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For each polynomial $g$, take the real analytic adjoint solution on the common cap from the preceding part. On the flat boundary face the full boundary jet of $u$ vanishes; on the curved face that of $v$ vanishes. The complete boundary-jet condition for formal adjoints, proved under its corrected hypothesis, therefore applies. The two corner points have zero boundary measure, and the cap is a bounded piecewise smooth domain. It follows thatBy the Weierstrass approximation theorem, polynomials uniformly approximate $\bar{u}$ on the compact cap closure. Taking the limit gives ${\int }_{{\omega }_{\epsilon }}|u{|}^2=0$. Continuity then gives $u=0$ pointwise in the cap. For real $u$, simply approximate $u$ itself.

A positive cap alone is not a neighborhood of the origin. Apply the same argument after reflecting $x\mapsto -x$; the reflected operator remains real analytic and non-characteristic. Shrinking the two parameters to their minimum proves vanishing onThis set is open and contains the origin, including the flat face where the value data vanish. The argument is the parabolic-cap proof of Holmgren uniqueness, and it establishes Holmgren uniqueness theorem for nonanalytic $C^m$ solutions here.

True. Define $h(u)=g(1/u)$ for $0<u\le 1$ and $h(0)=0$. The hypothesis at infinity makes $h$ continuous on $[0,1]$. The Weierstrass approximation theorem gives polynomials $R_n$ with ${\sup }_{0\le u\le 1}|R_n(u)-h(u)|\to 0$. Since $t\mapsto 1/t$ maps $[1,\infty )$ onto $(0,1]$,

True. By the Weierstrass approximation theorem, choose a polynomial $Q_n$ with ${\sup }_{|t|\le n}|Q_n(t)-f(t)|<1/n$. For each fixed $t$, this bound applies once $n\ge |t|$, so $Q_n(t)\to f(t)$. This proves pointwise convergence; the same construction even gives uniform convergence on every closed interval.

For $f\in C[0,1]$, define its Bernstein polynomialIf $X_{n,x}\sim \mathrm{Bin}(n,x)$, thenLet $M=\Vert f{\Vert }_{\infty }$. By uniform continuity, for any $\epsilon >0$ choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon /2$ whenever $|u-v|<\delta$. Splitting the expectation according to that event and using Chebyshev's inequality gives, uniformly in $x$,For sufficiently large $n$ this is below $\epsilon$. Thus $B_{n}f\to f$ uniformly on $[0,1]$. An affine change of variable treats every compact interval, proving the Weierstrass approximation theorem.

True. For each $n$, the Weierstrass approximation theorem gives a polynomial $P_n$ such thatFor each fixed $x\in \mathbb{R}$, this estimate applies once $n\ge |x|$, so $P_n(x)\to f(x)$ pointwise.

True. First choose polynomials $Q_n\to f$ uniformly by the Weierstrass approximation theorem. Let $L_1,\dots ,L_m$ be the Lagrange interpolation polynomials for the nodes $x_1,\dots ,x_m$, so $L_j(x_i)={\delta }_{ij}$. SetThen $P_n(x_j)=f(x_j)$ for every $j$, while

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.

Choose $M<\infty$ such that $|X|,|Y|\le M$ almost surely. Equality of the positive moments, together with equality of the zeroth moments $\mathbb{E}[1]=1$, givesfor every polynomial $p$. By the Weierstrass approximation theorem, every continuous function on $[-M,M]$ is a uniform limit of polynomials. Expectations preserve this limit, so $X$ and $Y$ have equal expectations for every continuous function on their common compact support. The moment determinacy on a compact interval therefore gives equality of their laws.

Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\le M$. Given $ϵ>0$, choose $\delta >0$ such thatThenThe bound is independent of $p$. Taking the supremum, then $n\to \infty$, gives a limit superior at most $ϵ$. Since $ϵ$ is arbitrary,This is the Bernstein polynomial proof of the Weierstrass approximation theorem.

Let $C\subset \mathbb{R}$ be compact and let $M=\Vert f{\Vert }_{\infty }$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is uniformly continuous on the compact setGiven $\epsilon >0$, choose $\delta >0$ so thatwhenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\ge 0$,Property 3 makes the final term smaller than $\epsilon /2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an approximate identity.

For the second part, extend $g$ to a continuous function $\tilde{g}$ on $\mathbb{R}$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. DefineThese nonnegative kernels have integral one. For every $\delta >0$, their mass outside $[-\delta ,\delta ]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.

For $x\in [0,1]$, the convolution isBecause $|x-y|\le 1$ on the square $[0,1{]}^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a polynomial in $x$. The first part, applied to the compact interval $[0,1]$, givesThis proves the Weierstrass approximation theorem for functions with the stated endpoint values.

Given $ϵ>0$, the Weierstrass approximation theorem supplies a polynomial $Q$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For all sufficiently large $n$, the quadrature is exact on $Q$. Positivity of the weights and ${\sum }_j{A}_j=\int r$ then giveLetting $ϵ\downarrow 0$ proves convergence.

Fix a continuous function $f$ and $\epsilon >0$. By the Weierstrass approximation theorem, choose a polynomial $p\in {\mathcal{P}}_m$ such thatFor every $n\ge m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial giveConsequentlySince $\epsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the convergence of positive quadrature rules.

Statement (iv) is true. Apply the Weierstrass approximation theorem separately to the real and imaginary parts of the continuous function $f$ on the compact interval $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging uniformly to $f$ on $I$.

Assume first that $X_n\stackrel{d}{\to }X$. Because all variables take values in the compact interval $[-C,C]$, each monomial $x^k$ agrees there with a bounded continuous function on $\mathbb{R}$. The bounded moment criterion for weak convergence in this case begins withfor every natural number $k$.

Conversely, suppose all moments converge. Let $f$ be bounded and continuous. By the Weierstrass approximation theorem, for every $\epsilon >0$ there is a polynomial $P$ withMoment convergence implies $\mathbb{E}[P(X_n)]\to \mathbb{E}[P(X)]$, whileTaking the limit superior and then $\epsilon \downarrow 0$ proves $\mathbb{E}[f(X_n)]\to \mathbb{E}[f(X)]$, which is convergence in distribution.

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.