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A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Work in one fixed inertial frame, and take positive velocity in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of mass $-m^{\prime }(\tau )d\tau$. Its exhaust has inertial velocity $v(\tau )-u$, by the Galilean transformation of velocities. Once expelled, that material retains its velocity because it experiences no external force. The system's momentum is therefore the rocket momentum plus the sum of all these exhaust momenta:The minus sign comes from the positive expelled mass being $-m^{\prime }(\tau )d\tau$.

Differentiate, applying the product rule to the rocket momentum and the fundamental theorem of calculus to the integral:The momentum conservation law thus gives the nonrelativistic rocket equation

The same bookkeeping for kinetic energy givesIts derivative iswhere the rocket equation cancels the first two terms. Thus kinetic energy increases during mass ejection, with $\boxed{K^{\prime }=-u^2{m}^{\prime }/2>0}$ when $u>0$ and $m^{\prime }<0$. This does not violate conservation of energy: the increase comes from stored internal energy.

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.

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion

Set $h(x)=f(x)-yx$. ThenThe extreme value theorem gives a minimum of $h$ on $[a,b]$. If it occurs in $(a,b)$, the two-sided difference quotient gives $h^{\prime }(c)=0$. If it occurs at $a$, either $h^{\prime }(a)=0$ or a negative right derivative contradicts minimality; the endpoint $b$ is analogous. Hence some $c\in [a,b]$ satisfiesThis proves the Darboux theorem for derivatives in the stated case without assuming that $f^{\prime }$ is continuous.

Because a differentiable function is continuous, defineThe fundamental theorem of calculus gives $F^{\prime }(x)=f^{\prime }(x)+f(x)$. At the endpoints,whileApplying the result just proved to $F$ yields $d\in [a,b]$ with

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

Construct a mean-square derivative of a Gaussian process on the original probability space before choosing its continuous modification. The real covariance function satisfies $K(h)=K(-h)$, because covariance is symmetric. Three-times differentiability implies that $K^{\prime \prime }$ is continuous everywhere and differentiable at zero. Since $K^{\prime \prime }$ is an even function,This uses differentiability of $K^{\prime \prime }$ at zero, not a bound on $K^{\prime \prime \prime }$ throughout a neighborhood.

Let $D_{h}X(t)=(X(t+h)-X(t))/h$ for $h\ne 0$. Twice applying the fundamental theorem of calculus to the covariance function givesOriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K^{\prime \prime }(0)$ as $h,k\to 0$. In particular,By completeness of L2 space, the mean-square derivative of a Gaussian processexists. Finite vectors consisting of values of $X$ and $Y$ have a centered multivariate normal distribution: approximate the $Y$ coordinates by difference quotients and use mean-square convergence and characteristic functions. Thus $Y$ is a centered Gaussian process jointly constructed with $X$, withIn particular $-K^{\prime \prime }(0)\ge 0$, and $-K^{\prime \prime }$ is a positive-semidefinite kernel because it is an actual covariance function. Merely constructing an unrelated Gaussian process with that covariance function would not yet construct a modification of a stochastic process of $X$.

The increment variance of $Y$ isFor small $|t-s|$, differentiability of $K^{\prime \prime }$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K^{\prime \prime }$ enlarges $C$ to give the same bound. The fourth moment of a centered normal distribution is three times its variance squared, soThe one-dimensional Kolmogorov continuity theorem states that $\mathbb{E}|U(t)-U(s){|}^p\le C|t-s{|}^{1+\eta }$ with $p,\eta >0$ gives a continuous modification, with sample-path Hölder continuity of every order less than $\eta /p$. Apply it here with $p=4,\eta =1$. We obtain a continuous modification $\tilde{Y}$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.

Define, on this same probability space,The path integral exists because its integrand is a continuous function. It agrees in L2 space with the integral of $Y$: Riemann sums converge in L2 space because $Y$ is mean-square continuous; sums of $\tilde{Y}$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the Bochner integral of the L2 space-valued map $s\mapsto Y(s)$.

To verify the modification of a stochastic process property directly, put $A_t=X(t)-X(0)$ and $I_t={\int }_0^{t}Y(s)ds$. The preceding covariance identities and the fundamental theorem of calculus giveTherefore $\mathbb{E}(A_t-I_t{)}^2=0$ and $\tilde{X}(t)=X(t)$ almost surely for every fixed $t$. This is also the mean-square fundamental theorem of calculus in this stationary setting. If a version on all of $\mathbb{R}$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.

The ordinary fundamental theorem of calculus now gives the simultaneous pathwise conclusionThus the requested differentiable modification of a stationary Gaussian process exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.

The range of the Volterra integration operator is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a Sobolev space element that is an absolutely continuous function. Indeed $Ku$ has weak derivative $u$ and zero initial trace; conversely the fundamental theorem of calculus reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.

The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its distributional derivative is a Dirac delta distribution, not an $L^2$ function. ThusFor a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative norm is $\sqrt{n}$, illustrating unstable differentiation.

The supplied SVD gives $K({\sigma }_j{u}_j)={\sigma }_j^2{v}_j$, or more simply $K^{†}({\sigma }_j{v}_j)=u_j$. Since ${\sigma }_j{v}_j\to 0$ but $\Vert u_j\Vert =1$, the Moore–Penrose inverse is discontinuous. Its domain is the dense, nonclosed range above, and there it is the weak derivative.

The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincare inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

The fundamental theorem of calculus says first that if $h$ is continuous on $[a,b]$ and $H(x)={\int }_a^{x}h(t)dt$, then $H^{\prime }(x)=h(x)$. Indeed,whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F^{\prime }$ shows that $F(x)-{\int }_a^x{F}^{\prime }(t)dt$ has zero derivative and is constant. Hence

A derivative need not be Riemann integrable. Define $f(0)=0$ and $f(x)=x^2\sin (x^{-2})$ for $x\ne 0$. Then $f$ is differentiable at zero with $f^{\prime }(0)=0$, while for $x\ne 0$,which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, $g=f^{\prime }{|}_{[0,1]}$ need not be Riemann integrable.

For a coordinate curve $v=v_0$, the length between $u_1$ and $u_2$ iswhere $E$ is a coefficient of the first fundamental form. Thus $\partial E/\partial v=0$ makes the two $u$-directed sides of every coordinate rectangle equally long. Similarly,so $\partial G/\partial u=0$ makes the two $v$-directed sides equally long.

Conversely, suppose every pair of opposite sides has equal length. Fix $v_0,v_1$. Equality of the first pair for every interval $[u_1,u_2]$ givesThe fundamental theorem of calculus implies that the continuous integrand vanishes, so $E$ is independent of $v$. The same argument shows that $G$ is independent of $u$. Hence the stated equal-length property is equivalent to $E_v=G_u=0$.

The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

True. By the Weierstrass approximation theorem, choose polynomials $q_n$ with $q_n\to f^{(m)}$ uniformly. Define $P_n$ by integrating $q_n$ exactly $m$ times and choose the $m$ integration constants so thatThen $P_n^{(m)}=q_n$, and repeated use of the fundamental theorem of calculus givesEvery derivative therefore converges uniformly as required.

LetThe fundamental theorem of calculus and the Cauchy-Schwarz inequality giveThus the one-half Hölder seminorm is at most $m_R^{-1/2}{E}_R(h{)}^{1/2}$.

Writing $Z_R={\int }_{-R}^{R}w>0$ and using the zero weighted mean, for each $x$ we haveSince $|x-y|\le 2R$, the preceding estimate yieldsConsequently the required inequality holds, for example, with

For $f,g\in X$, the Lipschitz condition givesChoose $c$ so that $c+1>Mb$. Then $\varphi$ is a contraction mapping. The space $X$ is complete in the sup norm, and therefore also in the equivalent weighted norm.

For initial value $y_0$, defineIt has the same contraction constant. The Banach fixed-point theorem gives a unique fixed point $f$, which satisfies the integral equation. The fundamental theorem of calculus then givesConversely every solution satisfies that integral equation, proving uniqueness on $[1,b]$. This weighted-norm argument is the Bielecki norm method.

Write $u_Q=|Q{|}^{-1}{\int }_{Q}u$. Since $u-u_Q$ has arithmetic mean zero,and the pairwise-difference identity givesFirst take $u$ smooth. Join $x$ to $y$ by changing one coordinate at a time and apply the Cauchy-Schwarz inequality:For a one-dimensional slice $v$, the fundamental theorem of calculus yieldsIntegrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\Vert D_{i}u{\Vert }_{L^2(Q)}^2$. HenceThe density of smooth functions in a Sobolev space extends the estimate to every $u\in H^1({\mathbb{R}}^n)$. Thus

Define $F(x)={\int }_a^{x}f(t)dt$. By the fundamental theorem of calculus, $F^{\prime }=f$. The chain rule givesIntegrating from $\alpha$ to $\beta$ and using $g(\alpha )=a$, $g(\beta )=b$ yields the change of variables formulaThe argument also covers a decreasing $g$ because both integrals are oriented.