Suppose that \(f\in C([0,1])\) and
that \(\displaystyle \int_0^1
f(x)x^n\,dx=0\) for all \(n=99,100,101,\ldots\). Show that \(f\equiv 0\).
Note: Many variations on this problem exist. See June 2012 6b and others.
Solution
1. The hypotheses give \(\int_0^1 f(x)\,x^{99}p(x)\,dx = 0\) for every polynomial \(p\).
\(x^{99}p(x)\) is a finite linear combination of powers \(x^{99}, x^{100}, \ldots\), each of which integrates to \(0\) against \(f\).
2. Consequently \(\int_0^1 f(x)\,x^{99}h(x)\,dx = 0\) for every \(h \in C([0,1])\).
2.1. Polynomials are dense in \(C([0,1])\) (Weierstrass), so \(p_k \to h\) uniformly.
Weierstrass approximation theorem.
2.2. \(x^{99}p_k \to x^{99}h\) uniformly.
\(|x^{99}p_k - x^{99}h| \le \|p_k - h\|_\infty \to 0\).
2.3. Q.E.D.
\(\int f\,x^{99}h = \lim_k \int f\,x^{99}p_k = 0\) by uniform convergence under the integral.
3. Take \(h = f\): \(\int_0^1 x^{99}f(x)^2\,dx = 0\).
2 with \(h = f \in C([0,1])\).
4. \(f \equiv 0\).
\(x^{99}f(x)^2 \ge 0\) is continuous with integral \(0\), hence identically \(0\); so \(f = 0\) on \((0,1]\), and continuity at \(0\) gives \(f(0) = 0\).