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A risky-asset portfolio $\phi$ financed through the risk-free asset has zero-cost discounted gainIt is an arbitrage when $G_{\phi }\ge 0$ almost surely and $\mathbb{P}(G_{\phi }>0)>0$.

True. This follows from the maximum principle for harmonic functions. A direct reduction to part (i) is also possible: for $\epsilon >0$, setThen $\Delta {\varphi }_{\epsilon }=4\epsilon >0$, soBecause $\Omega$ is bounded, ${\varphi }_{\epsilon }\to \varphi$ uniformly as $\epsilon \downarrow 0$. Taking the limit yields

False. On the open unit disk take, for $0<\epsilon <1$,Its relevant fourth derivatives satisfyNevertheless, since $x^4+y^4\le (x^2+y^2{)}^2$, every point with $0<x^2+y^2\le 1$ hasThus the maximum lies strictly inside $\Omega$.

False. On the open unit disk, letThen $\Delta \varphi =-4<0$, but the unique maximum is $phi(0,0)=0$, whereas $phi=-1$ on the boundary.

True. Continuity and compactness give a maximum of $\varphi$ on $\overline{\Omega }$. At an interior maximum the Hessian matrix is negative semidefinite, so its trace satisfies $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. The maximum must therefore occur on $\partial \Omega$. This is the strict form of the maximum principle for subharmonic functions.

Let $h=n^{-1/3}$. Away from the boundary, the estimator isIts variance is at mostand the bias of a kernel density estimator is bounded by $h\Vert f^{\prime }{\Vert }_{\infty }/2$, so its squared bias is at most $h^2/4$. This gives the claimed bound, with room to spare, whenever the full window lies in the support.

At $x=0$, however, the assertion is false as printed. For examplesatisfies $\sup (|f|+|f^{\prime }|)=3/4<1$, butwhereas $f(0)=1/2$. The mean-square error therefore tends to $1/16$, contradicting a bound tending to zero. The intended statement needs $x\ge h/2$, a boundary correction, or a condition such as $f(0)=0$.

Split $Z_1,\dots ,Z_{mn}$ into $m$ blocks of size $n$. Compute the same Kolmogorov-Smirnov statistic against $F_0$ for the observed $X$ sample and for every $Z$ block. Under $H_0$ the $m+1$ statistics are exchangeable and, because the distribution is continuous, ties have probability zero. Reject exactly when the statistic from the $X$ sample is the largest. Its null rejection probability is exactly

The Kolmogorov-Smirnov theorem states that, for continuous $F$,where $B$ is a Brownian bridge; the finite-sample law is also independent of continuous $F$. To test $H_0:F=F_0$, computeand reject when $\sqrt{n}{D}_n$ exceeds the appropriate null quantile.

For the first proposed generator,the state-dependent factor cancels when the jump probabilities are normalized. Hencedepends only on $y-x$ and is symmetric. The jump chain is therefore a one-dimensional symmetric random walk with finite variance, and is recurrent. By part (b), the corresponding continuous-time process is recurrent. This proves that at least one of the two generators has the required property.

An irreducible continuous-time chain is recurrent when, after leaving any state, it returns there almost surely. Non-explosion ensures that returns of $X$ correspond exactly to returns of its embedded jump chain, so one is recurrent precisely when the other is.

The expected total occupation time of zero isEach visit contributes an independent holding time of mean $1/q_0$. Thus this expectation is $(1/q_0)$ times the expected number of visits. If the integral is infinite, the jump chain has infinitely many expected visits; for an irreducible chain this is equivalent to recurrence.

Put $q_x=-q_{x,x}={\sum }_{y\ne x}{q}_{x,y}$. The jump chain records the successive states visited by $X$ and has transition probabilitiesBy the exponential holding-time construction, conditional on the current state $x$, the destination of the next jump has this distribution independently of the earlier path. Hence the jump chain has the Markov property.

Take the extension from part (a), and choose $M>N+n+1$. The positive functionbelongs to $X_{N,k}$. Positivity extends from nonnegative Schwartz functions to nonnegative elements obtained as $X_{N,k}$ limits. Sincepositivity givesTaking $K=\tilde{v}[w]$ proves that every positive distribution has order zero.

Continuity of $u$ at zero supplies finitely many Schwartz seminorms and a constant $C$ bounding $|u[\varphi ]|$. Increasing their two indices to common maxima $N,k$ givesThus $u$ is bounded for the $X_{N,k}$ norm. Since $\mathcal{S}$ is dense in the Banach space $X_{N,k}$, it has a unique continuous extension $\tilde{u}\in X_{N,k}^{\prime }$.

DefineThe hypothesis makes $T$ everywhere defined. The dual $V^{\ast }$ is Banach even when $V$ is incomplete. If $f_j\to f$ in $V^{\ast }$ and $T{f}_j\to a$ in ${\ell }^1$, then coordinatewise convergence gives $a_n=f(x_n)$, so the graph of $T$ is closed. The closed graph theorem therefore givesTaking the supremum over the unit ball proves the required finite bound.

The identity mapis a continuous linear bijection. Since both spaces are Banach, the bounded inverse theorem makes $I^{-1}$ continuous, giving $\Vert x{\Vert }_1\le C_2\Vert x{\Vert }_2$.

Completeness is essential. On the space $c_{00}$ of finitely supported sequences, $\Vert x{\Vert }_2\le \Vert x{\Vert }_1$, but for $x=(1,\dots ,1,0,\dots )$ the ratio $\Vert x{\Vert }_1/\Vert x{\Vert }_2=\sqrt{n}$ is unbounded. Both normed spaces are incomplete.

If a subspace $W\subset V$ contains a ball about some $w\in W$, translation gives a ball about zero in $W$, and scaling then gives every vector of $V$. Thus every proper subspace has empty interior.

If a complete infinite-dimensional $V$ had a countable Hamel basis $(e_n)$, thenEach finite-dimensional subspace is closed and, being proper, nowhere dense. This contradicts the Baire category theorem. The polynomials have the countable Hamel basis $1,x,x^2,\dots$ and are infinite-dimensional, so no norm can make that whole vector space complete.

Label the four vertices cyclically by $\mathbb{Z}/4$. Reading the arrows in the diagram, an $a$-edge moves one step in one direction and a $b$-edge one step in the other. Thus the monodromy homomorphism is(up to simultaneously reversing the chosen numbering). By the Galois correspondence for covering spaces, the subgroup associated with the chosen vertex above $x_0$ is

A covering map induces an injection on fundamental groups. There is no injection ${\mathbb{Z}}^2\hookrightarrow \mathbb{Z}/2$, and there is no injection $\mathbb{Z}/2\hookrightarrow {\mathbb{Z}}^2$ because ${\mathbb{Z}}^2$ is torsion-free. Thus neither proposed covering exists.

The Seifert-van Kampen theorem says that if a path-connected space is the union of path-connected open sets $U,V$ with path-connected intersection, thenUsing the usual polygonal cell decompositions, the torus attaches a two-cell to $S^1\vee S^1$ along $ab{a}^{-1}{b}^{-1}$, while the projective plane attaches one along $a^2$. Hence

Let $K=\ker \chi$, which is nontrivial. Every tensor power affording ${\chi }^n$ is trivial on $K$, so each of its irreducible constituents has $K$ in its kernel. Some irreducible representation is nontrivial on $K$; otherwise the regular representation, which contains every irreducible, would also be trivial on $K$. Choose its character $\psi$. Then $\psi$ is a constituent of no ${\chi }^n$, and