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A compact interval in the real numbers is either the empty set or a closed bounded interval $[a,b]$ with $a\le b$. Its compactness follows from the Heine-Borel theorem.

Let $a_m\ge 0$ and $a_m\to 0$. The set $K=\{x=(x_m):x_m\in \mathbb{R},|x_m|\le a_m\}$ is a compact metric space for the supremum norm metric. Every member is a sequence tending to zero.

For any sequence in $K$, successively extract a convergent subsequence in each compact interval $[-a_m,a_m]$, then take diagonal indices. The coordinate limits define $x\in K$. For any $M$, coordinatewise differences in the tail are bounded by $2{\sup }_{m>M}{a}_m$, independently of the sequence index. This tends to zero as $M\to \infty$, while finitely many initial coordinates converge together. Hence the diagonal subsequence converges in the supremum norm. This proves sequential compactness, equivalent to compactness for metric spaces.

On $[0,1]$, for $n\ge 2$, these continuous functions have support $[0,2/n]$, height $n$, and integral one. Their pointwise limit is zero: at zero every value vanishes, while every fixed positive argument eventually leaves the support. Thus pointwise convergence of continuous functions on a compact interval does not suffice for convergence of their integrals. The mass becomes more concentrated as its height grows.

Near $x=0$, separation of variables gives $dy/y=dx/(1-x)$, so $\log |y|=-\log (1-x)+C$. The initial condition selectsDirect substitution verifies both the ordinary differential equation and the initial condition. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.

For $|x|\le h<1$ and any two admissible values $y,\tilde{y}$,This is a Lipschitz condition in the dependent variable, uniform in $x$, with Lipschitz constant $L=(1-h{)}^{-1}$. Also $|f(x,y)|\le (1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the Picard-Lindelöf theorem hold.

Integrating from the initial point gives the integral equationConversely, differentiating this equation using the fundamental theorem of calculus recovers the initial value problem. Start Picard iteration with the constant initial approximation $y_0(x)=1$ and setThe first three approximations areTo prove the general expression by mathematical induction, put $\ell (x)=\log (1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the substitution rulegivesThus the formula follows at every stage.

For every fixed real $x<1$, these are partial sums of the exponential series evaluated at the finite real number $-\log (1-x)$. ThereforeThis range is larger than $|x|<1$: the expansion is in powers of $\log (1-x)$, not a Taylor series in $x$. For example it converges at $x=-2$. It also converges uniformly on every compact interval contained in $(-\infty ,1)$, because $|\log (1-x)|$ is bounded there and the exponential series supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is Picard iteration for a simple-pole linear differential equation.

For real $s$, the Sobolev space $H^s(\mathbb{R})$ consists of those tempered distributions $u$ whose Fourier transform is represented by a locally square-integrable function and for whichThis is the Sobolev norm used here; inserting the fixed factor $(2\pi {)}^{-1}$ gives an equivalent convention.

For the constant function $c$, Fourier inversion at zero givesA nonzero Dirac delta distribution cannot be represented by a locally square-integrable function. Indeed, if $G$ represented $2\pi c{\delta }_0$, choose a test function $\rho$ with $\rho (0)=1$, and put ${\rho }_{\epsilon }(\xi )=\rho (\xi /\epsilon )$. On a fixed compact interval containing their supports, Cauchy-Schwarz inequality givesThus $c=0$. Every Sobolev space requires local square integrability of the Fourier transform, regardless of how negative $s$ is. ThereforeThe zero constant function belongs to every Sobolev space.

By the Heine-Cantor theorem, a continuous function on the compact interval $[a,b]$ is uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s,t\in [a,b]$ and $|s-t|<\delta$.

Choose a finite partition of an interval $a=t_0<t_1<\cdots <t_m=b$ with every width below $\delta$. Define the step functionFor each $t<b$, its distance from the chosen left endpoint is below $\delta$, so $|f(t)-\varphi (t)|<\epsilon$; at $b$ the error is zero. HenceA step function may have separately assigned values at its finitely many partition points; these values do not affect its Riemann integral. If $a=b$, simply take $\varphi =f$ on that singleton.

Let $f_n$ be the continuous triangular bumpand set $f_1=0$. At $x=0$ every term is zero, and each fixed $x>0$ eventually lies outside the support, so $f_n\to 0$ pointwise. The triangle has base $2/n$ and height $n$, henceThis concentrating triangular bump shows that pointwise convergence does not preserve integrals, even for continuous functions on a compact interval.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.

Yes. Let $(x^{(k)}{)}_{k\ge 1}$ be any sequence in $A$, where $x^{(k)}=(x_m^{(k)}{)}_{m\ge 1}$. At coordinate $m$, the values lie in the compact interval $[-1/m,1/m]$. By part (a)(ii) we can successively extract a convergent subsequence for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for whichThe closed interval bound passes to the limit, so $|x_m|\le 1/m$. Hence $x_m\to 0$ and $x=(x_m)\in A$.

Coordinate convergence alone would not imply supremum norm convergence. Here the common tail bound supplies the missing estimate. Given $ϵ>0$, choose $M$ with $2/(M+1)<ϵ$. For $m>M$,For the finitely many $m\le M$, coordinate convergence makes all differences less than $ϵ$ once $j$ is large. Therefore $d(x^{(k_j)},x)<ϵ$, andThis is an instance of compactness from uniformly vanishing sequence tails. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.

Every quotient class has a representative in $[0,1]$: the collapsed class has representatives $0,1$, and the others are the points of $(0,1)$. Thus $p{|}_{[0,1]}$ is a continuous surjection from a compact interval onto $X_A$. It follows that $X_A$ is compact, despite the noncompact original space.

Define the continuous mapThe formulas agree at both junctions, and its fibres are exactly the required equivalence classes. It induces a continuous bijection $X_A\to S^1$, which is a homeomorphism by compactness and the Hausdorff condition on the circle.The quotient notation means that the displayed entire subset becomes one point.

For real constants $a,c$, apply Picard iteration to the initial value problem $y^{\prime }=ay/(1-x)$, $y(0)=c$, starting with $y_0(x)=c$. The recurrence isSince ${\int }_0^x{\log }^k(1-t)/(1-t)dt=-{\log }^{k+1}(1-x)/(k+1)$, mathematical induction gives the displayed partial sums. The exponential series converges at every finite argument, so $y_n(x)\to c(1-x{)}^{-a}$ for every real $x<1$, uniformly on each compact interval in that range. This is not a Taylor series in $x$, so its convergence is not restricted to $|x|<1$. The differential equation's coefficient is singular at $x=1$ even when a special solution has a removable singularity there.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

A continuous function on a compact interval is uniformly continuous by the Heine-Cantor theorem. Choose a partition of an interval whose widths lie below the input tolerance for a desired output error $\epsilon$. On each half-open partition subinterval, use the value of the function at its left endpoint, and assign the rightmost endpoint its own value. This defines a step function whose error is strictly below $\epsilon$ at every input. Finitely assigned partition-point values are allowed and do not alter the Riemann integral.

Every continuous real-valued function on a compact interval can be approximated by polynomials with arbitrarily small error everywhere on the interval. More precisely, if $a<b$ and $f:[a,b]\to \mathbb{R}$ is continuous, then for every $\epsilon >0$ there is a polynomial $p$ withThe error is measured in the supremum norm, so one approximation works at every point. The degenerate interval $a=b$ needs only a constant function. A change of variable $t=(x-a)/(b-a)$ reduces the nondegenerate case to $[0,1]$.^[ref]

This is useful because a polynomial is described by finitely many coefficients and can be evaluated with addition and multiplication. Its surprise is that the target need not be differentiable: the sharp corner of an absolute value function can be approximated by smooth polynomials. The Bernstein approximation of a corner below makes this visible. Approximating values, however, does not by itself approximate derivatives; Weierstrass approximation does not preserve derivatives gives a counterexample.