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A continuous real function on $[0, 1]$ whose every moment $\int_{0}^{1} x^{n} f$ vanishes is identically zero

Statement

Let $f : [0, 1] \to R$ be continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) and suppose that

$\int_{0}^{1} x^{n} f (x) d x = 0 for every n \in N .$

Then $f (x) = 0$ for every $x \in [0, 1]$ .

The hypothesis includes $n = 0$ , which reads $\int_{0}^{1} f = 0$ . Continuity is doing real work here rather than tidying: the last step of the proof is A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ , and its companion FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero shows that a merely integrable nonnegative function with integral $0$ need not be identically zero.

Facts & Assumptions

Given: A continuous $f : [0, 1] \to R$ with $\int_{0}^{1} x^{n} f (x) d x = 0$ for every $n \in N$ .

[L1]

For every $f \in C ([0, 1], R)$ and $ε > 0$ , there is a polynomial $p$ with $\sup_{x \in [0, 1]} ∣ p (x) - f (x) ∣ < ε$ (Polynomials are uniformly dense in $C ([0, 1], R)$ ).

[L2]

Sums, scalar multiples and products of functions continuous at a point are continuous at that point; and, with no hypothesis at all, every constant function, the identity, every $x \mapsto x^{n}$ for $n \in N$ , and every polynomial function with real coefficients are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L3]

For reals $a < b$ , a continuous $g : [a, b] \to R$ is bounded and Riemann integrable on $[a, b]$ (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L4]

For reals $a < b$ , integrable $g, h : [a, b] \to R$ and reals $λ, μ$ , the function $λ g + μ h$ is integrable on $[a, b]$ and $\int_{a}^{b} (λ g + μ h) = λ \int_{a}^{b} g + μ \int_{a}^{b} h$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L5]

For reals $a < b$ and integrable $g : [a, b] \to R$ : if $g (x) \geq 0$ for every $x \in [a, b]$ then $\int_{a}^{b} g \geq 0$ ; and if $m \leq g (x) \leq M$ for every $x \in [a, b]$ with $m, M$ real, then $m (b - a) \leq \int_{a}^{b} g \leq M (b - a)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L6]

For reals $a < b$ and integrable $g, h : [a, b] \to R$ , the functions $g h$ and $g^{2}$ are integrable on $[a, b]$ (If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ ).

[L7]

For reals $a < b$ , if $g : [a, b] \to R$ is continuous with $g (x) \geq 0$ for every $x \in [a, b]$ and $\int_{a}^{b} g = 0$ , then $g (x) = 0$ for every $x \in [a, b]$ (A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ ).

Proof

technique · direct

1.1givenL2L3

For each $n \in N$ the function $x \mapsto x^{n}$ is continuous on $[0, 1]$ , so $x \mapsto x^{n} f (x)$ is continuous on $[0, 1]$ as a product of continuous functions, and is therefore integrable; so each integral in the hypothesis is defined.

1.2givenL3choose

$f$ is bounded and integrable on $[0, 1]$ , so there is a real $M > 0$ with $∣ f (x) ∣ \leq M$ for every $x \in [0, 1]$ ; if the bound supplied is $0$ , replace it by $1$ .

1.3givenL2L3L5

$f^{2}$ is continuous on $[0, 1]$ as a product of continuous functions, hence integrable, and $f (x)^{2} \geq 0$ for every $x \in [0, 1]$ , so $\int_{0}^{1} f^{2} \geq 0$ .

2.1step 1.1L4givenalgebra

Let $p (x) = a_{0} + a_{1} x + \dots + a_{m} x^{m}$ be any real polynomial. Each $x \mapsto a_{j} x^{j} f (x)$ is integrable by step 1.1, and applying the linearity identity $m$ times to the finite sum gives $\int_{0}^{1} p f = \sum_{j \leq m} a_{j} \int_{0}^{1} x^{j} f$ , every summand of which is $0$ by hypothesis, so $\int_{0}^{1} p f = 0$ .

2.2step 1.2L1choose

Let $ε > 0$ . Choose a polynomial $p$ with $\sup_{x \in [0, 1]} ∣ p (x) - f (x) ∣ < ε / M$ , which is legitimate since $ε / M > 0$ by step 1.2.

3.1step 1.1step 2.1step 2.2L2L3L4L6algebra

The polynomial $p$ chosen in step 2.2 is continuous on $[0, 1]$ by [L2] and hence integrable by [L3]; so $f - p$ is integrable by [L4], and both $(f - p) f$ and $p f$ are integrable by [L6]. Since $f^{2} = (f - p) f + p f$ pointwise on $[0, 1]$ , [L4] gives $\int_{0}^{1} f^{2} = \int_{0}^{1} (f - p) f + \int_{0}^{1} p f$ , and the second term is $0$ by step 2.1, giving $\int_{0}^{1} f^{2} = \int_{0}^{1} (f - p) f$ .

3.2step 1.2step 2.2L5algebra

For every $x \in [0, 1]$ , $∣ f (x) - p (x) ∣ ∣ f (x) ∣ < (ε / M) \cdot M = ε$ by steps 1.2 and 2.2, so $- ε \leq (f - p) (x) f (x) \leq ε$ on $[0, 1]$ ; since $1 - 0 = 1$ , the two-sided bound gives $\int_{0}^{1} (f - p) f \leq ε$ .

4.1step 1.3step 3.1step 3.2

Combining, $0 \leq \int_{0}^{1} f^{2} \leq ε$ .

5.1step 4.1algebra

Step 4.1 holds for every $ε > 0$ , and the value $\int_{0}^{1} f^{2}$ does not depend on $ε$ ; were it positive, taking $ε$ to be half of it would contradict step 4.1, so $\int_{0}^{1} f^{2} = 0$ .

6.1step 1.3step 5.1L7algebra∎

$f^{2}$ is continuous on $[0, 1]$ , nonnegative there, and has integral $0$ by step 5.1, so $f (x)^{2} = 0$ for every $x \in [0, 1]$ , and hence $f (x) = 0$ for every $x \in [0, 1]$ .

Depends on

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Sources

J. M. Erdman, A Companion to Real Analysis, Proposition 21.2.9 (standard reference, not scraped)

A continuous real function on [0,1][0,1] whose every moment ∫01xnf∫01​xnf vanishes is identically zero