Exercises · 5. Approximation

1Hard

Let $f$ be continuous on $[0, 1]$ and suppose $\int_0^1 x^n f(x)\,dx = 0$ for every integer $n \ge 0$ . Use the Weierstrass approximation theorem to prove $f \equiv 0$ .

\int_0^1 x^n f(x)\,dx = 0 \quad \forall n \ge 0 \quad \Longrightarrow \quad f \equiv 0

\int_0^1 f(x)^2\,dx = \lim_{k \to \infty} \int_0^1 p_k(x) f(x)\,dx = 0

Weierstrass Approximation Theorem

Show solution

First extend the hypothesis to all polynomials. If $p(x) = \sum_{j=0}^m c_j x^j$ , linearity of the integral gives $\int_0^1 p(x) f(x)\,dx = \sum_{j=0}^m c_j \int_0^1 x^j f(x)\,dx = 0$ . So $\int_0^1 pf = 0$ for every polynomial $p$ .
By the Weierstrass approximation theorem, there are polynomials $(p_k)$ with $p_k \to f$ uniformly on $[0, 1]$ : $\|p_k - f\|_\infty \to 0$ . Put $M = \max_{[0,1]} |f|$ (a continuous function on a compact interval attains its maximum). Then
$\left|\int_0^1 p_k f - \int_0^1 f^2\right| = \left|\int_0^1 (p_k - f) f\right| \le \|p_k - f\|_\infty \cdot M \longrightarrow 0.$
Since each $\int_0^1 p_k f = 0$ , passing to the limit gives $\int_0^1 f(x)^2\,dx = 0$ . As $f^2$ is continuous and nonnegative, $f^2 \equiv 0$ : if $f(c)^2 = \eta > 0$ at some $c$ , continuity would force $f^2 \ge \eta/2$ on an interval, making the integral positive. Hence $f \equiv 0$ .