True. The preceding distributional argument alone shows that almost-sure convergence to zero cannot hold with probability one. To establish almost-sure nonconvergence, rather than merely failure of a probability-one convergence claim, use the Kolmogorov zero-one law.

Let $A=\{B_n/\sqrt{n}\to 0\}$. For every fixed $m$, subtracting $B_m/\sqrt{n}\to 0$ shows thatThus $A$ belongs to the tail sigma-algebra of the independent unit-time increments; it is a tail event. The Kolmogorov zero-one law gives $\mathbb{P}(A)\in \{0,1\}$. If its probability were one, then $B_n/\sqrt{n}\to 0$ in probability, contrary to its standard normal distribution at every $n$. Therefore $\mathbb{P}(A)=0$. Continuous-time convergence would imply integer-time convergence, soThe more general Brownian fluctuations exceed the square-root scale result is consistent with this direct tail-event proof. No independence of the normalized values themselves has been assumed.

The nonzero quadratic residues modulo five are $1,4$. The values of $x^3+x+1$ and the numbers of possible $y$ areThere are eight affine points and one point at infinity. Thus $\#E({\mathbb{F}}_5)=9$ and part(ii) gives $\boxed{\mathrm{tr}(\varphi )=5+1-9=-3}$. The curve is nonsingular, since its short-Weierstrass discriminant $-16(4+27)$ is nonzero modulo five.

The dual operation reverses products, so $\widehat{{\varphi }^2}=(\hat{\varphi }{)}^2$. Also $\varphi \hat{\varphi }=[5]$ and $\varphi +\hat{\varphi }=[-3]$. Squaring the latter identity givesTherefore $\boxed{\mathrm{tr}({\varphi }^2)=-1}$. Over ${\mathbb{F}}_{25}$ the Frobenius is ${\varphi }^2$, of degree $25$, soMore generally the Frobenius trace recurrence has $t_0=2$, $t_1=t$ and $t_j=t{t}_{j-1}-q{t}_{j-2}$; the square calculation is its first nontrivial case.

The $q$-power Frobenius isogeny sends affine coordinates to $(x^q,y^q)$, has degree $q$, and fixes exactly the points defined over ${\mathbb{F}}_q$, including the origin. ConsequentlyThe degree is $q$ because the $q$-power Frobenius on a smooth curve has function-field inseparable degree $q$. By the permitted separability of $1-\varphi$, this separable isogeny has exactly $\mathrm{deg}(1-\varphi )$ geometric kernel points; there is no inseparable multiplicity to remove. Part(i) therefore givesThis relates the elliptic-curve point count over a finite field to the integer trace, rather than treating the point count as an independently assumed definition of the trace.

Apply the dual isogeny degree identity to $1-\varphi$. Its dual is $1-\hat{\varphi }$, soRearrangement givesThe right side is multiplication by an integer, proving the assertion. In particular the trace of an elliptic-curve endomorphism isThe argument includes $\varphi =0$ and $\varphi =1$ with the zero-map convention. Multiplying $\hat{\varphi }=[t]-\varphi$ by $\varphi$ also yields the useful characteristic relation

Suppose an integer right triangle with square area exists, and choose one with the least positive hypotenuse $c$. It is primitive. Indeed, dividing all sides by their common divisor $d$ gives an integer right triangle whose area is the square of the rational number $s/d$, where $s^2$ was the original area. Its area is an integer, since one of its legs is even. A rational number whose square is an integer is itself an integer, as follows by writing it in lowest terms. Thus division by $d>1$ would give another square-area triangle with smaller hypotenuse.

By the primitive Pythagorean parametrization, after ordering the legs the triangle has sides $n^2-m^2,2nm,n^2+m^2$, with coprime $n>m>0$ of opposite parity. Its area isThese four positive factors are pairwise coprime: the pair $n-m,n+m$ has gcd dividing $2m$ and is odd, and the other pairs have gcd one by $\gcd (n,m)=1$. Since the product is a square, each factor is a square. WriteBoth $w,z$ are odd. Odd squares are $1(\mathrm{mod}8)$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integersThey satisfyThus $(r,s,u)$ is another positive integer right triangle with square area, but its hypotenuse $u=\sqrt{n}$ is strictly smaller than $n^2+m^2=c$. This contradicts minimality. No integer right triangle has a nonzero square area. This is the Fermat right triangle theorem, with the actual infinite descent construction exhibited.

Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.

Complex conjugation preserves the real Kähler form and interchanges $\partial ,\bar{\partial }$ and their formal adjoints. Conjugating the permitted identity among the Kähler identities therefore yields $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$. The same nilpotence calculation as in part (a) gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$. Now expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$:Both mixed anticommutators vanish, proving $\boxed{{\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}}$.

With the usual exterior-boundary interpretation that the harmonic function remains bounded as $r\to \infty$, positive radial powers and $\log r$ are excluded. HenceThe series must converge locally for $r>1$, with any required finite boundary values at $r=1$. These are the interior harmonic Fourier expansions in planar concentric domains after the inversion $r\mapsto 1/r$.

If “finite” means only finite at each finite point, without a condition at infinity, growing solutions such as $r\cos \theta$ and $\log r$ are allowed. Under that literal reading the general exterior expansion iswith local convergence. Boundedness at infinity is the condition selecting the decaying exterior expansion.

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.

The fixed-point property also fails on the real line. The continuous function $\boxed{f(t)=t+1}$ maps $\mathbb{R}$ into itself but has no fixed point, since $t+1=t$ is impossible.

In a homogeneous Ohmic conductor, a steady electric current density obeys $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{j}=0$. Uniform axial current in a straight cylinder follows if its side is insulated and its flat end faces are equipotentials. The electric potential then solves Laplace's equation with those boundary conditions, and uniqueness makes it linear in the axial coordinate. Steadiness alone does not impose these contact conditions: nonuniform end potentials can produce nonuniform current even with an insulating side. Bending also generally redistributes current; in an annular conductor the azimuthal current density is proportional to the inverse radius.

For any measurable real path $x$, put $A_n(t)=n{\int }_0^t{1}_{\{|x(s)|<1/n\}}ds$. If $n_k\le n\le n_{k+1}$, nested windows give $(n/n_{k+1})A_{n_{k+1}}(t)\le A_n(t)\le (n/n_k)A_{n_k}(t)$. Consequently, when $n_{k+1}/n_k\to 1$ and the subsequence $A_{n_k}$ converges uniformly on bounded time intervals, the full sequence does also, to the same limit. This deterministic interpolation turns summable-subsequence First Borel-Cantelli lemma estimates into almost-sure convergence without incorrectly claiming that convergence in $L^2$ alone implies full-sequence almost-sure convergence.

Over an algebraically closed field of characteristic different from two, a nondegenerate quadratic form of rank at least three is irreducible, its cone has only the vertex singular, and its hypersurface ring is Cohen-Macaulay. Thus $(R_1)$ and $(S_2)$ in the Serre criterion for normality hold. In characteristic two, the reduced zero variety of ${\sum }_i{t}_i^2$ is instead the hyperplane ${\sum }_i{t}_i=0$, which is normal. The unreduced double-hyperplane scheme is not normal.

A singular point on a normal two-dimensional algebraic variety has an integrally closed local domain which is not a regular local ring. A rational double point is an important isolated example with a crepant minimal resolution.

If a transcendental meromorphic function had only finitely many simple preimages over each of five distinct finite values, their simple counting functions would be $O(\log r)$. The displayed multiplicity inequality and the Nevanlinna first main theorem would bound the truncated sum by $5T_f/2+O(\log r)$. The Nevanlinna second main theorem requires at least $3T_f$ up to its $o(T_f)$ error, a contradiction. Thus at least one value has infinitely many simple preimages; four other values can remain totally ramified.

For a nonconstant meromorphic function on the unit disc and $q\ge 3$ distinct spherical values, this version of the Nevanlinna second main theorem holds outside a set $E$ with ${\int }_{E}dr/(1-r)<\infty$, as $r\uparrow 1$. The truncated Nevanlinna counting function counts distinct preimages rather than multiplicity. The boundary-distance error is essential and need not be small relative to the characteristic. It follows from the disk Nevanlinna logarithmic derivative lemma and the growth lemma in the variable $\log (1/(1-r))$. The disk error estimate and weighted exceptional-set argument appear in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

For a nonnegative increasing continuously differentiable function not identically zero, choose $x_0$ with $\varphi (x_0)>0$. On $I=\{x\ge x_0:{\varphi }^{\prime }>{\varphi }^2\}$, $1<{\varphi }^{\prime }/{\varphi }^2$. Thus $|I|\le {\int }_{x_0}^{\infty }{\varphi }^{\prime }/{\varphi }^2\le 1/\varphi (x_0)$. Adding the finite initial interval controls all exceptions. For the zero function there are none. Taking logarithms turns derivative growth into logarithmic auxiliary-function growth in value-distribution estimates.

The truncated counting function counts each distinct $a$-point of a meromorphic function once, rather than with its local multiplicity. It uses the same logarithmic radial weights as the Nevanlinna integrated counting function, including an origin contribution $\log r$ when appropriate. The difference $N-\overline{N}$ measures integrated excess multiplicity.

The rational composition law for the Nevanlinna characteristic gives $T_f(r)=3r/\pi +O(1)$. Its zeros are precisely $2\pi i\mathbb{Z}$, all of multiplicity two, so $N(r;0)=2r/\pi +O(\log r)$ and $\overline{N}(r;0)=r/\pi +O(\log r)$. Hence both its Nevanlinna deficiency and Nevanlinna ramification index at zero equal one third.

For a degree-$d$ rational function $U$ of the Riemann sphere, choose relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit three-sphere has positive lower and finite upper bounds. Thus $\log \Vert (P,Q)(v)\Vert -d\log \Vert v\Vert$ is bounded. Averaging over a circle using a nonvanishing entire homogeneous lift of a meromorphic function proves the displayed law for the spherical characteristic. Jensen's formula and $0\le \frac{1}{2}\log (1+x^2)-{\log }^{+}x\le \frac{1}{2}\log 2$ show that it also holds for the ordinary Nevanlinna characteristic.