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Multiply the representation in part (c) by ${\int }_0^{\infty }{\alpha }_{u}d{W}_u$ and take expectations. This Itô integral has mean zero, and the bilinear form of the Itô isometry givesEquating this with part (b), and writing both ordinary integrals with the same time variable, yieldsAll terms are integrable by the Cauchy-Schwarz inequality, the assumed square integrability of $\alpha$, and boundedness of ${\varphi }^{\prime }$. This is an orthogonality statement against predictable processes; its second term need not itself be predictable.

For $n>m$, define $r_j=\lfloor j{2}^{m-n}\rfloor 2^{-m}$ and $h_j=X_{j{2}^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j{2}^{-n}$, so the summands in the given difference formula are orthogonal martingale transforms. ThereforeLetPath continuity on the compact interval gives ${\omega }_m\to 0$ almost surely, and ${\omega }_m\le 2C$. Since $0\le j{2}^{-n}-r_j<2^{-m}$, $|h_j|\le {\omega }_m$. Thus, by the Cauchy-Schwarz inequality and part (b),The last step is the dominated convergence theorem. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence the terminal martingale transforms are Cauchy in $L^2$.

For the unit-width box kernel, the convolution and kernel density estimator areandIn particular $\mathbb{E}{\hat{f}}_{n,h}^K=g_h$. The scaling gives $\Vert K_h{\Vert }_2^2=h^{-1}\Vert K{\Vert }_2^2=h^{-1}$. Even if $f$ is not square-integrable, Young's convolution inequality gives $g_h\in L^2$ because $f\in L^1$ and $K_h\in L^2$.

Using independence to eliminate the cross terms in the centred estimator, and Tonelli theorem to integrate the nonnegative variance, gives the exact integrated variance of a kernel density estimator:The Cauchy-Schwarz inequality now yieldsThis constant comes from the unit $L^2$ norm of the unscaled unit-width box kernel.

For the piecewise constant probability density function, integrability forces the constants on both unbounded outer intervals to be zero. Thus $f$ is bounded, has compact support, and has finitely many jumps. Let ${\Delta }_r=f(x_r+)-f(x_r-)$ denote the jump at $x_r$. For $h$ smaller than the least gap between consecutive breakpoints, the smoothing regions $[x_r-h/2,x_r+h/2]$ do not overlap. Outside these regions the convolution equals $f$.

Within a region, put $v=x-x_r$. For $-h/2<v<0$, the bias is ${\Delta }_r(1/2+v/h)$; for $0<v<h/2$, it is $-{\Delta }_r(1/2-v/h)$. Endpoint values do not affect an $L^2$ norm. Integrating these two triangular errors givesThis is the box-kernel bias of a piecewise constant density. By the triangle inequality, with $C_f=({\sum }_r{\Delta }_r^2/12{)}^{1/2}$,Balancing this bias-variance tradeoff with $h_n=n^{-1/2}$ gives, for all sufficiently large $n$,

For the unit-width box kernel, the order-zero local polynomial estimator minimizes, over constants $a$,A common factor such as $h^{-1}$ in these weights does not change the minimizer. Put $I_x=\{i:|i/n-x|\le h/2\}$ and $N_x=|I_x|$. If $N_x>0$, differentiating the quadratic gives its unique minimizer:The last expression is the Nadaraya–Watson estimator. Every $x\in [0,1]$ is within $1/n$ of some design point $i/n$. Since $nh\ge 2$ gives $h/2\ge 1/n$, that point is in the closed kernel window. Thus $N_x\ge 1$ and the equality holds throughout the domain, including $x=0$ and $x=1$.

For the error bound, first take the usual bandwidth range $0<h\le 1$. The clipped window $[x-h/2,x+h/2]\cap [0,1]$ has length at least $h/2$. Any closed interval of length $L$ in $[0,1]$ contains at least $\lfloor nL\rfloor$ points of the design $\{1/n,\dots ,1\}$. Consequentlywhere the last inequality uses $nh/2\ge 1$. This is a window occupancy for an equally spaced regression design bound that remains valid at the boundary.

Let $L_m=\Vert m^{\prime }{\Vert }_{\infty }$, the Lipschitz constant supplied by the bounded-derivative assumption. The bias is bounded byIndependence and the variance bound giveBy the Cauchy-Schwarz inequality and the triangle inequality,Thus $\kappa =2$ works uniformly in $x$ and the design size in this bandwidth range. The same clipped-length proof in fact works for $0<h\le 2$.

There is a genuine omission in the unrestricted formulation: an upper restriction on bandwidth is necessary for the displayed $1/\sqrt{nh}$ variance scale. Take $n=1$, $m\equiv 0$, $Y_1\sim N(0,{\sigma }^2)$ and $h\ge 2$. The window contains the sole observation for every $x\in [0,1]$, and $nh\ge 2$ holds. Its mean absolute error is $\sigma \sqrt{2/\pi }$, whereas the proposed variance term is $\kappa \sigma /\sqrt{h}$. No fixed $\kappa$ can satisfy this for unbounded $h$.

A valid statement for every $h>0$ satisfying $nh\ge 2$ uses $q=\min (h,1)$. The clipped window has length at least $q/2$, so $N_x\ge nq/2-1$, while the preceding coverage argument gives $N_x\ge 1$. If $nq\ge 4$, the first bound gives $N_x\ge nq/4$; if $nq<4$, the second does. Therefore the same bias and variance calculation proves the unrestricted correctionThis mean absolute error of local constant regression bound reduces to the requested rate for ordinary small bandwidths and saturates its stochastic term when the window covers the whole design.

The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincare inequality makes this a Hilbert space norm equivalent to the usual $H^1$ norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous $p,q$ on the closed interval with $p>0$ and $q\ge 0$, setIf $p_{\ast }=\min p>0$, then Cauchy-Schwarz inequality and the Poincare inequality giveBy the Riesz representation theorem, a unique bounded linear operator $T:H\to H$ satisfies $(Tu,v{)}_H=a(u,v)$. Symmetry of $a$ makes $T$ self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by $a$: for smooth functions, integration by parts giveswith the boundary term zero. A forcing $f\in L^2$ becomes the bounded functional $v\mapsto \int fv$, or its Riesz representative in $H$.

Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on $L^2(-1,1)$ with domain $H^2\cap H_0^1$. Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator $T$ used above.

The PDF states sign conditions without coefficient regularity. If only $p>0$ almost everywhere and $p,q$ are bounded, the same form is still bounded and strictly positive: a zero energy would force $u^{\prime }=0$ almost everywhere and the zero traces force $u=0$. Uniform ellipticity, however, requires a positive lower bound. For instance $p(x)=x^2$ for $x\ne 0$, $p(0)=1$, and $q=0$ satisfy literal pointwise positivity. Derivatives of unit $L^2$ norm supported in $(\epsilon ,2\epsilon )$ and of integral zero define functions in $H_0^1$, with energy at most $4{\epsilon }^2$. Thus pointwise positivity without regularity does not imply coercivity in this $H$ norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.

Use the weighted Hilbert structure of velocity-reset relaxation, with inner product $⟨f,g⟩=\int f\overline{g}/M$ and norm $\Vert f{\Vert }_H$. The Cauchy-Schwarz inequality givesThus $\rho (f)$ is an absolutely convergent integral and a bounded linear functional. Since $\Vert M{\Vert }_H^2=\int M=1$, the normalized velocity-reset collision operator obeysThis establishes boundedness. The subsequent orthogonal decomposition improves the operator norm to exactly one.

The absolute Fourier convergence from a square-integrable derivative uses more than the pointwise estimate $|\hat{f}(n)|=O(1/|n|)$. Periodic integration by parts givesBy Bessel's inequality,Now apply the Cauchy-Schwarz inequality:Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has $f^{\prime }\in L^2$, all hypotheses of part (iii) hold and its Fourier series converges uniformly.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.

Part (i) and the Cauchy-Schwarz inequality showDirect integration of this finite Fourier partial sum givesMoreover Bessel's inequality puts both coefficient sequences in ${\ell }^2$, so their product series is absolutely convergent by the Cauchy-Schwarz inequality. Consequently the cross form of Parseval's identity isTaking $g=f$ also gives equality of the squared function norm and the squared coefficient norm.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.

The converse is false. Take $x_n=n^{-3/2}$. Each block $2^j\le n<2^{j+1}$ contributes at most $2^j{2}^{-3j/2}=2^{-j/2}$, and the resulting geometric series converges. But $n^2{x}_n^2=1/n$, whose harmonic series diverges. This proves the counterexample without assuming the full P-series criterion.

(A) For a finite point set $P$ and a family $\mathcal{C}$ of distinct unit circles in ${\mathbb{R}}^2$, the Szemerédi–Trotter theorem for unit circles stateswith an absolute constant $C$. Here $I$ counts point-circle incidences. Distinctness matters: repeated copies of the same circle are not separate members of this geometric family. A dilation gives the same bound for circles of any one fixed positive radius, with the same constant.

(B) Write $N=|P|$ and $D=|\Delta (P)\setminus \{0\}|$. The comparison is a bound on the cardinality of the distinct-distance set, rather than on the set itself. For each positive distance $\rho$, take the $N$ circles of radius $\rho$ centred at points of $P$. Their incidences between points and curves count exactly the ordered pairs $(p,q)$ at distance $\rho$. After dilation by $1/\rho$, part (A) bounds this number byEvery ordered pair of distinct points contributes to exactly one of these counts. Thus the unit-circle method for a distinct-distance lower bound givesFor $N\ge 2$, $N-1\ge N/2$, soFor $N=1$ the distance set is $\{0\}$ and the conclusion holds after adjusting the absolute constant; the empty set causes no difficulty.

(C) A direct incidence bound from two-point multiplicity suffices. Put $M=|\mathcal{L}|=N^2$ and $k_{\gamma }=|P\cap \gamma |$. Count unordered pairs of distinct points on each curve. By double counting,Writing $I={\sum }_{\gamma }{k}_{\gamma }$, this givesThe Cauchy-Schwarz inequality now yieldsIf $x^2\le ax+b$ with $x,a,b\ge 0$, then $x\le a+\sqrt{b}$. ConsequentlyIn fact the argument proves the stronger $O(N^{9/4})$ bound. The two-point multiplicity hypothesis alone controls these incidences between points and curves; the algebraic degree bound is not needed for the requested estimate.

Use the normalization in which a projective line has area $2\pi$. On the affine chart of Complex projective space where $Z_0\ne 0$, set $w_j=Z_j/Z_0$ and $S=1+{\sum }_j|w_j{|}^2$. Define the Fubini-Study form byOn another chart the corresponding potential differs by $\log |h{|}^2$ for a nowhere-zero holomorphic function $h$, whose $\partial \bar{\partial }$ is zero. Thus these local differential forms glue to a global form. It is real and closed. Its Hermitian coefficient matrix is positive definite, since for $v\ne 0$ the Cauchy-Schwarz inequality givesHence it is a Kähler form and in particular a symplectic form.

On a complex projective line with $w=x+iy$ this is $2(1+|w{|}^2{)}^{-2}dx\wedge dy$, whose total area is $2\pi$. Thus $[{\omega }_{\mathrm{FS}}/(2\pi )]$ is the positive generator, equivalently $c_1(\mathcal{O}(1))$. Another common normalization uses half this form and gives line area $\pi$; the scale must be carried consistently into symplectic reduction.

A height function measures arithmetic size, which is what makes an otherwise infinite descent terminate. For a number field $K$, normalize absolute values to extend the standard real and $p$-adic ones, and write $n_v=[K_v:{\mathbb{Q}}_v]$. The logarithmic projective height isThe product formula makes it independent of the chosen homogeneous coordinates, and the local-degree normalization makes it independent of the number field containing them. Over $\mathbb{Q}$, $h([a:b])=\log \max (|a|,|b|)$ for coprime integer coordinates. Set $h_x(P)=h(x(P))$ on an elliptic curve, with $h_x(O)=0$.

Two features are essential. First, the Northcott theorem says that points of bounded projective height and bounded field degree form a finite set. For points on a fixed elliptic curve over $K$, each $x$-coordinate has at most two preimages, so bounded $h_x$ gives finitely many points. Second, a degree-$d$ morphism of the projective line satisfies $h(f(z))=dh(z)+O(1)$, uniformly in $z$. The upper bound comes from evaluating its homogeneous polynomials; for the lower bound, their lack of a common zero gives a resultant identity bounding the input coordinates by the output coordinates at each place. Summing the local bounds gives the asserted uniform constant.

The duplication map on the $x$-line has degree four. Nonsingularity ensures that its numerator and denominator have no common projective zero. ThereforeTelescoping defines the canonical height of an elliptic curveThe error in successive terms is bounded by a geometric series, proving convergence and the uniform bounded difference. It also gives $\hat{h}(2P)=4\hat{h}(P)$ and nonnegativity. The usual addition formula gives the approximate height parallelogram identity for $h_x$; equivalently, the unordered pair of sum and difference on the $x$-line has bidegree $(2,2)$. Applying that identity to $2^{j}P,2^{j}Q$ and passing to the limit givesIn particular $\hat{h}(nP)=n^2\hat{h}(P)$. Its polarization is the canonical height pairing, a positive semidefinite bilinear form even before finite generation has been proved. The Cauchy-Schwarz inequality for this pairing givesAlso $\hat{h}(P)=0$ precisely for torsion points of an elliptic curve: one direction follows from periodic multiples, and in the other direction all multiples have bounded $h_x$, so the Northcott theorem makes two multiples equal. Bounded canonical height of an elliptic curve likewise gives a finite set of $K$-rational points.

Now the Weak Mordell-Weil theorem gives finitely many representatives $R_1,\dots ,R_s$ for $E(K)/2E(K)$. Put $H={\max }_i\hat{h}(R_i)$. Write any point as $P=2Q+R_i$. ThenRepeatedly applying this height descent lemma eventually reaches height at most $H+1$: after $j$ steps the height is at most $H+2^{-j}(\hat{h}(P)-H)$. The set of points with height at most $H+1$ is finite. Reading the relations $P=2Q+R_i$ backwards shows that this finite set together with the $R_i$ generates $E(K)$. ConsequentlyHeights turn weak Mordell-Weil finiteness into the Mordell-Weil theorem. The canonical height pairing subsequently equips the free part with a positive definite quadratic form, useful for bounding searches and measuring independent generators; this interpretation is a consequence of the proof, not an assumption used in the descent.

The appropriate abstract object is a finite reduced crystallographic root system $R$ in a real inner-product space $E$. Its axioms are: $R$ is finite, spans $E$, and does not contain zero; for $\alpha \in R$, $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; each root reflectionpermutes $R$; and every Cartan integer $n_{\alpha \beta }=2(\alpha ,\beta )/(\beta ,\beta )$ is an integer. The restriction to a reduced root system and crystallographic integrality distinguishes roots of complex semisimple Lie algebras from more general reflection configurations.

For any two roots of a root system, the Cauchy-Schwarz inequality givesIf the inner product is zero, both Cartan integers vanish. Otherwise their signs agree, and the absolute value of each is a positive integer. Dividing their product by an integer of absolute value at least one provesFor nonproportional roots the product is strictly less than four. In a reduced root system, proportional roots are just $\pm \alpha$ and have Cartan integers $\pm 2$, so the printed bound is intentionally looser than the resulting bound of three.

A fundamental system of a root system is a basis $\Delta$ of $E$ made of roots, such that each root is an integer combination of $\Delta$ with either all coefficients nonnegative or all nonpositive. Its members are the simple roots. Suppose distinct $\alpha ,\beta \in \Delta$ had $n_{\alpha \beta }>0$. Thenhas a positive coefficient of $\alpha$ and a negative coefficient of $\beta$, contradicting the defining sign condition. Thus $n_{\alpha \beta }\le 0$. Distinct simple roots are linearly independent, so their Cartan-integer product is strictly less than four. Combining integrality and the sign condition givesHere nonpositive is the intended sense of the printed convention that includes zero among “negative” numbers; orthogonal simple roots really do give zero.

To form a Dynkin diagram, place a vertex at each simple root. Join two vertices by $n_{\alpha \beta }{n}_{\beta \alpha }$ bonds, hence zero, one, two or three. A multiple bond has an arrow toward the short root. Indeed $n_{\alpha \beta }/n_{\beta \alpha }=(\alpha ,\alpha )/(\beta ,\beta )$ determines the squared length ratio, and the diagram with the Cartan matrix reconstructs the angles and relative lengths. A single bond joins equal-length roots.

The connected finite Dynkin diagrams are the following. The descriptions include bond multiplicities and arrow directions, so distinguish dual diagrams:

There are no other connected finite Dynkin diagrams. Low-rank conventions also identify $B_1=C_1=A_1$ and $D_3=A_3$; $D_2$ is disconnected, so introduces no further connected type. Affine diagrams are outside this finite classification.

Finally suppose the underlying graph contained a cycle on $m\ge 3$ distinct simple roots ${\alpha }_1,\dots ,{\alpha }_m$. Put $u_i={\alpha }_i/\Vert {\alpha }_i\Vert$. Any bonded pair hasand all other distinct pairs have nonpositive inner products. The cycle contributes at least $m$ bonded pairs, soBut the simple roots, and hence these normalized vectors, are linearly independent, making the displayed sum nonzero. Positive definiteness gives a contradiction. Thus the underlying graph of a finite Dynkin diagram has no cycle. This acyclicity of a finite Dynkin diagram argument also excludes cycles with extra chords or multiple bonds; multiple bonds themselves are not treated as two-edge cycles.

The Vinogradov mean value isBy orthogonality of integer Fourier modes, it counts the ordered integer solutions of ${\sum }_{i=1}^k{x}_i^j={\sum }_{i=1}^k{y}_i^j$ for $1\le j\le r$, with every coordinate between one and $\lfloor Z\rfloor$. The diagonal solutions $y_i=x_i$ give $J_{k,r}(Z)\ge \lfloor Z{\rfloor }^k$.

Put $D=r(r+1)/2$. The moment vector $({\sum }_i{x}_i^j{)}_{j=1}^r$ has at most ${\prod }_{j=1}^r(k\lfloor Z{\rfloor }^j+1){\ll }_{k,r}{Z}^D$ possible values. If $R(v)$ counts the tuples with vector $v$, then ${\sum }_{v}R(v)=\lfloor Z{\rfloor }^k$ and $J_{k,r}(Z)={\sum }_{v}R(v{)}^2$. The Cauchy-Schwarz inequality gives $J_{k,r}(Z){\gg }_{k,r}{Z}^{2k-D}$. Combining the two lower bounds, with one common positive constant for large $Z$, yields

Here is an explicit way that upper bounds enter the Vinogradov mean-value method for a bilinear exponential sum. Write $U={\sum }_{x,y\le Z}e({\sum }_j{\alpha }_j{x}^j{y}^j)$ and $L_j=k\lfloor Z{\rfloor }^j$. Two applications of the Holder inequality, followed by grouping equal differences of moment vectors, giveAt integer ${\alpha }_{j}v$ the minimum is defined as $2L_j+1$; $\Vert u\Vert$ means distance to the nearest integer. To explain the mean-value factor, let $r(v)$ count pairs of $k$-tuples with prescribed moment difference. It is an autocorrelation of $R$, so $r(v)\le {\sum }_{w}R(w{)}^2=J_{k,r}(Z)$ by Cauchy-Schwarz inequality. The first Holder inequality groups the $y$ tuples, and the second groups the $x$ tuples, providing the two factors $J$. The remaining sums over moment differences are bounded by the displayed finite geometric series estimates. Good Vinogradov mean value upper bounds, together with rational approximation or spacing bounds for ${\alpha }_j$, therefore give cancellation in $U$. The mean-value estimate alone does not force cancellation for arbitrary coefficients: when all ${\alpha }_j$ are integers, $U=\lfloor Z{\rfloor }^2$. For the paper take $Z=N^{2/5}$ and the specified ${\alpha }_j$.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).