The integrand is nonnegative. Define the Lipschitz triangular wavesThey have zero endpoint values, slope $\pm 2$ almost everywhere, and height $1/N$. Hence $(u_N^{\prime 4}-16{)}^4=0$ almost everywhere. On each of the $N$ equal cells, direct integration of the fourth power of the two affine pieces gives $1/(5N^5)$. ThereforeThis is an oscillating minimizing sequence with no minimizer.

An admissible $\bar{u}$ is a global minimizer ifIn particular it attains the infimum; no closeness condition on competing admissible functions is imposed.

A strictly convex function satisfies the same inequality strictly whenever $x\ne y$ and $0<t<1$.

For the unlabelled Hessian criterion, convexity implies $F(z_0+th)+F(z_0-th)-2F(z_0)\ge 0$. Divide by $t^2$ and let $t\to 0$ to obtain $h^T{D}^{2}F(z_0)h\ge 0$ for every $h$. Thus the Hessian matrix is positive semidefinite.

Conversely, if every Hessian is positive semidefinite, Taylor's integral formula with $h=z-z_0$ givesThe allowed supporting-plane characterization then provesFor the integral inequality put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting tangent at $m$ gives $F(u(x))\ge F(m)+F^{\prime }(m)(u(x)-m)$ almost everywhere. Integrating cancels the linear term, soThis is Jensen's inequality. It remains valid with value $+\infty$ on the right: the affine lower bound has integrable negative part, so the extended integral is well-defined.

A convex function satisfiesThus its value on every line segment lies below the chord joining the endpoint values.

Let $z_0\ne 0$ be an equality point and set $c=g(z_0)$, so $|c|=1$. For any $|z_0|<r<1$, the same Brownian exit identity givesPart (i) gives $|g|\le 1$, so the real random variable in the expectation is at most one. Equality of its mean to one forces $g(Z_{T_r})=c$ almost surely. By the scaled version of the exit density from Question 2, every nonempty arc of the exit circle has positive probability. Continuity of $g$ on that circle therefore implies $g=c$ at every boundary point of the radius-$r$ disk.

For any $w$ inside that circle, the Brownian exit identity now gives $g(w)={\mathbb{E}}_{w}g(Z_{T_r})=c$. Letting $r$ increase to one makes this true throughout $D$. Writing $c=e^{i\theta }$ yieldsThis completes the equality case of the Brownian proof of the Schwarz lemma.

Since $f(0)=0$, the function $g(z)=f(z)/z$ extends analytically to zero by $g(0)=f^{\prime }(0)$. It is bounded on the disk: outside $|z|<1/2$, $|g(z)|\le 2$, while on the closed half-radius disk its analytic extension is continuous and hence bounded.

Fix $z\ne 0$ and choose $|z|<r<1$. Let $T_r$ be the first exit time of Brownian motion started at $z$ from the radius-$r$ disk. The bounded analytic martingale identity in part (a) givesOn this exit circle, $|g(Z_{T_r})|=|f(Z_{T_r})|/r\le 1/r$. The triangle inequality gives $|g(z)|\le 1/r$. Letting $r\uparrow 1$ proves $|g(z)|\le 1$, and thereforeAt zero the claim holds by the assumed zero value. This is the Schwarz lemma, obtained from Brownian optional stopping rather than an analytic maximum principle.

Start a planar Brownian motion $B$ at zero and apply part (a) to the entire function $f(z)=e^z$. ThenThe left side never vanishes. The clock is continuous, starts at zero, and tends to infinity almost surely, so its range is all nonnegative times. Consequently the complex Brownian motion $1+{\tilde{Z}}_s$ never reaches zero at any time $s$. Every planar Brownian motion started at one has that same law, henceThis is the complex exponential construction of planar Brownian motion and proves that a fixed point is a polar point for planar Brownian motion. The divergence of the clock is essential: it makes this a conclusion about all Brownian times.

Apply part (a) to the entire function $f(z)=e^{i\theta }z$. Its derivative has modulus one, so its conformal Brownian clock is $A_t=t$. Thuswhich is a planar Brownian motion starting from the rotated initial point. In particular, if the initial point is zero, its law is unchanged. This is rotational invariance of planar Brownian motion.

Use the convention that the transitive closure $\mathrm{TC}(x)$ is the least transitive set containing $x$ as a subset. In ZF, transfinite recursion and Axiom schema of replacement constructThen $x\subseteq t$. If $a\in b\in t$, some $n$ has $b\in x_n$, whence $a\in x_{n+1}\subseteq t$. Thus $t$ is transitive. If another transitive set $s$ contains $x$, induction gives $x_n\subseteq s$ for every $n$, and hence $t\subseteq s$. This proves existence and uniqueness of the transitive closure.

Let $W(w)$ be the bounded definition of omega above. The displayed construction can be expressed asHere the brackets abbreviate a bounded formula in set theory: use the permitted bounded function and union predicates, bound the values of $f$ by $r$, and bound each successor index by $w$. Ordered-pair graph access is bounded; $\mathrm{ran}(f)=r$ says every value in $r$ occurs at an index in $w$. Any auxiliary sets used for the graph coding may be put in the initial existential block. Thus this is a Sigma-one formula in set theory defining $y=\mathrm{TC}(x)$.

Denote this formula by $T(x,y)$. Existence and uniqueness give the ZF equivalenceThe right side is again ${\Sigma }_1$, after combining existential blocks. Consequently $T$ is a Delta-one formula modulo ZF, so Delta-one absoluteness applies. Therefore transitive closure is Delta-one definable and is ZF-absolute. If closure is instead required to contain $x$ as an element, apply exactly this construction to $\{x\}$; the absoluteness conclusion is unchanged.

There is a bounded definition of omega. Transitivity has the bounded definition $\mathrm{Trans}(x)\equiv \forall u\in x\forall v\in u(v\in x)$. An ordinal is a transitive set of transitive sets on which membership is a strict total order; this too has a bounded formula in set theory, since ZF supplies well-foundedness by the Axiom of foundation.

Define the bounded predicateFor an ordinal, this says that it is a nonzero limit ordinal. It follows in ZF thatsince $\omega$ is the least nonzero limit ordinal. The right side is a bounded formula in set theory; consequently it is absolute between the specified transitive models. This also shows directly that those models have the same $\omega$.

Two facts about a transitive set $A$ drive the transfinite induction. First, $A\subseteq \mathcal{P}(A)$: every $x\in A$ is a subset of $A$. Second, $\mathcal{P}(A)$ is a transitive set: if $x\in y\in \mathcal{P}(A)$, then $x\in A$, hence $x\subseteq A$, hence $x\in \mathcal{P}(A)$.

Starting with $V_0=\varnothing$, these facts prove the successor step. At a limit ordinal $\lambda$, the union of the earlier increasing transitive sets is transitive: if $x\in y\in V_{\lambda }$, some $\alpha <\lambda$ contains $y$ in $V_{\alpha }$, and transitivity puts $x$ there too. The union contains every earlier stage. Thus the induction gives both transitivity and $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.

Finally $V_{\alpha }\subseteq V_{\alpha }$, so $V_{\alpha }\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. The Axiom of foundation gives $V_{\alpha }\notin V_{\alpha }$. This witnesses strictness:This is the transitivity and strict growth of the cumulative hierarchy.

The cumulative hierarchy is the transfinite recursionHere $\mathcal{P}$ is the power set. The class $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$ consists of all sets in ZF, by the Axiom of foundation and Axiom schema of replacement.

For a line $p(x)=ax+b$, the error $e^x-ax-b$ is strictly convex. Its two endpoint maxima must agree, giving $a=e-1$. Its unique interior minimum is at $c=\log a\in (0,1)$. Equating its magnitude to the endpoint error givesThe errors at $0,c,1$ are respectively $E_1,-E_1,E_1$. The equioscillation theorem therefore certifies this best uniform approximation directly.

For degree above one, subtracting a general polynomial does not preserve convexity of the error. The required $n+2$ alternating extreme points include additional unknown interior points, and their derivative and equal-height equations form a coupled nonlinear problem. The simple endpoint-and-single-minimum calculation therefore does not extend as a direct higher-degree recipe.

Here the exchange point lies between adjacent reference points and has the same, positive error sign as the point it replaces. Hence the new set remains spatially ordered, and the old error at its points alternates in sign. Its magnitudes are $|h_0|$ everywhere except at $\eta$, where $M=u(\eta )-q(\eta )>|h_0|$.

Set $w_r={\prod }_{j\ne r}(z_r-z_j{)}^{-1}$. Apply the order-$(n+1)$ divided difference to $q$ and $v$; both vanish because their degrees are at most $n$. In the new alternation equations, all products $(-1{)}^r{w}_r$ have the same sign. Thus the single-point exchange in polynomial minimax approximation givesThe Remez algorithm repeatedly solves for the reference polynomial and common signed height, locates a larger error extreme, and performs an order-preserving exchange. The reference height bounds the optimum from below and the global error maximum bounds it from above. Their gap provides an optimality stopping certificate. A general implementation also needs the endpoint exchange rule when the new extreme lies outside the reference set.

Put $m={\min }_r|u(x_r)-q(x_r)|$ and $M={\max }_r|u(x_r)-q(x_r)|$. If the best uniform approximation had error below $m$, the values of $p_n^{\ast }-q$ would strictly alternate in sign at all $n+2$ reference points. The intermediate value theorem would give at least $n+1$ distinct roots of a polynomial of degree at most $n$, a contradiction. If $m=0$ the lower bound is immediate.

At least one reference point attains the global error maximum of $q$, so $M=\Vert u-q{\Vert }_{\infty }$. Optimality of $p_n^{\ast }$ gives the upper bound, and thereforeThis is the unequal-height version of the alternating-reference error bound.

For $n\ge 1$, ${\varphi }_n$ is orthogonal to every polynomial in $P_{n-1}$, so $(u,{\varphi }_n)=(u-p_{n-1}^{\ast },{\varphi }_n)$. The Cauchy-Schwarz inequality givesUsing the tail identity from part (b) and shifting the index,For the stated power law and $n>1$, the decreasing-function integral test givesConsequentlyThe assumption $a>1$ is sufficient; this summation estimate itself only needs $a>1/2$.

The moments are ${\int }_0^1{x}^{j+2}dx=1/(j+3)$. The Gram-Schmidt process first gives the constant ${\varphi }_0=\sqrt{3}$. Removing the constant component of $x$ gives $r_1=x-3/4$, with squared norm $1/80$, hence ${\varphi }_1=4\sqrt{5}(x-3/4)$.

For the quadratic, imposing orthogonality of $r_2=x^2+ax+b$ to one and $x$ gives $1/5+a/4+b/3=0$ and $1/6+a/5+b/4=0$. Thus $r_2=x^2-4x/3+2/5$, whose squared norm is $1/1575$. Taking positive leading coefficients, the first three orthonormal polynomials are

The orthogonal projection is $p_n={\sum }_{r=0}^n(u,{\varphi }_r){\varphi }_r$. Subtract it from $u$ and use the orthonormal basis relations to obtainThis is the requested Bessel inequality.

For the infinite tail, use completeness of the orthonormal polynomial family in the usual weighted L2 space. The relevant density facts are that continuous functions are dense for the finite measure $w(x)dx$, and the Weierstrass approximation theorem makes polynomials uniformly dense in continuous functions on the compact interval. Since $\int w<\infty$, uniform approximation also gives weighted L2 norm approximation. Parseval's identity therefore givesHere finiteness of the weight integral is also necessary for the constant polynomial to belong to the stated Hilbert space. Without completeness of the polynomial family, an additional squared component orthogonal to every polynomial would remain; Bessel inequality alone does not justify equality.

Let $e=u-p$. If $p$ minimizes the squared L2 norm error, then for any $q\in A$ the real quadratic $\Vert e-tq{\Vert }_2^2=\Vert e{\Vert }_2^2-2t(e,q)+t^2\Vert q{\Vert }_2^2$ has its minimum at $t=0$. Its derivative there is zero, giving $(e,q)=0$.

Conversely, if $e$ is orthogonal to $A$, then for every $q\in A$, $p-q\in A$ and the Pythagorean theorem for inner product spaces givesThis proves the equivalence and identifies $p$ as an orthogonal projection. It characterizes a minimizer whenever one exists; a general nonclosed vector subspace need not admit a projection for every $u$.

A cubic spline that vanishes for $x\le 0$ and is twice continuously differentiable can be written using truncated power functions as $s(x)={\sum }_{j=0}^k{a}_j(x-jh{)}_{+}^3$. Vanishing for $x\ge kh$ imposes the four moment equations ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Their Vandermonde matrix has rank $\min (4,k+1)$, so a nonzero coefficient vector first exists at $\boxed{k=4}$.

A normalized compact cubic cardinal B-spline isIt has nodal values $0,1/6,2/3,1/6,0$ at $0,h,2h,3h,4h$, so their sum is one. The alternating binomial coefficients meet the four moment equations, and the factor $1/6$ fixes the requested normalization.

For interpolation on $[-1,1]$, centre the translates as ${\beta }_j(x)=B_3(x/h-j+2)$, with $j=-n-1,\dots ,n+1$. Their restrictions form a basis for the cubic $C^2$ spline space on these knots. Write $S={\sum }_j{c}_j{\beta }_j$. The $2n+1$ interpolation equations areThe recurrence determines all remaining coefficients after choosing $c_{-n-1},c_{-n}$, giving exactly two free parameters. The two natural cubic spline conditions becomebecause $S^{\prime \prime }(rh)=h^{-2}(c_{r-1}-2c_r+c_{r+1})$.

For uniqueness, take the difference $s$ of two candidates. It vanishes at every knot and has $s^{\prime \prime }(\pm 1)=0$. Integrating by parts twice on each cubic piece gives ${\int }_{-1}^1(s^{\prime \prime }{)}^2=0$: terms involving $s$ vanish at the knots, and those involving $s^{\prime \prime }{s}^{\prime }$ cancel at internal knots and vanish at the endpoints. Thus $s^{\prime \prime }=0$; the remaining affine function vanishes at both endpoints and is zero. The square linear system for the coefficients is therefore nonsingular, giving existence as well as uniqueness of the natural cubic spline interpolant.