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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 Arzelà-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 Arzelà-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 Arzelà-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.

A locally compact Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset \overline{V}\subset U$, with $\overline{V}$ compact. The compact Hausdorff space $\overline{V}$ is normal, so the Urysohn lemma gives a continuous function $g:\overline{V}\to [0,1]$ equal to $1$ on $K$ and $0$ on $\overline{V}\setminus V$. Extend $g$ by zero outside $\overline{V}$. The boundary values make the extension continuous, and its support is a compact subset of $U$.

For the sum $F=f+g$:

For the product $G=fg$, the answer is no in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

On the unit ball, part 1 controls the $L^2$ norm. Outside it, $|x{|}^{\alpha }\ge 1$, soHence $\Vert u{\Vert }_2\le C\Vert u{\Vert }_{\Sigma }$, proving that the embedding $\Sigma \hookrightarrow L^2$ is continuous.

For compactness, let $(u_n)$ be bounded in $\Sigma$. On each ball, it is bounded in $H^1$, so the Rellich-Kondrachov compactness theorem gives a subsequence convergent in local $L^2$. The tail estimateis uniform in $n$ and tends to zero as $R\to \infty$. A diagonal argument therefore gives convergence in all of $L^2$. This is the compact embedding of a confining-potential energy space.

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

Applying the distributional Leibniz rule twice givesEliminating the middle term provesChoose $\rho \in \mathcal{D}(\mathbb{R})$ with $\rho (0)=1$. Parts i and ii and the identity just proved, with $\phi =\rho$ and $u=xH$, yieldThus explicit continuous functions of compact support are

Changing variables givesThe average of a continuous function over $[y,y+h]$ tends to $f(y)$ as $h\to 0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated limit.

Fix a continuous function $f$ and $\epsilon >0$. By the Weierstrass approximation theorem, choose a polynomial $p\in {\mathcal{P}}_m$ such thatFor every $n\ge m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial giveConsequentlySince $\epsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the convergence of positive quadrature rules.