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The Lévy continuity theorem states that if characteristic functions ${\phi }_n$ of probability distributions ${\mu }_n$ converge pointwise on $\mathbb{R}$ to $\phi$, and $\phi$ is continuous at zero, then $\phi$ is the characteristic function of a unique probability distribution $\mu$ andHere the arrow denotes weak convergence of probability measures, equivalently convergence in distribution of associated random variables. Conversely, weak convergence of probability measures implies pointwise convergence of their characteristic functions to the characteristic function of the limit. The continuity-at-zero condition is essential to ensure that the pointwise limit is a characteristic function.

Write $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $\phi (t)=\mathbb{E}{e}^{it{X}_1}$. The given two-sided limit $|f(u)|/u\to 0$ implies $|f(u)|/|u|\to 0$, so $f(u)=o(|u|)$. For each fixed real $t$, independence and the characteristic function of a sum of independent variables giveSet $z_n=iat/n+f(t/n)$. For $t\ne 0$, $n{z}_n\to iat$ and $z_n=O(n^{-1})$. The local Taylor expansion of the complex logarithm at $1$ givesThus $\mathbb{E}{e}^{it{A}_n}\to e^{iat}$; at $t=0$ the identity is immediate. The limit is continuous at zero and is the characteristic function of the constant random variable $a$. The Lévy continuity theorem proves the weak law from a characteristic-function expansion:No integrability hypothesis has been used. The conclusion is also convergence in probability by (d)'s convergence in distribution to a constant implies convergence in probability. If the word “constant” were to allow complex $a$, the characteristic function symmetry $\phi (-t)=\overline{\phi (t)}$ forces $a=\overline{a}$, so it is necessarily real.

Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

The Neumann boundary conditions give Fourier modes $q=n\pi /L$, including the spatially uniform mode $q=0$. In the frame translating with the mean deposition height, that mode has no restoring force and obeysHere $W_t$ is Brownian motion, independent of the initial height. More generally $⟨h_0(t{)}^2⟩=⟨h_0(0{)}^2⟩+{\sigma }^{2}t$ when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for ${\sigma }^2>0$. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, hence vanish for every $k\ne 0$, contradicting its continuity at zero and $\hat{\mu }(0)=1$. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has ${\dot{h}}_q$; the TeX's ${\hslash }_q$ is a transcription error.

For $d=1$ and $X\sim N(0,{\sigma }^2)$, the characteristic function isFor a multivariate normal distribution with covariance matrix ${\sigma }^2{I}_d$, the coordinates are uncorrelated and hence independent by independence of uncorrelated jointly normal variables. ThereforeThis is the characteristic function of an isotropic Gaussian vector.

The integrands are absolutely integrable because $f$ has compact support, $g\in L^1(\mathbb{R})$, and $|{\phi }_Z|\le 1$. Fubini's theorem therefore permits us to insert ${\phi }_Z(s)=\mathbb{E}[e^{isZ}]$ and interchange the expectation and integrals:With $r=\epsilon s$, the inner integral in $s$ isNow set $t=(Z-x)/\epsilon$. The change of variables $x=Z-\epsilon t$ giveswhich is the required identity.

Choose any $g$ with $g,\hat{g}\in L^1$, $g(0)\ne 0$, and for which Fourier inversion theorem applies, for example a Gaussian function. The left side factors through the Fourier transform of $f$ asOn the right side, $f(Z-\epsilon t)\to f(Z)$ andThe dominated convergence theorem and Fourier inversion at zero giveConsequently the Fourier-cutoff recovery of expectations from a characteristic function isThe right side depends only on ${\phi }_Z$. Equal characteristic functions therefore give equal expectations for every compactly supported continuous $f$, and part a yields the uniqueness theorem for characteristic functions.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

A finite linear combination of independent Gaussian random variables is Gaussian, soSince $v_n\to v:={\sum }_{k\ge 1}{a}_k^2<\infty$, the characteristic functions converge pointwise:The Lévy continuity theorem yields convergence in distribution toOn the given common probability space, the same variance calculation actually proves convergence in $L^2$ as well.

Put ${\lambda }_i=z|V_i|$. Thenand ${\lambda }_i\to \infty$. The Poisson central limit theorem givesSince $\sqrt{{\lambda }_i}=\sqrt{z}\sqrt{|V_i|}$,

Equivalently, the characteristic function of the left-hand side iswhich is the characteristic function of that centered normal distribution.

Let $V_1,\dots ,V_d$ be independent standard Cauchy random variables. Their characteristic functions and independence giveTaking real parts yieldsFor each realization of $V$, both products are rank-one positive-semidefinite kernels. Their sum and then their expectation remain positive semidefinite by the closure property of positive-semidefinite kernels. This is a Random Fourier feature representation of the Laplace kernel.

Let $(Y_n)$ be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function isIts first absolute moment diverges sinceOn the other hand, its characteristic function is $\phi (t)=e^{-|t|}$. The characteristic function of a sum of independent variables therefore givesThus the stability of the Cauchy distribution implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,where $Z$ is standard Cauchy and $\stackrel{d}{\to }$ denotes convergence in distribution.

The $d$-dimensional Lévy characterization of Brownian motion says that a continuous local martingale $X$ with $X_0=0$ is a standard $d$-dimensional Brownian motion exactly whenfor all $i,j$ and $t$.

One direction follows directly from independent Gaussian increments. Conversely, fix $\theta \in {\mathbb{R}}^d$. Applying Itô formula and the bracket assumption shows thatis a complex local martingale. After stopping $X$ on leaving large balls it is bounded, so optional sampling and then dominated convergence give, for $s<t$,This is the characteristic function of $N(0,(t-s)I_d)$ and is deterministic. Thus each increment is Gaussian with the required covariance and independent of the past. Together with continuity, these are precisely the defining properties of standard $d$-dimensional Brownian motion.

The phase screen adds the phase accumulated across its thickness, so the reduced field just after the screen isFor a zero-mean random variable $w$ with a normal distribution and variance ${\sigma }^2$, its characteristic function givesThe same expression is approximately valid for a non-Gaussian weak fluctuation: the cumulant expansion begins with $-k^2{\xi }^2{\sigma }^2/2$, while higher cumulants give higher-order corrections.

For $x>0$, every realization obeys the parabolic wave equationLinearity permits ensemble averaging, soThe initial mean is independent of $z$, hence diffraction does not change it and $⟨E(x,z)⟩=\exp (-k^2{\xi }^2{\sigma }^2/2)$ for every $x\ge 0$.

If $N_{\lambda }\sim \mathrm{Pois}(\lambda )$ and $\lambda \to \infty$, thenIndeed, its characteristic function is

For a Poisson random measure on $E$ with finite intensity $\nu (E)$, its integral of a finite-valued measurable mark function is the finite sum of the marks at its random atoms. Its characteristic function is $\mathbb{E}\exp (iu\int fdN)=\exp (\int (e^{iuf}-1)d\nu )$. On a time-mark space with intensity $ds\nu (dz)$ and finite mark intensity, restriction to each bounded time interval has finite intensity and defines a Compound Poisson process. Zero-valued marks do not affect the integral and need not appear in its Lévy measure.

If $Z_1,Z_2$ are independent $N(0,1)$ random variables, their product has characteristic function $(1+t^2{)}^{-1/2}$ and density $K_0(|x|)/\pi$, where $K_0$ is a modified Bessel function. A scale multiple $a{Z}_1{Z}_2$ has the corresponding scaled distribution.

An independent unit-scale Rayleigh distribution variable $R$ and standard normal distribution variable $Z$ give $RZ$ with Laplace distribution of density $e^{-|x|}/2$. Indeed its characteristic function is $\mathbb{E}{e}^{-t^2{R}^2/2}=1/(1+t^2)$.

The standard Cauchy distribution is $\mathrm{Cauchy}(0,1)$, with density $1/[\pi (1+x^2)]$ and characteristic function $e^{-|t|}$.