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Let $f$ be continuous on $[0,1]$ and suppose $\int_0^1f(x)x^ndx=0$ for all $n\in\mathbb{N}_0$. Prove that $f=0$.

*Let $\epsilon\geq 0$. By Weierstrass theorem, we can find a polynomial $p(x)$ such that $|p(x)-f(x)|<\epsilon$, $\forall x\in[0,1]$.

Since $\int_0^1 f(x)x^ndx=0$, then $\int_0^1 f(x)p(x)dx=0$ (Why is this true? Could someone help me with details of rewriting it in terms of a polynomial?)

Since $f$ is continuous on $[0,1]$, $f$ is bounded, so assume $|f(x)|\leq M$ on $[0,1]$.

$|\int_0^1 f(x)^2 dx|=|\int_0^1 f(x)(f(x)-p(x))dx|\leq M\times\epsilon\times(1-0)$ so by letting $\epsilon\rightarrow 0$, we can have $\int_0^1 f(x)^2dx=0$. By continuity of $f(x)^2$, we can conclude that $f(x)=0$ therefore.

Seems like as long as the interval is finite, not only we can apply Weierstrass theorem but also after applying that we can have $\int_a^bf(x)^2 dx=0$ for all $a,b\in\mathbb{R}$, right? What about negative domain?

secondaccount
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  • The first question follows from the linearity of the integral. Second question, yeah why not? The only thing that changes is the $1-0$ to $b-a$. –  Aug 02 '17 at 23:13
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    Since $\int_0^1 f(x)x^ndx=0$, then $\int_0^1 f(x)p(x)dx=0$ (Why is this true? Could someone help me with details of rewriting it in terms of a polynomial?)$,,$ Every polynomial is a linear combination of monomials. – zhw. Aug 02 '17 at 23:21

2 Answers2

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Any polynomial can be written as $p(x) = \displaystyle\sum_{i=0}^n a_ix^i$, where the $a_i$ are the respective coefficients, and $n$ is the degree of the polynomial if $a_n \neq 0$.

Now, if $f$ is as given in the question, $$ \int_0^1 f(x)p(x) = \int_0^1 f(x) \left(\displaystyle\sum_{i=0}^n a_ix^i\right) = \int_{0}^1 \sum_{i=0}^n a_i f(x)x^i = \sum_{i=0}^n a_i\left(\int_0^1 f(x)x^i\right) = 0 $$

Which is simply written as : the integral is linear, so you can inspect the behaviour of monomials to understand the behaviour of polynomials under the integral.

As for the question of finite intervals, and intervals with negative end points, the theorem still holds true, because the Weierstrass theorem still holds true for these intervals. Depending on the proof you have seen for the Weierstrass theorem, you will be able to adapt the proof for the general case. For example, you can scale the Bernstein polynomials appropriately for a general interval, when it is given for $[0,1]$. The rest of this proof is as you have written.

Sarvesh Ravichandran Iyer
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  • At which step do you use Weierstrass Approximation Theorem? – Mariana Jul 11 '21 at 16:38
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    @Mariana Good point. Actually, I complete only the start of the OP's argument, so I don't use the theorem yet. What the OP does, after getting my answer, is that they approximate $f(x)$ uniformly by polynomials $p(x)$, in the sense that for a given $\epsilon>0$ they find a polynomial $p(x)$ such that $|f(x)-p(x)| < \epsilon$ for all $x\in [0,1]$. That requires the Weierstrass theorem. Then they prove from $\int_{0}^1 f(x)p(x) dx = 0$, that $\int_0^1 f(x)^2 dx \leq M\epsilon$ (where $M$ is the supremum of $f$). So I don't use the Weierstrass theorem ,but the rest of the argument does. – Sarvesh Ravichandran Iyer Jul 11 '21 at 16:47
  • I am still confused. The original question is to show f is 0. Does your answer here complete the proof? How can we get f = 0 from $\int f(x)^2 dx \leq M \epsilon$? – Mariana Jul 11 '21 at 16:57
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    @Mariana Yes. Let me tell you the entire proof in a nutshell. I prove this in my post : if $\int_0^1 f(x)x^n = 0$ for all $n$, then $\int_0^1 f(x)p(x) = 0$ for all polynomials $p$. Now, OP does this : let $\epsilon >0$, and let $|f(x)| \leq M$ for all $x \in [0,1]$. Then, the OP picks a specific polynomial $p(x)$ such that $|f(x)-p(x)|<\epsilon$ for all $x \in [0,1]$ (they actually did not make this clear, but I am making this clear for you). Now, they use the estimate $\int_{0}^1 f(x)^2 - \int_0^1 f(x)p(x) \leq \sup_{x \in [0,1]} f(x) \sup_{x \in [0,1]} |f(x)-p(x)|$... – Sarvesh Ravichandran Iyer Jul 11 '21 at 17:01
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    ... to prove that $\int_0^1 f(x)^2dx < M \epsilon$, because I proved that $\int_0^1 f(x)p(x) = 0$. Because $\epsilon$ is arbitrary, we get that we can let $\epsilon$ drop to zero, and therefore $\int_0^1 f(x)^2dx = 0$. So $f(x)^2$ is a non-negative function with integral zero : by a standard argument, it is proved that $f(x)^2 = 0$. Then of course $f=0$. – Sarvesh Ravichandran Iyer Jul 11 '21 at 17:03
  • @Mariana The details on the standard argument in my last comment is presented here. So unfortunately I didn't actually complete the entire argument, but apparently I wasn't asked to, because I felt then that the OP had given the rest of the details. Anyway, now I have filled them in. – Sarvesh Ravichandran Iyer Jul 11 '21 at 17:13
  • @Mariana Let me take a look., thank you for your clarifications on this question, of course. – Sarvesh Ravichandran Iyer Jul 11 '21 at 17:18
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Suppose diferent, then there exists $c \in [0,1]$ with $f(c) \not = 0$, suppose w ithout loss of generality $f(c) > 0$, by continuity, there exists $[a,b] \subset [0,1]$ with $c \in [a,b]$ such that $f(x) > 0$ in $[a,b]$. Take $P_{n}(x) = (1+(x-a)(b-x))^{n}$. Taking $x \in [0,a] \cup [b,1]$ we have $0 \leq P_{n}(x) \leq 1$. Now:

$|\int_{0}^{a} f(x)P_{n}(x)dx| \leq \int_0^a |f(x)| \leq \int_0^1 |f(x)|$ and $|\int_{b}^{1} f(x)P_{n}(x)dx| \leq \int_b^1 |f(x)| \leq \int_0^1 |f(x)|$. Also $(x-a)(b-x) \geq 0$ with $x \in [a, b]$ therefore for all $x \in [a,b]$ and $n \geq 1$ we have $P_{n} \geq 1+ n(x-a)(b-x)$. From this, follow: $\int_a^b f(x)P_{n}(x)dx \geq \int_a^b f(x)dx + n\int_a^bf(x)(x-a)(b-x)$, but as $\int_a^bf(x)(x-a)(b-x) \geq 0$ (by hypothesis $f(x) > 0$ in $[a, b]$). We have: $\lim_{n \to \infty} \int_a^b f(x)P_{n}(x)dx = \infty$. From this, there exist $n_{0}$ such that $\int_0^1 f(x)P_{n_{0}}(x)dx > 0 $. Here you use the hypothesis, this integral must be 0.

Thiago Nascimento
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