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Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

Fix $z_0\notin \Gamma$ and put $\delta =\mathrm{dist}(z_0,\Gamma )>0$. If $|z-z_0|<\delta /2$, thenuniformly for $w\in \Gamma$. Multiplication by the bounded continuous function $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv 1$. By the Cauchy integral formula, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. Integrals over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours ${\Gamma }_j\subset \Omega \setminus K$, and their sum isfor every $z\in K$.

Pointwise, $\tanh (nx)$ tends to $-1,0,1$ according as $x<0,x=0,x>0$. This limit is discontinuous, whereas a uniform limit of continuous functions is continuous. Convergence is therefore not uniform.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.

Chebyshev's equal-ripple criterion says that a polynomial $q$ of degree at most $m$ is a best uniform approximation to a continuous function $F$ if and only if the error $F-q$ attains alternating extrema of maximum magnitude at $m+2$ ordered points.

Since $T_n$ has leading coefficient $2^{n-1}$, the polynomial $2^{1-n}{T}_n$ is monic and alternates between $\pm 2^{1-n}$ at $t_j=\cos (j\pi /n)$. Therefore

At each $t_j$, both $T_n-f$ and $T_n+f$ have sign $(-1{)}^j$, because $|f(t_j)|<1$. Each has a root in every interval between consecutive extrema, hence all its at most $n$ roots lie in $(-1,1)$. For $t>1$ both polynomials remain positive, giving $|f(t)|<T_n(t)$. For $t<-1$ both have sign $(-1{)}^n$, giving $|f(t)|<|T_n(t)|$. On $[-1,1]$ the hypothesis gives $|f|<1$. Thus

False. Continuous functions can converge pointwise to the non-Riemann-integrable Dirichlet function; for example enumerate the rationals and use continuous narrow bumps whose first $n$ peaks cover the first $n$ rationals while widths shrink suitably pointwise off $\mathbb{Q}$.

A standard Brownian motion $(B_t{)}_{t\ge 0}$ satisfies $B_0=0$ almost surely, has almost surely continuous sample paths, and has independent stationary increments such that

The evaluation map $f\mapsto f(1)$ is continuous on $C[0,1]$ in the uniform norm. The continuous mapping theorem applied to part (c) therefore givesThis recovers the central limit theorem from the functional version.

For the level-$N$ dyadic partition, writeThe dyadic partitions are nested, so the triangle inequality makes $V_N(t)$ nondecreasing in $N$, and $\Vert f_t\Vert ={\lim }_N{V}_N(t)$.

Fix $t_1<t_2$. As the mesh tends to zero, the last dyadic point before $t_1$ approaches $t_1$. Refining from there to $t_2$, the triangle inequality says that the added variation is at least $|f(t_2)-f(t_1)|$ minus the two endpoint errors, which tend to zero by continuity. ThereforeSince $\Vert f\Vert <\infty$, both limits are finite and may be subtracted, giving

No. The point evaluations ${\delta }_x$ satisfy $\Vert {\delta }_x-{\delta }_y\Vert =2$ for $x\ne y$ (choose a continuous function taking values $1,-1$ at the two points). Thus $C[0,1{]}^{\ast }$ is nonseparable, whereas ${\ell }_1$ is separable.

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 independent site percolation or bond percolation on a rooted locally finite graph, $\theta (p)$ is right continuous on $[0,1)$. Finite-radius connection events have polynomial probabilities in $p$ and decrease to infinite connection. Thus $\theta$ is their infimum by continuity from above of a measure. This infimum of continuous functions is upper semicontinuous; parameter monotonicity from the monotone coupling of Bernoulli percolation then gives right continuity.

For a fixed spline space and admissible interpolation sites, this linear operator sends a continuous function to its unique interpolating spline. Here $Rf$ is the vector of sampled values and $A$ is the B-spline collocation matrix. It reproduces every spline in its range. Its sensitivity to perturbing the data is quantified by the B-spline interpolation operator norm.