Suppose \(f:[-1,1]\to\mathbb R\) is bounded and continuous at \(0\). Let \[ \alpha(x)= \begin{cases} -1,&x\in[-1,0],\\ 1,&x\in(0,1]. \end{cases} \] Prove that \(f\in\mathcal R(\alpha)[-1,1]\) and \[ \int_{-1}^1 f\,d\alpha=2f(0). \]
Let \(g:[0,1]\to\mathbb R\) be continuous and suppose \[ \int_0^1 g(x)x^{3k+2}\,dx=0 \qquad(k=0,1,2,\ldots). \] Prove that \(g\equiv0\).
Solution
1. Prove the Riemann–Stieltjes assertion.
The integrator \(\alpha\) is nondecreasing and has a single jump of size \(2\) immediately to the right of \(0\).
Fix \(\varepsilon>0\). By continuity of \(f\) at \(0\), choose \(r\in(0,1)\) such that \[ |f(x)-f(0)|<\frac{\varepsilon}{4} \qquad(0\le x\le r). \] Take any partition containing the points \[ -1<0<r<1. \] All increments of \(\alpha\) are zero except the one on the subinterval \([0,r]\), where \[ \alpha(r)-\alpha(0)=1-(-1)=2. \] Consequently, if \(M\) and \(m\) are the supremum and infimum of \(f\) on \([0,r]\), then \[ U(f,P,\alpha)-L(f,P,\alpha)=2(M-m). \] Since \(|f(x)-f(0)|<\varepsilon/4\) on \([0,r]\), \[ M-m<\frac{\varepsilon}{2}, \] and hence \[ U(f,P,\alpha)-L(f,P,\alpha)<\varepsilon. \] Thus \(f\in\mathcal R(\alpha)[-1,1]\).
Moreover, \[ 2m\le \int_{-1}^1f\,d\alpha\le2M, \] and both \(m\) and \(M\) can be made arbitrarily close to \(f(0)\) by shrinking \(r\). Therefore \[ \boxed{\int_{-1}^1f\,d\alpha=2f(0)}. \]
2. Extend the moment identities from monomials to polynomials in \(x^3\).
For every polynomial \(p\), \[ x^2p(x^3) \] is a finite linear combination of the functions \(x^{3k+2}\). Hence the hypothesis gives \[ \int_0^1 g(x)x^2p(x^3)\,dx=0. \]
3. Approximate \(g\) by polynomials in \(x^3\).
Define \[ h(t):=g(t^{1/3}),\qquad t\in[0,1]. \] Then \(h\) is continuous. By the Weierstrass approximation theorem there are polynomials \(p_n\) such that \[ p_n\to h \] uniformly on \([0,1]\). Substituting \(t=x^3\) gives \[ p_n(x^3)\to g(x) \] uniformly on \([0,1]\).
Since \(x^2g(x)\) is bounded, we may pass to the limit in the identities from Step 2: \[ 0 =\lim_{n\to\infty}\int_0^1 g(x)x^2p_n(x^3)\,dx =\int_0^1x^2g(x)^2\,dx. \] The integrand is continuous and nonnegative, so it vanishes identically. Thus \(g(x)=0\) for every \(x\in(0,1]\), and continuity gives \(g(0)=0\). Hence \[ \boxed{g\equiv0}. \]