Analysis example: int f(x) x^n dx = 0 implies f = 0

By | April 6, 2015

Problem. Let \(f\) be continuous on \([0,~1]\) and let \[\int_{0}^{1} f(x) x^n ~ dx =0 \] for every nonnegative integer \(n . \) Prove that \(f=0\) on \( [0,~1] .\)

We first observe two lemmas. These lemmas are for beginners, so you may skip them.

Lemma 1. Under the same condition, \[\int_{0}^{1} f(x) p(x) ~ dx =0\] holds for any polynomial \(p(x) .\)

Proof. If \(p(x)\) is expressed as\[p(x) = a_0 + a_1 x^1 + a_2 x^2 + \cdots + a_n x^n \] where \(a_i \mathrm{ ' s}\) are coefficients, we have \[\begin{eqnarray} \int_{0}^{1} f(x)p(x) ~ dx &=& \int_{0}^{1} f(x) \sum_{i=0}^{n} ~ a_i x^i ~ dx \\ &=& \sum _{i=0}^{n} \left( a_i \int_{0}^{1} f(x) x^i ~ dx \right) \\ &=& \sum_{i=0}^{n} \left( a_i \cdot 0 \right) =0 . \end{eqnarray} \]

Lemma 2. If \(f\) is continuous on a nondegenerate interval \([a,~b]\) satisfying \[\int_{a}^{b} ~ f^2 (x) ~ dx =0 \] then \(f\) is zero on \([a,~b]\) identically.

Proof. Trivial, by continuity.

Solution to the Main Problem. We will show that the integral of \(f^2\) over \([0,~1]\) is zero. This will show that \(f\) is zero by Lemma 2.

Using the Stone-Weierstrass theorem (also called Weierstrass’ approximation theorem), we can approximate uniformly by a sequence of polynomials \(\left\{ p_n \right\}\) so that \[\left\Vert p_n -f \right\Vert < \frac{1}{n} \] where \(\Vert f \Vert\) represents the supremum norm of \(f\) over \([0,~1].\) The given condition obviously implies that the integral of \(f(x)p(x)\) is zero for any polynomial \(p\) by Lemma 1. Now we have \[ \begin{eqnarray} \left| \int_{0}^{1} f(x)f(x) ~ dx \right| &=& \left| \int_{0}^{1} f(x) p(x) ~ dx - \int_{0}^{1} f(x) p_n (x) ~ dx \right| \\ & \le & \int_{0}^{1} \left| f(x) \right| ~ \left| f(x) - p_n (x) \right| ~ dx \\ & \le & \frac{1}{n} \int_{0}^{1} |f(x)| ~ dx . \end{eqnarray} \] The last integral is constant, so taking \(n\) arbitrarily large completes the proof.

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