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Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.

There is an injection from the power set of $\omega$ into the real numbers given byIf $S$ and $T$ first differ at $k$, the contribution at $k$ has magnitude $2/3^{k+1}$, while the entire later tail has magnitude at mostTheir real values are therefore different. This proves injectivity directly and avoids any ambiguity of numeral expansions.

Conversely, map a real number $x$ to the set $\{q\in \mathbb{Q}:q<x\}$. Distinct real numbers give distinct sets because the rational numbers are dense in the real line. Fixing a bijection between $\omega$ and the countable set $\mathbb{Q}$ turns this into an injection $\mathbb{R}\to \mathcal{P}(\omega )$. Cantor-Schröder-Bernstein theorem now gives

Write $C(\mathbb{R},\mathbb{R})$ for the set of continuous real-valued functions on the real line. Constant functions give an injection of $\mathbb{R}$ into this set. In the other direction, restriction to the rational numbers is injective. Indeed, if continuous functions $f,g$ agree at every rational and $x$ is real, choose rational numbers $q_n\to x$. Continuity givesThus restriction embeds $C(\mathbb{R},\mathbb{R})$ into ${\mathbb{R}}^{\mathbb{Q}}$, which has the same cardinality as the real-sequence space because $\mathbb{Q}$ is countable. The previous part gives the upper bound $2^{{\aleph }_0}$, while constant functions give the matching lower bound. HenceThis computes the cardinality of continuous real functions. All five sets consequently have the same cardinality. The continuity assumption is decisive for restriction to a dense subset to determine a function.

Use a self-financing portfolio with ${\Delta }_t$ units of the traded asset and the remainder in the continuous-time bank account. If its value equals $V(S_t,t)$, its self-financing condition isMeanwhile Itô's formula givesMatching the Brownian motion coefficients gives the delta hedge ${\Delta }_t=V_S$. Matching drifts then cancels the physical drift $\alpha$, leaving the pricing equation for a linear asset with additive noise:The domain is the full real line: additive noise can take the asset through zero. The diffusion coefficient is $\sigma$, so the second-derivative term has no factor $S^2$. As usual, the replication argument assumes suitable differentiability and an admissible trading strategy.

For $V\in C^{1,2}$, Itô's formula givesMatch the Brownian coefficient with the self-financing portfolio by choosing the option delta $\varphi =V_s$ and bond value $\psi B=V-s{V}_s$. Equating drifts cancels the physical drift $\mu$ and givesThis is the pricing equation for a linear asset with additive noise. Its spatial domain is the whole real line, because the asset can be negative. In particular the diffusion term is ${\sigma }^2{V}_{ss}/2$, not ${\sigma }^2{s}^2{V}_{ss}/2$. The derivation is delta hedging and does not impose a boundary condition at $s=0$.

Let $x$ have a dense forward orbit $O=\{f^j(x):j\ge 0\}$, and suppose that $f$ is surjective. Every tail $O_k=\{f^j(x):j\ge k\}$ is then dense. Indeed, continuity givesand the left side equals $f^k(X)=X$ by surjectivity.

Given nonempty open sets $U,V$, choose $k\ge 0$ with $f^k(x)\in U$. The dense tail $O_{k+1}$ supplies $j\ge k+1$ with $f^j(x)\in V$. Taking $n=j-k\ge 1$ gives $f^j(x)\in f^n(U)\cap V$. This proves the required positive-time intersection property.

The result can fail without surjectivity. Take the compact metric spacewith the metric induced from the real line. All nonzero points are isolated, so continuity there is automatic; at zero, $f(1/j)\to 0=f(0)$ proves continuity. The forward orbit of $1$ is $\{1,1/2,1/3,\dots \}$, whose closure is $X$, so this is point transitivity. But $f(X)=X\setminus \{1\}$. The singletons $U=\{1/2\}$ and $V=\{1\}$ are nonempty open sets, and for every $n\ge 1$,The isolated initial point can be visited only at the start of the dense forward orbit; deleting that initial segment destroys density.

Let $S_N(x)={\sum }_{n=1}^N[a_n\cos (nx)+b_n\sin (nx)]$. Under the first coefficient bounds,This proves the uniform Cauchy condition on the whole real line and hence uniform convergence to $f$. It is the Weierstrass M-test applied with the summable majorant $2c{n}^{-1-\epsilon }$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequalityproves continuity. Also $S_N(x+2\pi )=S_N(x)$ for every $N$, so taking limits gives $f(x+2\pi )=f(x)$.

Under the stronger coefficient bounds, the derivative series has summable uniform majorant $2c{n}^{-1-\epsilon }$. Thus $S_N^{\prime }$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the fundamental theorem of calculus gives $S_N(x)-S_N(0)={\int }_0^x{S}_N^{\prime }(t)dt$. Uniform convergence permits passage to the limit inside this finite integral, since the error is at most $|x|\Vert S_N^{\prime }-g{\Vert }_{\infty }$. Therefore $f(x)-f(0)={\int }_0^{x}g(t)dt$, and continuity of $g$ gives $f^{\prime }=g$. This justifies Termwise differentiation of a Fourier series rather than assuming it:

Parametrize the logarithmic spiral byThe map is continuous into the ambient plane, hence into $X$ with its subspace topology, and it is onto by the definition of the spiral. It is injective because its radius $|h(t)|=e^t$ determines $t$ uniquely, even when the angular coordinate winds through repeated turns.

Every point of $X$ is nonzero. The inverse is thereforethe restriction of a continuous function on the punctured plane. Hence $h$ is a homeomorphism and the spiral is homeomorphic to the real line. Its accumulation at the origin causes no failure of this argument: the origin is not included in $X$.

Write $R(v)={\int }_{\mathbb{R}}v(u{)}^{2}du$. The kernel density estimator with bandwidth $h>0$ isSufficient conditions are that $K$ be a nonnegative symmetric probability density function with $R(K)<\infty$ and finite nonzero second moment ${\mu }_2(K)=\int u^{2}K(u)du$, and that $h\to 0$, $nh\to \infty$. Symmetry and the second-moment assumption give $\int uK(u)du=0$. The expectation is a convolution:Taylor's theorem with integral remainder givesThe zeroth-order term integrates to $f(x)$, and the first-order term vanishes. The second derivative is bounded and continuous, so dominated convergence, with dominating function proportional to $u^{2}K(u)$, yieldsCombining this with the supplied leading variance gives the asymptotic mean squared errorWhen $f(x)>0$ and $f^{\prime \prime }(x)\ne 0$, its derivative is $-R(K)f(x)/(n{h}^2)+{\mu }_2(K{)}^2{f}^{\prime \prime }(x{)}^2{h}^3$. It changes sign exactly once, giving the pointwise optimal kernel bandwidthPositivity of $f(x)$ is needed for this positive interior optimum. The printed condition $f^{\prime \prime }(x)\ne 0$ alone does not ensure it. For instance, $f(x)=x^2\varphi (x)$ is a normalized bounded probability density function with bounded continuous square-integrable second derivative, but $f(0)=0$ and $f^{\prime \prime }(0)=2\varphi (0)>0$. At this point the displayed leading variance vanishes and the two-term AMSE has no positive interior minimizer; finer variance terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.

For integrated error, the same expansion holds in $L^2$: translation continuity of $f^{\prime \prime }\in L^2$ applied to the integral remainder givesIndeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^{2}K(u){\int }_0^1(1-s)\Vert f^{\prime \prime }(\cdot -shu)-f^{\prime \prime }{\Vert }_{2}dsdu$; each translated difference tends to zero and is bounded by $2\Vert f^{\prime \prime }{\Vert }_2$. This proves the claimed integrated bias expansion without assuming a uniform pointwise remainder over the whole real line.

If $K_h(x)=h^{-1}K(x/h)$, direct integration of the variance givesThe last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and convolution with a probability density function does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the asymptotic mean integrated squared error and its minimizer areHere $R(f^{\prime \prime })>0$: otherwise the continuous second derivative vanishes everywhere, making $f$ affine, impossible for a probability density function on the whole real line.

Now take the standard normal density $f=\varphi$. Differentiation gives ${\varphi }^{\prime \prime }(x)=(x^2-1)\varphi (x)$. Gaussian integration givesThe kernel constants cancel from the bandwidth ratio. At $x\ne \pm 1$,Put $t=x^2$. The logarithmic derivative of $e^{t/2}/(t-1{)}^2$ is $1/2-2/(t-1)$. On $0\le t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the global minimizers are $x=\pm \sqrt{5}$, and the normal-density local-to-global bandwidth ratio isThe ratio diverges at the two inflection points $\pm 1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

The Volterra integration operator is injective and has dense range. Its range of the Volterra integration operator is $\{f\in H^1(0,1):f(0)=0\}$, and its Moore–Penrose inverse of an operator is differentiation on this domain. Thus the error estimate for $K^{†}f$ implicitly requires $f(0)=0$, in addition to the printed $H^2$ assumption. For an arbitrary $H^2$ function the same calculation bounds error against $f^{\prime }$, but that function need not be in $\mathcal{D}(K^{†})$.

Put $a=(1-\alpha )/2$, $b=(1+\alpha )/2$ and $h=\alpha /2$. The three stencils are forward, central finite differences, and backward on $[0,a)$, $[a,b)$ and $[b,1]$, respectively. All shifted evaluation points lie in $[0,1]$. For $g\in L^2(0,1)$, apply $|A-B{|}^2\le 2(|A{|}^2+|B{|}^2)$ and change variables in each of the six evaluation terms. The resulting integration intervals arePair $[0,a]$ with $[a,1-\alpha ]$, $[\alpha ,b]$ with $[b,1]$, and the two middle intervals. Each pair has disjoint interiors, so almost every point is counted at most three times. This overlap multiplicity bound for piecewise difference operators provesFor the bias on the middle interval, write the central finite difference as an average of $f^{\prime }$. The fundamental theorem of calculus giveswhere $\Vert k_{\alpha }{\Vert }_1=\alpha /4$. Extend $f^{\prime \prime }$ by zero to the real line. Young's convolution inequality bounds the middle-interval $L^2$ error by $\alpha c/4$. Combining its squared error with the supplied outer-interval estimate gives the central-difference noise-bias boundThe triangle inequality and the operator norm estimate now giveFor $c>0$, balancing noise and approximation bias minimizes this upper bound atfor sufficiently small $\delta$ that $\alpha <1/2$. Capping the parameter at $1/4$ supplies an admissible rule for all positive noise levels. More generally, $\alpha \to 0$ and $\delta /\alpha \to 0$ suffice; $\alpha =\sqrt{\delta }$ avoids needing $c$ or dividing by $c=0$.

To prove convergence for every admissible datum, not just $H^2$ data, let $f=Ku$ with $u\in L^2(0,1)$ and extend $u$ by zero. Each stencil is a forward, centred or backward average of $u$. Translation continuity in Lp makes each averaged difference from $u$ tend to zero in $L^2(\mathbb{R})$. Restricting those three bounds to their respective intervals proves $R_{\alpha }Ku\to u$. The noise-bias decomposition for linear regularization, with $\delta /\alpha \to 0$, finishes the convergent regularization of an inverse problem.

The wavefunction is a plane wave and a simultaneous formal eigenfunction of the momentum operator and the time-translation energy operator:Its momentum and energy are therefore definite, while its probability density is spatially uniform:For a particle of mass $m$, the probability current is $j=(\hslash /m)\mathrm{Im}(\overline{\varphi }{\partial }_x\varphi )=\hslash k{A}^2/m$, directed according to the sign of $k$.

On the whole real line, a nonzero constant-amplitude plane wave is not square-integrable and cannot be normalised to total probability one. It represents an idealised momentum eigenstate, interpreted through generalized eigenfunctions or as a limit of wave packets. Also, arbitrary $k,E$ do not automatically solve the Schrödinger equation for a specified potential energy: for a free particle, $E={\hslash }^2{k}^2/(2m)$, or this value plus the constant potential energy for a constant-potential region. If $A=0$, the zero function is not a physical quantum state.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

The Schwartz space on the real line isThese seminorms define its Fréchet space topology. The space of tempered distributions is its continuous dual space,

With the angular-frequency convention, the Fourier transform of a Schwartz function isIt maps $\mathcal{S}$ continuously to itself. The transform of $u\in {\mathcal{S}}^{\prime }$ is defined through the dual pairing:up to the fixed reflection and $2\pi$ factor if the inverse-transform convention is used for the test function.

Yes. By uniform convergence, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in \mathbb{R}$. ThenThe polynomial $p_n-p_N$ is bounded on the whole real line. The supplied growth property of nonconstant polynomials implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the limit of a sequence in $p_n(x)=p_N(x)+c_n$ to obtainThis is a polynomial. The uniform limits of polynomials on the real line therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.

For $c>0$ and $L>0$, D'Alembert's formula for the wave equation on the full real line isFor the specified initial data, define the cumulative initial-velocity integralThen $\boxed{y(x,t)=[A(x+ct)-A(x-ct)]/(2c)}$. The initial velocity jumps at $\pm L$, so the resulting solution is interpreted piecewise classically, or as a weak wave-equation solution across the corresponding characteristic curves.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

False. Removing a point from the real line leaves two disjoint nonempty open rays, so the remainder is disconnected. Removing a point from the Euclidean plane leaves a path-connected space: join any two remaining points by a line segment if it avoids the removed point, or by two segments through a point not on either of the two lines through the removed point and the chosen endpoints.

A proposed homeomorphism would restrict to a homeomorphism between these punctured spaces and preserve connectedness, which is impossible. Hence

The inverse tangent $T(x)=\arctan x$ is a homeomorphism from the real line with its Euclidean metric onto $I=(-\pi /2,\pi /2)$ with its subspace topology: $T$ and its inverse $\tan$ are continuous. It is also an isometry from $(\mathbb{R},\rho )$ onto $I$, because $\rho (x,y)=|T(x)-T(y)|$. Consequently, for $U\subseteq \mathbb{R}$,For a direct open ball interpretation, $B_{\rho }(x,r)=T^{-1}((T(x)-r,T(x)+r)\cap I)$ is $d$-open by continuity of $T$. Conversely, continuity of $\tan$ at $T(x)$ supplies a sufficiently small $\rho$-ball inside any prescribed $d$-ball about $x$. Thus both metrics generate exactly the same open sets.