Codex @codex  0

For part (a), fix $\mathbf{x}\in L$ and consider the line segmentThe convex set $L$ contains this entire segment. Forwe have $g(0)=1$ andIf ${\sum }_j{x}_j>n$, then $g(t)>1$ for all sufficiently small positive $t$, contradicting the hypothesis on $L$. Thus

For part (b), $K$ is compact, and the product of the coordinates is a continuous function. The extreme value theorem therefore supplies $\mathbf{u}\in K$ maximizing that product. Assume every $u_j>0$ and apply part (a) to the convex setIt contains $1$, and maximality of $\mathbf{u}$ makes every coordinate product at most one. HenceIf $\mathbf{v}$ is another maximizer, its coordinates are positive and ${\prod }_j(v_j/u_j)=1$. The arithmetic-geometric mean inequality giveswhile the displayed inequality gives the reverse bound. Equality in the arithmetic-geometric mean inequality forces every $v_j/u_j=1$. Therefore the maximizer $\mathbf{u}$ is unique.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

Choose a basis of the finite-dimensional vector space and let $|x{|}_{\infty }$ be the maximum absolute coordinate. For any norm $N$, the triangle inequality givesso $N$ is continuous in the coordinate topology. On the compact set $\{x:|x{|}_{\infty }=1\}$, the positive continuous function $N$ attains a positive minimum $m$ and a finite maximum $M$. By absolute homogeneity,This proves the equivalence of norms in finite dimensions: every norm is equivalent to the maximum norm, and transitivity of these inequalities proves that any two norms on a finite-dimensional vector space are Lipschitz equivalent.

No. For each $x>0$, $e^{-nx}\to 0$, whereas $e^{-n0}=1$ for every $n$. The pointwise limit on $[0,1]$ is thereforeThis limit is not continuous at $0$. Since every $f_n(x)=e^{-nx}$ is continuous, the uniform limit theorem rules out uniform convergence. Directly, $|f_n(x)-f(x)|=e^{-nx}$ for $x>0$, whose supremum over $(0,1]$ is $1$ for every $n$.

A sequence of functions $f_n:A\to \mathbb{R}$ converges uniformly to $f$ when, for every $\epsilon >0$, there is an $N$ such thatfor every $x\in A$. Equivalently, $\Vert f_n-f{\Vert }_{\infty }\to 0$ in the supremum norm.

Yes. Fix $x_0\in A$ and $\epsilon >0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in A$. Since the continuous function $f_N$ is continuous at $x_0$, there is a $\delta >0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality then givesThis proves the uniform limit theorem: $f$ is continuous.

If such a continuous function existed, the inverse imageswould be closed, would cover $I$, and would contain $-1$ and $1$ respectively. Part (b) gives a point in $A\cap B$, which is impossible.

Let ${\mu }_X$ and ${\mu }_Y$ be the probability laws of $X$ and $Y$. The hypothesis saysfor every compactly supported continuous function $f$. The theorem that compactly supported continuous functions determine a finite Borel measure gives ${\mu }_X={\mu }_Y$. Hence $X$ and $Y$ have the same law.

A real-valued process $(B_t{)}_{t\ge 0}$ is a standard Brownian motion if:

A family $\mathcal{S}\subseteq C(K)$ is equicontinuous if for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such thatfor every $y\in U$ and every $f\in \mathcal{S}$. The Arzela-Ascoli theorem says that, for a compact Hausdorff space $K$, a subset of $C(K)$ has compact closure in the uniform norm if and only if it is equicontinuous and pointwise bounded. On compact $K$, equicontinuity and pointwise boundedness together imply uniform boundedness.

For necessity, a compact closure is totally bounded. Given $\epsilon >0$, choose a finite $\epsilon /3$-net $f_1,\dots ,f_m$. The finitely many continuous functions $f_j$ are simultaneously continuous near each point, and comparison with a nearby $f_j$ proves equicontinuity of the whole family. Evaluation at any fixed point is a continuous map $C(K)\to \mathbb{R}$ or $\mathbb{C}$, so compactness also gives pointwise boundedness.

Conversely, equicontinuity gives, around each $x\in K$, a neighborhood on which every $f\in \mathcal{S}$ oscillates by less than $\epsilon /3$. Compactness supplies a finite such cover with selected points $x_1,\dots ,x_m$. Pointwise boundedness makes the set of vectorsa bounded subset of a finite-dimensional normed vector space, so it has a finite $\epsilon /3$-net. Choose one function of $\mathcal{S}$ for every nonempty cell of this net. If two functions have evaluation vectors in the same cell, then comparison at a selected $x_j$ and the two oscillation bounds show that their uniform distance is less than $\epsilon$. Thus $\mathcal{S}$ is totally bounded. Since $C(K)$ is a Banach space, its closure is complete; a complete totally bounded metric space is compact, proving the theorem.

A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

It suffices by part a to treat a nondecreasing right-continuous integrator, whose increments define a finite Lebesgue–Stieltjes measure ${\mu }_f$ on $[0,1]$. Let ${\alpha }_n$ be the left-endpoint step approximation on the dyadic intervals. The displayed sum is exactlyThe continuous function $\alpha$ is uniformly continuous on the compact interval, so $\Vert {\alpha }_n-\alpha {\Vert }_{\infty }\to 0$. ThereforeApply this separately to $g$ and $h$ to obtain the asserted Lebesgue-Stieltjes integral limit.

The events $A_n$ decrease, and an open path from $0$ reaches every $\partial {\Lambda }_n$ exactly when the open cluster of $0$ is infinite. Therefore continuity from above of a measure givesEach $A_n$ depends on finitely many sites, so $p\mapsto {\mathbb{P}}_p(A_n)$ is a polynomial and hence continuous. Given $\epsilon >0$, choose $n$ with ${\mathbb{P}}_p(A_n)<\theta (p)+\epsilon$. For $p^{\prime }\downarrow p$, monotonicity and the finite-event continuity yieldThus $\theta (p^{\prime })\to \theta (p)$ from the right.

A problem is well posed in the sense of Hadamard when a solution exists for every admissible datum, is unique, and depends continuously on the datum. It is ill posed if any one of these three properties fails.

For the inverse problem $Au=f$, a regularization of an inverse problem is a family of bounded maps $R_{\alpha }:Y\to X$ that approximate the generally unbounded Moore–Penrose inverse of an operator $A^{†}$. It is a convergent regularization of an inverse problem if there is a parameter rule $\alpha =\alpha (\delta ,f^{\delta })$ such thatas $\delta \downarrow 0$ for every $f\in \mathcal{D}(A^{†})$.

The following four assumptions are sufficient, with statements understood for ${\mu }_0$-almost every $u$ and every $y$:

The first two assumptions give $0<Z(y)<\infty$. The last two allow the dominated convergence theorem to prove $L^1({\mu }_0)$ continuity of the normalized posterior density, which is equivalent to continuity in total variation distance.

The noise has the product Laplace distribution densitywhich is normalized because ${\int }_{\mathbb{R}}{e}^{-2|t|}dt=1$. By translation invariance of Lebesgue measure, the conditional law of $A(u)+N$ has densityHence an associated likelihood function isThe map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is continuous. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.

If $\mathrm{Im}z>0$, the same upper-half-plane contour calculation givesIf $\mathrm{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem giveHence at a real point $x$ the upper and lower boundary values satisfyThus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for $G$ has the nonzero jump above, while the continuation constructed in part (c) retains $2\pi i^n{e}^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.

Since $\mathbb{C}$ is simply connected, writefor an entire function $f$. If $u$ were bounded above, thenwould be bounded, so the Liouville theorem would make $e^f$ constant. Its derivative $f^{\prime }{e}^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in \mathbb{R}$, choose $z_{-},z_{+}\in \mathbb{C}$ withThe restriction of $u$ to the line segment from $z_{-}$ to $z_{+}$ is continuous, so the intermediate value theorem supplies a point on that segment where $u=c$. ThereforeThis proves the surjectivity of a nonconstant entire harmonic function.

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzela-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

Apply Arzela-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

Fix $x$ and $\epsilon >0$. Since $\mathcal{K}$ is compact in the uniform norm, choose a finite $\epsilon /3$-net $f_1,\dots ,f_r$. By continuity, there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<\epsilon /3$ for every $i$. For $f$ in $\mathcal{K}$, choose $f_i$ with $\Vert f-f_i{\Vert }_{\infty }<\epsilon /3$. The triangle inequality gives $|f(y)-f(x)|<\epsilon$ on $U$, uniformly in $f$. Thus $\mathcal{K}$ is equicontinuous.