Is $\{\phi \in \mathcal C ([0,1], \mathbb R) \mid \forall k \in\mathbb N^+:\int^1_0 x^k\phi(x) =0\}$ a singleton?

Question

Let $$A=\{\phi \in \mathcal C ([0,1], \mathbb R) \mid \forall k \in\mathbb N^+:\int^1_0 x^k\phi(x) =0\}$$

Clearly, the function $[0,1] \to \mathbb R, \, x \mapsto 0$ belongs to $A$. I would like to ask if $A$ contains any other element.

Thank you so much for your help!

This is the only function. You can prove it for polynomials and then extend to all continuous functions by way of uniform convergence. — Cameron Williams, Oct 08 '19 at 21:58
Hi @CameronWilliams, please shed me some light on how to prove in case of polynomials! — Akira, Oct 08 '19 at 22:02
Easy.. Taking linear combinations, this is true for polynomials, not just monomials. Making a wise choice, you end up with $\int_0^1 p(x)^2 , dx =0.$ — Cameron Williams, Oct 08 '19 at 22:20
Hi @CameronWilliams, I meant $\mathbb N^{+}$, not $\mathbb N$. — Akira, Oct 08 '19 at 23:51

score 3 · Accepted Answer · answered Oct 09 '19 at 00:02

3

Let $g(x)=x\phi(x)$. Then $\int_0^{1} x^{k} g(x)dx=0$ for $k=0,1,2...$. Using Wierstrass approximation it follows that $\int_0^{1} g(x)^{2} dx=0$ so $g(x)=0$ for all $x$. Hence $x\phi(x)=0$ for all $x$. This gives $\phi (x)=0$ for $x \neq 0$ but continuity gives $\phi(0)=0$.

answered Oct 09 '19 at 00:02

Kavi Rama Murthy

311,013

I meant $\int_0^{1} x^{k} g(x)dx=0$ for only $k \ge 1$. Could you please have a check? – Akira Oct 09 '19 at 00:06
1

@MadnessForMATH $x^{k}g(x)=x^{k+1}\phi(x)$. $g$ is not the same as $\phi$. – Kavi Rama Murthy Oct 09 '19 at 00:10
Thank you so much :)) – Akira Oct 09 '19 at 00:10
Great answer! +1 – triple_sec Oct 09 '19 at 01:14
+1: Nice observation. – copper.hat Oct 09 '19 at 13:55

triple_sec · Answer 2 · 2019-10-09T01:10:46.447

Suppose $\phi\in A$. Let $K>0$ be such that $|\phi(x)|\leq K$ for each $x\in[0,1]$ (Weierstrass). Fix $\varepsilon>0$ and find a polynomial $$p(x)=a_0+a_1 x+\cdots+a_n x^n\quad\text{for each $x\in[0,1]$},$$ where $n$ is a non-negative integer, such that $$|\phi(x)-p(x)|<\frac{\varepsilon}{K}\quad\text{for each $x\in[0,1]$}$$ (another Weierstrass). Then, \begin{align*} \int_0^1\phi(x)^2\,\mathrm dx&=\int_0^1\phi(x)[\phi(x)-p(x)]\,\mathrm dx+\underbrace{\int_0^1\phi(x)p(x)\,\mathrm dx}_{=0}\\ &\leq\left|\int_0^1\phi(x)[\phi(x)-p(x)]\,\mathrm dx\right|\\ &\leq\int_0^1|\phi(x)||\phi(x)-p(x)|\,\mathrm dx\leq K\frac{\varepsilon}K=\varepsilon. \end{align*} Since $\varepsilon>0$ can be arbitrarily small, it follows that $$\int_0^1\phi(x)^2\,\mathrm dx=0.$$ Exploit continuity and you’re done.

UPDATE: If it is not assumed that $\int_0^1\phi(x)\,\mathrm dx=0$ (that is, if $\mathbb N^+$ is taken to be $\{1,2,3,\ldots\}$, excluding $0$), then replace $\phi$ with $\widehat\phi$ in the proof above, where $\widehat\phi(x)=x\phi(x)$ for $x\in[0,1]$. The conclusion is that $\widehat\phi$ is identically zero and the proof can be finished as suggested in Kavi Rama Murthy’s answer.

If $\mathbb N^{+}$ stands for the set of positive integers then this is not correct. The constant term in the polynomial creates a problem. — Kavi Rama Murthy, Oct 08 '19 at 23:46
@KaviRamaMurthy You’re right. I included a small modification at the end along the same lines as in your answer. — triple_sec, Oct 09 '19 at 01:17

Is $\{\phi \in \mathcal C ([0,1], \mathbb R) \mid \forall k \in\mathbb N^+:\int^1_0 x^k\phi(x) =0\}$ a singleton?

2 Answers2