}_i=\kappa$. Since $a_i<\kappa$, their successor cardinals satisfy $a_i^{+}\le \kappa$. The König theorem for cardinal numbers therefore givesThis König bound for the Gimel function applies to regular and singular cardinals alike; choosing cofinal ordinals also covers successor cardinal numbers.

For an infinite cardinal number $\kappa$, its cofinality isviewing $\kappa$ as an initial ordinal. Equivalently it is the smallest cardinal number which is the domain size of a cofinal map into $\kappa$.

Put $\lambda =2^{\kappa }$ and $\mu =\mathrm{cf}(\lambda )$. Suppose $\mu \le \kappa$, and choose cofinal ordinals ${\alpha }_i<\lambda$ for $i<\mu$. Put $a_i=|{\alpha }_i|<\lambda$. Their union is $\lambda$, so infinite cardinal arithmetic givesApply the König theorem for cardinal numbers to $a_i<a_i^{+}$. Each successor cardinal $a_i^{+}$ is at most $\lambda$, whenceHere the last equalities use the currying law for cardinal exponentiation and square of an infinite cardinal. This contradiction proves the cofinality of a power-set cardinal bound

Use the usual cardinal arithmetic with the axiom of choice. The König theorem for cardinal numbers states that for any index set $I$ and cardinal numbers ${\kappa }_i<{\lambda }_i$,Represent the cardinal numbers as initial ordinals. Let $A=\{(i,\alpha ):i\in I,\alpha <{\kappa }_i\}$ and $B={\prod }_{i\in I}{\lambda }_i$. There is an injection $A\to B$: send $(i,\alpha )$ to the function whose $i$th coordinate is $\alpha +1$ and whose other coordinates are zero. The unique nonzero coordinate identifies $i$, and its value identifies $\alpha$. All values are allowed because $\alpha +1\le {\kappa }_i<{\lambda }_i$.

For any proposed surjection $F:A\to B$, its $i$th row has at most ${\kappa }_i$ elements, soChoose $b_i$ to be the least omitted ordinal. The function $b=(b_i{)}_{i\in I}$ belongs to $B$ and differs from $F(i,\alpha )$ at coordinate $i$, for every $(i,\alpha )\in A$. Thus $F$ is not surjective. Cardinal comparison now gives the strict inequality. If $I=\varnothing$, the conclusion is $0<1$, also valid.

Begin with bounded formulas in set theory. Atomic membership and equality have the same interpretation on shared elements of $U_1\subseteq U_2$, each a transitive class. Boolean connectives preserve agreement. For a bounded quantifier with bound $a\in U_1$, every member of $a$ lies in $U_1$ by transitivity. Its quantifier range is therefore exactly the same in both structures. Structural induction proves set-theoretic absoluteness for every bounded formula in set theory.

A Sigma-one formula in set theory is upward absolute: witnesses in $U_1$ are still witnesses in $U_2$, and the bounded matrix agrees. A Pi-one formula in set theory is downward absolute, by negation or by restricting its universal quantifiers.

For the given Delta-one formula modulo ZF $\phi$, select a ${\Sigma }_1$ equivalent $\sigma$ and a ${\Pi }_1$ equivalent $\pi$. Both structures satisfy ZF, so both equivalences hold in both structures. Upward absoluteness of $\sigma$ proves the forward implication, and downward absoluteness of $\pi$ proves the reverse implication. Thus Delta-one absoluteness gives

A bounded formula in set theory, or ${\Sigma }_0$ formula in the Lévy hierarchy, uses only bounded quantifiers $\exists u\in v$ and $\forall u\in v$, with the usual Boolean connectives. A Sigma-one formula in set theory has the formwhere $\theta$ is a bounded formula in set theory; harmless logically equivalent rearrangements are allowed. A Pi-one formula in set theory is defined with a universal unbounded block instead.

A Delta-one formula modulo ZF is a first-order formula $\phi (\mathbf{x})$ for which there are such ${\Sigma }_1$ and ${\Pi }_1$ formulas $\sigma ,\pi$ withEquivalently, both $\phi$ and $\neg \phi$ have Sigma-one formulas in set theory provably equivalent to them in ZF. These are levels of the Lévy hierarchy, with bounds by sets.

Every finite ordinal belongs to $V_{\omega }$, so $\omega \subseteq V_{\omega }$ and $\omega \in \mathcal{P}(V_{\omega })=V_{\omega +1}$. However, $\omega \notin V_{\omega }$, whose elements are the hereditarily finite sets.

Suppose $y\in V_{\omega +1}$ were an internal power set of $\omega$. Internally $\omega \subseteq \omega$, and $\omega$ is an available element of the structure. Thus the Axiom of power set would require $\omega \in y$. But $y\in \mathcal{P}(V_{\omega })$ means $y\subseteq V_{\omega }$, which would put $\omega \in V_{\omega }$, a contradiction. ThereforeThis proves power-set failure at the first rank above V omega by an internal obstruction, including the subset $\omega$ itself.

Fix $x\in V_{\delta }$. Since $\delta$ is a limit ordinal, choose $\beta <\delta$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Every subset $u\subseteq x$ therefore belongs to $V_{\beta +1}=\mathcal{P}(V_{\beta })$, and consequentlyThe last inclusion uses $\beta +2<\delta$, since a finite number of successor steps remains below a limit ordinal. In particular every subset of $x$ is also an element of the transitive set $V_{\delta }$.

Take $y=\mathcal{P}(x)$. For any $u\in V_{\delta }$, transitivity ensures that the internal statement $u\subseteq x$ agrees with the external one. HenceSuch a $y\in V_{\delta }$ exists for every $x\in V_{\delta }$, so $(V_{\delta },\in )$ satisfies the Axiom of power set. This power-set axiom at limit ranks actually holds for every nonzero limit ordinal; the stated restriction $\delta >\omega$ is sufficient.

The bilinear square finite element contains the terms $1,x,y,xy$. In anticlockwise corner order beginning at $(0,0)$, its shape functions areThe mass matrix has entries $M_{ij}={\int }_0^h{\int }_0^h{N}_i{N}_{j}dxdy$. Each product separates into one-dimensional factors, with ${\int }_0^1(1-s{)}^{2}ds={\int }_0^1{s}^{2}ds=1/3$ and ${\int }_0^{1}s(1-s)ds=1/6$. Diagonal, adjacent-corner and opposite-corner entries are respectively $h^2/9$, $h^2/18$, $h^2/36$. ThusNumber global grid nodes by $k=i+(n+1)j$, with $x_i=ih$, $y_j=jh$ and $0\le i,j\le n$. For an interior node, four elements contribute to its diagonal, two to each axial neighbour and one to each diagonal neighbour. Its global row therefore has the nine-point stencilIn the stated ordering the axial entries occur at columns $k\pm 1$ and $k\pm (n+1)$; the four diagonal ones occur at $k\pm (n+1)\pm 1$. No boundary degrees of freedom have been removed.

For general valid quadrilaterals, use an isoparametric finite element. Map the unit square to an element by $F(\xi ,\eta )={\sum }_{i=1}^4{\mathbf{x}}_i{N}_i(\xi ,\eta )$ and define physical shape functions $N_i\circ F^{-1}$. Require a one-to-one nondegenerate map and compatible common edges. On a shared edge the map and the trace of each shape function are linear between the same endpoints; the neighbouring traces therefore agree. The assembled functions are continuous and lie in $H^1$, giving a conforming finite element space. On a non-parallelogram the pulled-back functions are generally not bilinear polynomials in the physical coordinates themselves.

Take the affine trial space $V_1=\{v\in H^1(a,b):v(a)=1\}$ and the test space $V_0=\{v\in H^1(a,b):v(a)=0\}$. Multiplication by $v\in V_0$ and integration by parts incorporate the right Neumann boundary condition as a load:For the continuous piecewise-quadratic trial space, impose the left value exactly and use the same equation for every piecewise-quadratic test function vanishing at $a$. With a basis $({\psi }_j)$ of that test space and $u_h=1+{\sum }_j{c}_j{\psi }_j$, the finite element equations areThe Endpoint Poincaré inequality, reflected about the interval midpoint, makes the derivative norm a norm on $V_0$. Galerkin orthogonality then gives the exact best-approximation identity: for every admissible $v_h$,Therefore $u_h$ minimizes the derivative error over all admissible piecewise-quadratic functions.

To prove the requested error bound, assume $f\in H^1(a,b)$. Then $u\in H^3(a,b)$ and $u^{\prime \prime \prime }=-f^{\prime }$. We need an elementary Poincare-Wirtinger inequality on each element $I$ of length $h_I$: if $w\in H^1(I)$ has mean zero, its continuous representative has a zero somewhere, and the fundamental theorem of calculus and Cauchy-Schwarz inequality give $\Vert w{\Vert }_{L^2(I)}\le h_I\Vert w^{\prime }{\Vert }_{L^2(I)}$.

Choose an affine function $q_I$ whose slope is the average of $u^{\prime \prime }$ on $I$ and whose average equals the average of $u^{\prime }$. Then $w=u^{\prime }-q_I$ and $w^{\prime }=u^{\prime \prime }-q_I^{\prime }$ both have mean zero. Applying the just-proved bound twice yieldsDefine a quadratic interpolant on $I$ by integrating $q_I$ from the true left-endpoint value. The equal-average condition ensures that it reaches the true right-endpoint value, so these pieces join to an admissible continuous piecewise-quadratic function $I_{h}u$. Sum the squared estimates and use the best-approximation property:Only the maximum element length is used; equal spacing is unnecessary.

For the Aubin–Nitsche duality argument, let $e=u-u_h$ and solve $-z^{\prime \prime }=e$, $z(a)=0$, $z^{\prime }(b)=0$. Explicitly $z^{\prime }(x)={\int }_x^{b}e(t)dt$, so $\Vert z^{\prime \prime }{\Vert }_2=\Vert e{\Vert }_2$. Integration by parts and Galerkin orthogonality givewhere $J_{h}z$ is its nodal piecewise-linear interpolant, which belongs to the quadratic test space. On each element $(J_{h}z{)}^{\prime }$ is the average of $z^{\prime }$, so the same mean-zero inequality gives $\Vert z^{\prime }-(J_{h}z{)}^{\prime }{\Vert }_2\le h\Vert z^{\prime \prime }{\Vert }_2$. Canceling $\Vert e{\Vert }_2$ when nonzero proves the additional order:

The first-order Sobolev space is $H^1(\Omega )=\{v\in L^2(\Omega ):{\partial }_{1}v,{\partial }_{2}v\in L^2(\Omega )\}$, with derivatives understood as weak derivatives and norm squared ${\int }_{\Omega }(v^2+|\nabla v{|}^2)$.

The printed normal boundary condition must be distinguished from a conormal condition. Take $K$ symmetric positive definite, as required for the symmetric matrix requested later. Integration by parts givesThe natural boundary term contains the conormal derivative $\mathbf{n}\cdot K\nabla u$, not the ordinary normal derivative ${\partial }_{n}u$. On a smooth boundary, writing $\mathbf{t}$ for its unit tangent, the literal condition ${\partial }_{n}u=0$ leaves the term $b{\partial }_{\mathbf{t}}u$, where $b=\mathbf{n}\cdot K\mathbf{t}$. Its weak form therefore retains a boundary pairing:Here the tangential derivative of the Sobolev trace can be interpreted in $H^{-1/2}$ and paired against $H^{1/2}$ on a smooth boundary. This form is generally nonsymmetric; the requested usual symmetric energy construction does not follow from the literal boundary condition.

For an explicit failure of the claimed mean-zero compatibility, take the unit disk, $K=\mathrm{diag}(2,1)$ and $u=(x^2-y^2)[1-(x^2+y^2)/2]$. It has ${\partial }_{n}u=0$ on the circle, butThus replacing the printed normal condition by a zero conormal flux is not harmless.

For the intended zero-conormal derivative problem, the requested variational and finite element conclusions are as follows. Seek $u\in H^1(\Omega )$ withPutting $v=1$ gives ${\int }_{\Omega }f=0$. On a connected domain this condition permits a solution unique modulo constants; the Poincare-Wirtinger inequality gives coercivity after fixing the mean.

The energy is $J(v)=\frac{1}{2}a(v,v)-\ell (v)$. A minimizer satisfies $\frac{d}{dt}J(u+tv){|}_{t=0}=a(u,v)-\ell (v)=0$. Conversely, a weak solution satisfiesso it is a minimizer. Testing by $u$ gives the minimum valueThis value is independent of the arbitrary additive constant when $\int f=0$. If the compatibility fails, constant shifts make $J$ unbounded below.

Write $u_h={\sum }_j{c}_j{\varphi }_j$. The Galerkin method givesSymmetry of $K$ gives $A^T=A$. For $v_h={\sum }_j{z}_j{\varphi }_j$, $z^{T}Az={\int }_{\Omega }\nabla v_h\cdot K\nabla v_h\ge 0$. Equality forces $v_h$ constant by positive definiteness and connectedness. Since the basis sums to one, the coefficient vector $1=(1,\dots ,1{)}^T$ represents that constant, givingThus the stiffness matrix is symmetric positive semidefinite and singular; the displayed compatibility condition is necessary and sufficient for the finite system. Positive definiteness of $K$ controls every gradient direction, preventing additional zero-energy directions and supplying ellipticity. A degenerate coefficient can introduce further kernel modes, while an indefinite one destroys the minimizing-energy argument.