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Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

Assume the intended nondegenerate interval $a<b$ and real-valued continuous functions. The integral is bilinear and symmetric. Since $w>0$, it gives $⟨f,f⟩\ge 0$. If a continuous $f$ is nonzero at some point, continuity makes $f^{2}w$ strictly positive on a subinterval of positive length, so its integral is positive. Hence $⟨f,f⟩=0$ forces $f=0$, proving positive definiteness. Orthogonality means $\boxed{{\int }_a^{b}f(x)g(x)w(x)dx=0}$. If $a=b$, the displayed integral is identically zero and is not an inner product on the stated function space; that degenerate case must be excluded.

Define $L_{\xi }(f)={\lim }_N{A}_N(f)$. Limits preserve linearity and positivity, and $|L_{\xi }(f)|\le \Vert f{\Vert }_{\infty }$, $L_{\xi }(1)=1$. The Riesz-Markov-Kakutani representation theorem supplies a unique Radon probability measure ${\mu }^{\xi }$. By its definition,which is exactly weak-star convergence in $C(\mathbb{T}{)}^{\prime }$.

If $\xi /(2\pi )=p/q$ in lowest terms, the orbit has exactly $q$ points. Averaging a periodic sequence givesThis includes $q=1$. If $\xi /\pi$ is irrational, then $e^{ik\xi }\ne 1$ for every nonzero integer $k$. The preceding calculation gives $L_{\xi }(z^k)=0$ for $k\ne 0$ and $L_{\xi }(1)=1$, exactly the integrals against normalized Haar measure $m$. Density of trigonometric polynomials implies equality on all continuous functions, and uniqueness givesThese are the periodic-orbit and irrational-rotation cases of equidistribution.

For each $n\ge 0$, choose an open arc $I_n$ about $e^{in}$ of arclength $\pi 2^{-n-2}$, and put $G^1={\bigcup }_{n\ge 0}{I}_n$. It contains every point used by every empirical measure, whereasThus ${\mu }_N^1(G^1)=1$ for all $N$. Let $K=\mathbb{T}\setminus G^1$. This is closed and ${\mu }_N^1(K)=0$ for all $N$. But $1/\pi$ is irrational, so the weak-star limit is normalized arclength andTherefore the proposed implication is false. Weak-star convergence tests continuous functions and does not assert convergence of the indicator function of an arbitrary closed set.

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

Set $C={\sup }_k\Vert {\mu }_k\Vert$. Given $f\in C(\mathbb{T})$, choose $g\in X$ with $\Vert f-g{\Vert }_{\infty }<\epsilon$. ThenTake the upper limit in $k$ and then $\epsilon \downarrow 0$. Thus convergence on the dense test family suffices; no assumption that $X$ is a vector space is needed.

For nonempty compact $K$, put $f_j(x)=\max \{0,1-j\mathrm{dist}(x,K)\}$. These continuous functions decrease to $1_K$ and satisfy ${\mu }_k(K)\le \int f_{j}d{\mu }_k$. Consequently ${\mathrm{limsup}}_k{\mu }_k(K)\le \int f_{j}d\mu$ for every $j$. Dominated convergence for the finite positive measure $\mu$ yieldsThe empty-set case is immediate. Positivity is essential for the comparison with these cutoff functions.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

For $s>0$, the denominator $s(s+1)$ is positive, so every value of $f$ is positive. Conversely, for every $r>0$, the positive number $s=(\sqrt{1+4/r}-1)/2$ satisfies $f(s)=r$. Thus its image of a function, or range, isOne can also see the full range from continuity and monotonicity: $f$ is a continuous function with $f^{\prime }(s)=-(2s+1)/(s^2+s{)}^2<0$, and its endpoint limits are $+\infty$ as $s\to 0^{+}$ and $0$ as $s\to \infty$. The endpoints themselves are not attained.

It is surjective but not injective; there is no global inverse function. Noninjectivity follows fromFor surjectivity, restrict to $x<1$, whereThis restriction is a continuous function, tends to $-\infty$ as $x\to -\infty$ and to $+\infty$ as $x\to 1^{-}$. The intermediate value theorem therefore supplies a preimage for every real number. The additional branch $x>1$ destroys global injectivity but cannot destroy surjectivity.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

A Peano zero through order $r$ means that the function admits the zero Taylor polynomial with remainder $o(|t{|}^r)$ at $x_0$. This does not assert the existence of ordinary iterated derivatives on a neighborhood. For example, multiplying $t^{r+1}$ by a bounded continuous function with dense points of nondifferentiability preserves this estimate while destroying neighborhood differentiability.

For $2\pi$-periodic continuous functions, convergence of positive linear operators on continuous functions to the identity on $1,\sin x,\cos x$ implies uniform convergence on every function. Indeed $1-\cos (t-x)$ is nonnegative and has just one zero on the circle, so the compact Korovkin test space criterion applies.

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.

A piecewise smooth curve has a continuous function as a parametrization on a compact interval, with a finite partition into subintervals on each of which the parametrization is continuously differentiable. One-sided derivatives may differ at joining points, so polygonal contours are allowed. A contour integral is the sum of the integrals over those smooth pieces.

Choose $a_U$ in each bounded component of $\mathbb{C}\setminus K$. Every nonvanishing continuous function on $K$ extends nonvanishingly to a neighborhood. Use finite-hole neighborhoods of a planar compact set within that neighborhood, and subtract its finite list of winding obstructions by multiplying by $\prod (z-a_U{)}^{-k_U}$. The continuous logarithm lifting criterion gives the remaining exponential. Uniqueness follows by extending a purported logarithm of such a finite product to a smaller neighborhood: a small multiplicative error has a local logarithm, whereas its winding around the $U$th hole is $k_U$. Thus all $k_U$ vanish. The quotient by exponentials is the free abelian group on the bounded complementary components; each element has finite support.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

A point interaction concentrates a coefficient of a linear differential equation at a single interior point. On an interval $0<x<\ell$, with $0<c<\ell$, consider $L_{\alpha }y=y^{\prime \prime }+\alpha y(c){\delta }_c$. The domain consists of continuous functions that are twice differentiable away from $c$, with the required one-sided derivatives. Multiplication of the Dirac delta by $y$ means the unambiguous point value $y(c){\delta }_c$. It does not mean an undefined product of a delta with an arbitrary discontinuous function.

The equation is interpreted using distributional derivatives. Smooth forcing gives an ordinary equation on each side, with transmission conditions for a delta coefficient at the interaction. A delta potential in the one-dimensional Schrödinger equation is a physical example after choosing the corresponding sign and constants.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.