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Orient the unit circle counterclockwise and use the standard principal-value Cauchy operatorFor a Fourier/Laurent mode $t^n$ it gives $+t^n$ when $n\ge 0$ and $-t^n$ when $n<0$. This follows from the interior/exterior Cauchy projections, or by direct indentation of the pole. Decompose $\varphi ={\varphi }_{+}+{\varphi }_{-}$ into its nonnegative and negative modes, so $K\varphi ={\varphi }_{+}-{\varphi }_{-}$. LetThe printed equation reduces toPositive-power coefficients force ${\varphi }_{+}=J/2$, a constant; powers at most $-2$ force ${\varphi }_{-}=0$. The $t^{-1}$ coefficient then gives $-J=2$, and the integral definition is consistent with $c_0=-1$. ThusDirect substitution has $K\varphi =-1$, $J=-2$ and gives $2t^{-1}$ as required. The same projection argument on the homogeneous equation forces $J=0$ and both components zero, proving uniqueness in the usual Hölder or $L^2$ class for this singular-integral problem.

Choose a point of $K$ with both coordinates positive, so its product is positive. The set $K\cap [0,\infty {)}^2$ is nonempty and compact; the continuous function $xy$ therefore attains its maximum there. A maximizer cannot lie on either axis, where the product is zero. It consequently belongs to $K\cap P^{+}$ and also maximizes the product on that set.

For uniqueness, the function $\Phi (x,y)=\log x+\log y$ is strictly concave on the positive quadrant: its Hessian is $\mathrm{diag}(-1/x^2,-1/y^2)$, negative definite. If two distinct positive points attained the same maximal product, their midpoint would belong to $K\cap P^{+}$ by convexity and strict concavity would give a larger value of $\Phi$, hence a larger product. This contradiction proves existence and uniqueness of the positive product maximizer. This argument is positive-product maximization on a convex set; compactness should be applied to the closed quadrant intersection, not directly to the potentially noncompact set $K\cap P^{+}$.

For fixed terminal time and fixed endpoint, use the maximizing form of the Pontryagin maximum principle. For dynamics $\dot{q}=f(q,u,t)$ and running cost $L$, there are multipliers $(p,p_0)$, not all zero, with $p_0\ge 0$, and Hamiltonian of an optimal-control problem $\mathcal{H}=p\cdot f-p_{0}L$. Along an optimum, $\dot{q}={\mathcal{H}}_p$, $\dot{p}=-{\mathcal{H}}_q$, and the chosen control maximizes $\mathcal{H}$ over admissible controls almost everywhere. Fixed terminal state imposes no prescribed terminal value of $p$; fixed time imposes no free-time condition $\mathcal{H}(T)=0$. The normal case scales $p_0=1$; abnormal extremals have $p_0=0$.

For this problem the normal Hamiltonian of an optimal-control problem isMaximization gives $u=p_x+ky$, $v=p_y-kx$. The costate equations give ${\dot{p}}_x=kv$, ${\dot{p}}_y=-ku$, hence $\dot{u}=2kv$, $\dot{v}=-2ku$. Nonzero velocities rotate at constant angular rate, and rotational symmetry permits $u(0)=A$, $v(0)=0$. For a closed path, $2k=2\pi n$, with integer $n\ne 0$. Then $z(1)=A^2/(2\pi n)$, so the required positive endpoint has $n>0$, $A^2=2\pi n$ and cost $\pi n$.

To prove a global minimum rather than only classify stationary extremals, write $r=(x,y)$ and subtract its time average; this changes neither its derivative nor $\int (y\dot{x}-x\dot{y})dt$. The resulting closed mean-zero path satisfies the periodic Wirtinger inequality $\int |r{|}^2\le (4{\pi }^2{)}^{-1}\int |\dot{r}{|}^2$. This follows immediately by its Fourier expansion: each nonzero mode of the derivative has squared multiplier at least $4{\pi }^2$. The Cauchy-Schwarz inequality now givesThe argument extends to square-integrable controls by the periodic Sobolev/Fourier interpretation. Thus the cost is at least $\pi$ for every feasible control. Equality is attained byThese satisfy every endpoint and have cost $\pi$. Therefore the minimum is $\pi$, an area-constrained quadratic control cost. The global inequality also prevents an overlooked abnormal candidate from lowering it; in fact an abnormal finite maximizing Hamiltonian of an optimal-control problem would force both linear control coefficients to vanish and then force zero velocity, incompatible with $z(1)=1$.

An arbitrage is a portfolio with nonpositive initial cost, nonnegative terminal payoff almost surely, and a strict benefit either from a negative initial cost or a positive payoff with positive probability. An equivalent martingale measure is a probability law equivalent to the physical law under which discounted traded prices are martingales. The bond has constant price one, so no discount factor is needed here.

Order the four states as $UU,UD,DU,DD$. Equivalence requires positive probability in every state, since the physical product law has full support. Each stock has risk-neutral up probability $1/2$, obtained by solving $1=(1+ϵ)p+(1-ϵ)(1-p)$. Thus all equivalent martingale laws areThe two risky returns need not be independent under a pricing measure.

A replicating portfolio pays $b+\theta S_2+\phi S_3$, so its four payoffs satisfyConversely this condition suffices: take $\theta =(Y_{UU}-Y_{DU})/(2ϵ)$, $\phi =(Y_{UU}-Y_{UD})/(2ϵ)$, and choose the bond holding to match $Y_{DD}$. The relation then matches the fourth state as well. Its unique price is $(Y_{UU}+Y_{DD})/2=(Y_{UD}+Y_{DU})/2$.

For the same-direction claim, the payoff vector is $(1,0,0,1)$ and its price under $q$ is $2a$. Thus the possible strictly arbitrage-free prices fill $(0,1)$ and the lower arbitrage bound is the infimum zero. It is not achieved by an equivalent measure, and pricing the nonnegative nonzero claim at zero would itself permit arbitrage by buying it. Its payoff is not replicable, since the displayed payoff relation fails. A buyer may pay a positive amount for its insurance against joint downward moves, for a view about dependence, or according to risk preferences. Under the stated physical independent law its expected payoff is $13/25$, illustrating that a zero lower bound does not make the payoff economically worthless.