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Suppose the spectrum were empty. The resolventwould then be an entire operator-valued function. For $|\lambda |>\Vert T\Vert$,because the series converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\Vert R(\lambda )\Vert \le (|\lambda |-\Vert T\Vert {)}^{-1}$.

For fixed $x,y\in {\ell }^2$, the scalar function $⟨R(\lambda )x,y⟩$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its limit at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda )=0$, impossible for an inverse. Thus the spectrum is nonempty.

A compactly supported smooth function belongs to $H^s({\mathbb{R}}^n)$ for every real $s$. Part (a) therefore puts $u$ in $H^{s+4}$ for every $s$. Taking $s$ arbitrarily large and applying the Sobolev embedding theorem to every derivative shows that $u\in C^{\infty }({\mathbb{R}}^n)$.

For $s<t$,The increment is independent of the Brownian filtration at time $s$, and its exponential factor has mean one by the Gaussian moment-generating function. Hence the conditional expectation of the discounted time-$t$ price is the discounted time-$s$ price.

The function is continuous at $x$ exactly when ${\lim }_{t\to x}f(t)=f(x)$.

Both functions are continuous at every $x>0$. At zero, $f_1(1/(\pi /2+2\pi n))=1$ while $f_1(1/(3\pi /2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin (1/x)|\le x\to 0$, $f_2$ is continuous at zero and hence everywhere.

Let $S=\{0\}\cup \{1/n:n\ge 1\}$ and let $D$ be the Dirichlet function, equal to $1$ on rationals and $0$ on irrationals. Then $f(x)=d(x,S)D(x)$ is continuous exactly on $S$: the distance factor squeezes it to zero on $S$, while away from $S$ it is a positive continuous factor times an everywhere-discontinuous function.

For continuous $f$, the function $F(x)={\int }_a^{x}f(t)dt$ is differentiable with $F\prime =f$. Conversely, if $F\prime$ is continuous, then ${\int }_a^{b}F\prime (t)dt=F(b)-F(a)$.

The number of stops is $X\sim \mathrm{Bin}(n,1/n)$. For each fixed $k\ge 0$,the mass function of a Poisson random variable of mean one.

For radial functions, ${\nabla }^{2}h=h^{\prime \prime }+2h^{\prime }/r$. Since ${\nabla }^2{e}^{-r}=e^{-r}-2e^{-r}/r$ and ${\nabla }^2{r}^{-4}=12r^{-6}$, decay and the boundary value give

Uniform continuity means that for every $\epsilon >0$ there is a single $\delta >0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $x,y$. A $d^{\prime }$-Cauchy sequence converges uniformly pointwise to a bounded function; passing a uniform-continuity estimate through a sufficiently close member proves that the limit is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0({\mathbb{R}}^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A uniform limit of functions vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting sequence has pairwise sup distance one and no convergent subsequence.

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

The commutator follows directly on a test function:Expansion givesTo make $V_{-}=2m$, takeThenwhich tends to $2m$ at both ends.

Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hslash k-2mi$ at $-\infty$ and $\hslash k+2mi$ at $+\infty$, and no reflected wave. Their moduli agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m{\mathrm{sech}}^2$ potential is reflectionless:

Write the internal stage asThe two methods have $\theta =1/2$ and $3/4$. Expansion gives stability functionBoth therefore have order two; order three would require $\theta =1/3$.

A set is convex when it contains every segment joining two of its points. A function on a convex set is convex whenfor all $x,y$ and $0\le t\le 1$.

A euclidean domain is an integral domain with a function $d$ on nonzero elements such that for $a,b\ne 0$ there are $q,r$ with $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence

Rao–Blackwell says that if $U$ estimates a parameter and $T$ is sufficient, then $U^{\ast }=E(U\mid T)$ has the same mean and no larger variance, with strict improvement unless $U$ is already a function of $T$ almost surely. The mean statement is the tower property, whileproves the variance claim.

The joint mass factorizes through $T={\sum }_i{X}_i$, so $T$ is sufficient for $q$. If $\lambda =\sqrt{q}$, thenConditionally on $T$, $X_1\sim \mathrm{Bin}(T,1/n)$. Rao–Blackwell therefore gives the unbiased estimatorFor $n\ge 2$ the original estimator is not a function of $T$, so the variance reduction is strict.