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The average-cost Bellman equation isIf a constant $g$ and bounded relative cost function $\theta$ solve this equation and a stationary control attains the infimum in each state, that control is optimal and its long-run expected average cost is $g$. Indeed the Bellman inequality telescopes along any controlled trajectory; dividing by the number of steps removes the bounded endpoint terms and gives a lower bound $g$. The attaining policy makes the inequalities equalities. An unbounded $\theta$ requires the corresponding endpoint/transversality justification, which the question permits for this example.

To maximize occupation of zero, minimize $c(x,u)=-1_{\{x=0\}}$ and try $\theta (x)=\mu |x|$ with $\mu >0$. For $x>0$, the expected increment of $\theta$ is $\mu (2u-1)$, minimized at $u=\alpha$; for $x<0$, it is $\mu (1-2u)$, minimized at $u=1-\alpha$. At zero its expected next value is $\mu$, independent of $u$. Thus $g=\mu (2\alpha -1)$ away from zero and $g=-1+\mu$ at zero. Solving givesAn optimal control always biases a nonzero position towards zero: $u=\alpha$ on positive integers, $u=1-\alpha$ on negative integers, and any allowed $u$ at zero.

As an independent occupation check, fix the choice $u_0$ at zero. The stationary positive tail begins at ${\pi }_1={\pi }_0{u}_0/(1-\alpha )$ and has common ratio $\alpha /(1-\alpha )$; the negative tail begins at ${\pi }_{-1}={\pi }_0(1-u_0)/(1-\alpha )$ with the same ratio. Summing these geometric tails gives $1={\pi }_0[1+1/(1-2\alpha )]$, reproducing the boxed optimum. The chain has period two, but its time-averaged occupation still converges to this stationary proportion.

Interpret $\varphi (x)$ as the conditional rejection probability. The Type I error is $\alpha =E_0\varphi (X)$, the Type II error is $\beta =E_1[1-\varphi (X)]$, the size is $\alpha$, and the power is $1-\beta$. Their weighted sum isPointwise minimization over $0\le \varphi \le 1$ forces rejection where $p_1>c{p}_0$ and acceptance where $p_1<c{p}_0$, with arbitrary randomization on equality. Conversely those choices minimize every integrand. The equivalence holds almost everywhere, since changing a test on a common null set cannot change its errors. With both weights strictly positive, any simultaneous improvement of the two errors, strict in at least one, would lower this weighted sum. Every such positive-threshold likelihood ratio test is admissible for the two simple hypotheses.

For the specified densities the likelihood ratio isIt increases strictly on $(\theta ,1)$ from zero to $1/(1-\theta )$. For $0<c<1/(1-\theta )$, the likelihood-ratio test rejects when $x>\theta /[1-c(1-\theta {)}^2]$. Larger thresholds accept almost surely. Threshold zero rejects everywhere above $\theta$ and permits arbitrary rejection below $\theta$, where both weighted choices tie; values at individual endpoints do not affect errors. These describe all nonnegative-threshold likelihood ratio tests, including their limiting accept-all and reject-all cases.

The upper-tail test ${\varphi }_k$ has size $1-k^2$. If $\theta <k$, it is a positive-threshold likelihood ratio test and is admissible. If $\theta =k$, it rejects exactly the alternative's support, with power one; no test can retain power one with smaller size, so it remains admissible. If $\theta >k$, it still has power one but wastes rejection probability on $(k,\theta )$, where the alternative density vanishes. The test ${\varphi }_{\theta }$ has the same power and smaller size, soNevertheless ${\varphi }_k$ is a uniformly most powerful test at size $1-k^2$ against all $0<\theta <1$: for $\theta <k$, apply the Neyman-Pearson lemma; for $\theta \ge k$, its power is already the maximum one. It is also admissible against the composite alternative. Indeed choose any $0<\theta <k$. Strict monotonicity of its likelihood ratio makes ${\varphi }_k$ the unique positive-threshold minimizer up to null sets. A test dominating it for the composite problem would either improve the weighted simple risk at this $\theta$, impossible, or tie that risk and coincide almost everywhere. Since $p_0>0$ throughout $(0,1)$, such coincidence also precludes improvement at any other $\theta$.

A simple function is a measurable function taking finitely many values, expressible as $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ for disjoint measurable sets. For nonnegative coefficients define $\int sd\mu ={\sum }_j{a}_j\mu (A_j)$. The Lebesgue integral of a nonnegative measurable $f$ isSuppose $0\le g_n\uparrow f$ with each $g_n$ simple. Their integrals increase and are bounded above by $\int f$. For the reverse bound, fix any simple $0\le s\le f$ and $0<c<1$. The sets $E_n=\{g_n\ge cs\}$ increase to a set containing $\{s>0\}$. Hence $\int g_n\ge c\int s{1}_{E_n}$, whose right side tends to $c\int s$ by continuity from below of the measure on the finitely many level sets of $s$. Let $c\uparrow 1$ and then take the supremum over $s$. This proves $\int g_n\uparrow \int f$, including infinite limiting integrals.

The monotone convergence theorem states the same conclusion for any nonnegative measurable sequence $f_n\uparrow f$, without simplicity. The preceding lower-bound argument works unchanged: $E_n=\{f_n\ge cs\}$ and monotonicity of the integral give $\int f_n\ge c\int s{1}_{E_n}$. Taking limits and the same two suprema provesThe upper bound follows from $f_n\le f$. Finiteness of the measure makes all bounded simple integrals finite in this proof; the theorem in fact extends to arbitrary measure spaces.

A divisor on a compact connected Riemann surface is a finite formal sum $D={\sum }_p{n}_{p}p$ with integer coefficients; its degree is ${\sum }_p{n}_p$. A canonical divisor is the divisor of zeros and poles of a nonzero meromorphic differential. Two divisors are linearly equivalent when their difference is the divisor of a nonzero meromorphic function. Such principal divisors have degree zero, but equal degree does not generally imply linear equivalence of divisors.

For example take distinct points $p,q$ on a genus-one surface. If $p-q$ were principal, its function would define a degree-one meromorphic map to the sphere, since it has just one simple pole. A degree-one map between compact Riemann surfaces is a biholomorphism, contradicting their different genera. Thus these degree-one divisors are not equivalent.

Define $L(D)=\{f\text{ meromorphic}:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. The Riemann-Roch theorem statesSince all holomorphic functions on the compact surface are constant, $\ell (0)=1$. Setting $D=0$ gives $\ell (K)=g$. Multiplication by a fixed meromorphic differential identifies $L(K)$ with holomorphic differentials, so their dimension is $g$. Setting $D=K$ then givesFor $g=2$, choose a nonzero holomorphic differential, whose effective canonical divisor has degree two. Since $\ell (K)=2$, $L(K)$ contains a nonconstant meromorphic function. Its pole divisor is bounded by that effective $K$, proving existence of a nonconstant function with total pole multiplicity at most two.