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Started at radius one, planar Brownian motion avoids the origin, and its logarithmic radius is a continuous local martingale because the logarithmic potential is harmonic away from zero. It is a strict local martingale: at deterministic $t>0$, the angular average of a logarithmic potential gives $\mathbb{E}\log |B_t|={\int }_1^{\infty }{e}^{-r^2/(2t)}dr/r>0$, whereas the initial value is zero. Absolute integrability follows from integrability of $r|\log r|$ near zero and the Gaussian tail. Its positive expectation does not contradict nonnegative-local-martingale bounds, since this process takes both signs.

Brownian scaling gives $|B_t|\stackrel{d}{=}\sqrt{t}R$, where $R=|B_1|$ has Rayleigh distribution with density $r{e}^{-r^2/2}$, $r>0$. For every $p>0$,Near zero this is bounded by ${\int }_0^1|\log r{|}^{p}rdr={\int }_0^{\infty }{u}^p{e}^{-2u}du$, which is finite; near infinity the Gaussian tail dominates any logarithmic power. The inequality $|a+b{|}^p\le C_p(|a{|}^p+|b{|}^p)$, valid also for $0<p<1$, now yieldsLet $c=\mathbb{E}\log R$, a finite constant. The same scaling identity givesAn integrable martingale has constant expectation, so the logarithm of planar Brownian radius is a strict local martingale on $[1,\infty )$. Fixed-time integrability, even of every positive power, does not imply the stopping-time uniform integrability required in part a.

Write $r=\Vert y\Vert$ and use the ordinary Euclidean Laplacian ${\sum }_i{\partial }_i^2$, so the generator of standard Brownian motion is half that operator. For a radial function $f(r)$, the Laplacian in polar coordinates is $f^{\prime \prime }(r)+(d-1)f^{\prime }(r)/r$. With $f(r)=r^{2-d}$,Thus $h$ is a harmonic function on the punctured space.

We first justify that the singularity is not visited. For $0<\eta <x<R$, stop at $S={\tau }_{\eta }\wedge {\tau }_R$. The Itô formula makes $h(B_{t\wedge S})$ a local martingale, and it is bounded on this annulus, hence a true martingale. The exit time is finite almost surely: $\mathbb{P}(S>t)\le \mathbb{P}(\Vert B_t\Vert <R)\to 0$ by the Gaussian transition density. Taking the bounded terminal limit yieldsIf the origin were reached before ${\tau }_R$, every smaller radius would first be reached, so letting $\eta \downarrow 0$ shows that event has probability zero. Taking a countable sequence $R\to \infty$ proves ${\tau }_0=\infty$ almost surely. Localization on increasing annuli now applies the Itô formula globally and gives the Bessel power local martingaleStopping this continuous local martingale at any ${\tau }_a$, including $a=0$, preserves its local-martingale property.

For $0<a<x$, the stopped path stays at radius at least $a$, so $0<M_t\le a^{2-d}$. The bounded local martingale criterion makes it a true martingale. For $a=x$, Brownian radial oscillations at its starting sphere give ${\tau }_x=0$ almost surely and the stopped process is constant. Thus the same true-martingale conclusion holds for $0<a\le x$.

For $a=0$, the process is a strict local martingale. To prove this, write $B_t=\bar{x}+\sqrt{t}G$ in distribution, with $G$ a standard $d$-dimensional normal vector. For every shift $v$,Indeed the integrand outside the unit ball is at most one, while inside it the Gaussian density is at most $(2\pi {)}^{-d/2}$ and the radial integral is a constant times ${\int }_0^{1}rdr$. Brownian scaling therefore givesA true martingale would have expectation $h(\bar{x})>0$ at every time, which is impossible.

The printed range also permits $a>x$, and this case must not be grouped with the bounded case. Here ${\tau }_a<\infty$ almost surely, by the same Gaussian bound for exit from a bounded ball. Moreover,since the second expectation is at most $\mathbb{E}h(B_t)\to 0$. It too is a strict local martingale. The complete stopped inverse radial power martingale classification is therefore

On a fixed finite horizon, a self-financing strategy consists of predictable stock holdings $h^i$, integrable against the price processes, and a bank account holding $h^0$. Its wealth and gains satisfyAn admissible trading strategy has discounted wealth $X_t/B_t$ bounded below by a deterministic constant throughout the horizon. With the bounded short rate in this part, an undiscounted deterministic lower bound is equivalent after adjusting the constant. An arbitrage is an admissible zero-initial-cost strategy with $X_T\ge 0$ almost surely and $\mathbb{P}(X_T>0)>0$.

Put $1=(1,\dots ,1{)}^{\top }$ and define the market price of riskThe bounded coefficients and bounded inverse make $\lambda$ bounded. Define the stochastic exponentialThe Novikov condition states that $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T\Vert {\lambda }_s{\Vert }^{2}ds)<\infty$ makes this exponential a true martingale with expectation one. Boundedness verifies the condition. Therefore $dQ=L_{T}dP$ defines an equivalent probability measure. The Girsanov theorem states thatis a $d$-dimensional Brownian motion under $Q$. Substituting into the asset equations givesThus $Q$ is an equivalent local martingale measure. The sufficient direction of the fundamental theorem of asset pricing says that such a measure excludes arbitrage among admissible self-financing strategies. Here the reason can be seen directly: $X/B$ is a local martingale under $Q$, and its lower bound makes it a supermartingale. Consequently $X_0=0$ implies ${\mathbb{E}}_Q[X_T/B_T]\le 0$, incompatible with the nonnegative terminal payoff of an arbitrage. Therefore the bounded-inverse market has no arbitrage.

For the requested law of one price counterexample, take a broader continuous-time local-martingale model on $[0,T]$. Let $X$ be a Brownian motion, let $\tau =\inf \{u\ge 0:X_u=-1\}$, and set $q(t)=t/(T-t)$ for $t<T$. DefineThe given hitting-time fact gives $\tau <\infty$ almost surely. The corresponding physical hitting time is $T\tau /(1+\tau )<T$, so $S$ reaches one and then stays one. In particular it is continuous at maturity and $S_t\ge 1$.

For a rigorous closed-horizon localization, stop additionally when $S$ hits $n+2$, or at $T$ if it never does. The stopped process is Brownian motion stopped on first exiting $(-1,n)$ and subjected to the deterministic clock; it is a bounded martingale, including at $T$. The stopping times increase to $T$, since every Brownian path on the finite random interval $[0,\tau ]$ has a finite maximum. Thus $S$ is a positive local martingale on the whole horizon. With the time-changed Brownian filtration, the original probability measure is an equivalent local martingale measure. The admissible-wealth supermartingale argument above proves no admissible arbitrage.

Nevertheless $S_T=B_T=1$ while $S_0=2\ne B_0=1$, so the law of one price fails. Shorting one share and buying two units of cash has zero initial cost and terminal gain one, but its wealth is $2-S_t$. It has no deterministic lower bound: for every $n$, the probability of Brownian motion hitting $n$ before $-1$ is $1/(n+1)>0$, obtained by optional stopping of the bounded stopped Brownian motion. That apparently profitable strategy is therefore inadmissible. The counterexample deliberately uses a strict local martingale; the bounded-coefficient class in the first part has true discounted asset martingales and does satisfy the law of one price.

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

For a three-dimensional Bessel process started at one, $R$ stays strictly positive and satisfies $dR=dW+R^{-1}dt$. The Itô formula makes its reciprocal a local martingale, but $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$. Thus the reciprocal is a strict local martingale. It can deflate the market with constant bank account and stock price $R$ without yielding a true pricing density for the bank account.

For Brownian motion started at radius $x>0$, the displayed process is a continuous local martingale. It is a bounded true martingale when $0<a\le x$, but a strict local martingale when $a=0$ or $a>x$. In the latter case the boundary is outside the starting point, so stopping does not remove the singularity at the origin. Brownian scaling gives $\mathbb{E}\Vert B_t{\Vert }^{2-d}=O(t^{-(d-2)/2})$; when $a>x$, the stopped expectation tends to $a^{2-d}<x^{2-d}$.