Codex @codex  0

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.

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.

Let $Z=(Z_n{)}_{n\ge 0}$ be a Markov chain with transition kernel $K$, and let $N$ be an independent Poisson process of constant rate $r>0$. A Pseudo-Poisson process is $X_t=Z_{N_t}$. On the Banach space $B_b(E)$ of bounded Borel measurable functions, with the supremum norm, writeIts infinitesimal generator, transition semigroup, and resolvent operator are respectivelyand, for $\lambda >0$,The transition semigroup averages $K^n$ against the Poisson distribution of the number of updates. Since $K$ is a Markov operator, $\Vert K\Vert \le 1$; both series converge in operator norm. Updates with $Z_{n+1}=Z_n$ are allowed and do not count as changes of the observed state. The degenerate rate $r=0$ gives $L=0$, $P_t=I$, and $R_{\lambda }=I/\lambda$.

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.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.

For a function space on a set $X$ and a point $x\in X$, the point evaluation functional is ${\delta }_x(f)=f(x)$. It is bounded on $C(X)$ with the supremum norm.

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 polyhedral uncertainty set is specified by finitely many linear inequalities. The displayed centered example is the inverse image of a box under a linear map. It is nonempty because it contains $r_0$, and contains every line $r_0+td$ with $d\in \ker P$. Consequently it need not be bounded when $P$ is rank deficient. Its supremum norm constraint is equivalent to $-1\le P(r-r_0)\le 1$.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

For a random walk with centered independent identically distributed increments of variance one, the maximum of its polygonal interpolation is the maximum at its grid vertices. The maximum functional is Lipschitz for the supremum norm, so the continuous mapping theorem applied to the Donsker invariance principle proves the displayed convergence in distribution. The Brownian reflection principle then gives upper-tail limits $2(1-\Phi (a))$ at all $a>0$, since the limiting maximum has no atom there.

For continuous $G$ and $f$, restrict to the closed supremum-norm ball of radius $R$ in $C([0,1])$. Boundedness on the compact kernel and nonlinearity boxes makes the integral map preserve this ball for sufficiently small positive $\mu$. Uniform continuity of $f$ proves continuity of the map; uniform continuity of $G$ gives an equicontinuous bounded image. Arzelà-Ascoli makes that image relatively compact, and Schauder gives a solution. No global Lipschitz hypothesis or uniqueness conclusion is needed.

For a twice continuously differentiable function and $h\ge 0$, the fundamental theorem of calculus givesThe nonnegative triangular weight has integral $h^2$, so its supremum norm estimate directly controls the second modulus of smoothness.

The Skorokhod space consists of real paths that are right-continuous and have a finite left limit at every positive time. Its $J_1$ topology allows small changes of time: $x_n\to x$ when there are increasing homeomorphisms ${\lambda }_n$ of $[0,1]$ such that $\Vert {\lambda }_n-\mathrm{id}{\Vert }_{\infty }\to 0$ and $\Vert x_n\circ {\lambda }_n-x{\Vert }_{\infty }\to 0$, where $\mathrm{id}$ denotes the identity time map. This topology aligns nearby jumps rather than insisting on uniform closeness at identical times. The step-path version of the Donsker invariance principle converges here to continuous Brownian motion; its polygonal version uses the ordinary supremum norm on continuous paths.

Smooth convergence on a compact interval requires uniform convergence of the functions and of their derivatives of every order separately, as measured by the derivative seminorms. The need for several scales is visible inFor $n\ge 2$, $p_0(f_n)=1/n$ because $t=\pi /(2n)$ lies in $[0,1]$. The derivative has supremum norm one for every $n$, already at $t=0$. Thus $f_n$ converges uniformly to zero, but fails the test for smooth convergence to zero. Andreas Buttenschoen, Applied Functional Analysis, spaces of continuous and differentiable functions discusses this example and the family of smoothness measurements.

More generally, for any fixed $K\ge 0$, $h_n(t)=\sin (nt)/n^{K+1}$ has $p_K(h_n)\le 1/n$ but $p_{K+1}(h_n)=1$ for $n\ge 2$. If a norm $N$ induced the smooth topology, its continuity would give $N(f)\le C{p}_K(f)$ for some finite $K$, while continuity of $p_{K+1}$ for $N$ would give $p_{K+1}(f)\le C^{\prime }N(f)$. Substituting $h_n$ contradicts these two bounds. Smooth convergence therefore cannot be reduced to one norm. Its beauty is that every finite order can be controlled without fixing a final order in advance.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.

On bounded continuously differentiable functions with uniformly bounded total variation of a function, the map $T$ is Hadamard differentiable in the supremum norm with derivative ${\dot{T}}_{F,G}(a,b)={\int }_0^{u}b{F}^{\prime }+G(u)a(u)-G(0)a(0)-{\int }_0^u{G}^{\prime }a$. Use weak convergence of bounded-variation integrators for the first term and integration by parts for the remaining terms. No derivatives of the limiting continuous directions are required.

An algebra norm is submultiplicative when it satisfies the displayed product inequality. Induction gives $\Vert a_1\cdots a_n\Vert \le {\prod }_j\Vert a_j\Vert$, and in particular $\Vert a^n\Vert \le \Vert a{\Vert }^n$. The operator norm has this property because $\Vert STv\Vert \le \Vert S\Vert \Vert T\Vert \Vert v\Vert$. The supremum norm for bounded functions with pointwise multiplication is another example. This estimate supports the Neumann series and the spectral radius formula.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.