Is $f$ that satisfies $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$ a constant?

Question

My question is:

Define $f\in C^{0}[a,b]$ satisfying $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$. Must this function be a constant function?

I made a easy analysis, which can prove that $f$ has at least three points, as follows

we have $\int_{a}^{b}g(x)f\left(x\right)dx=0$ , in which $g(x)=mx^2+nx+p$. $\forall m,n,p \in \mathbb{R}$

If $f$ has less than three zero points, we can select appropriate $m$,$n$,$p$ satisfying $g(x)f(x)>0$ for all $x$ which is not the zero point of $f$. And it's contradictory to $\int_{a}^{b}g(x)f\left(x\right)dx=0$. Thus $f$ must have at least three zero points.

So can we make a further analysis that indicates $f$ is a constant function?
Any ideas would be highly appreciated!

Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $\int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera. — Jyrki Lahtonen, Nov 21 '18 at 15:19
Only if you add $\int_a^b x^n f(x){\rm }dx = 0$ for all $n\in\mathbb{N}$ can you guarantee that $f(x) \equiv 0$ (this follows from Weierstrass approximation theorem). — Winther, Nov 21 '18 at 15:26
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = \sin \left(\frac{2 \pi (x-a)}{b-a}\right)-2 \sin \left(\frac{4 \pi (x-a)}{b-a}\right)$. — Winther, Nov 21 '18 at 15:36
@HenningMakholm Yes, it's a very standard argument. See e.g. If $\int_0^1 f(x)x^n \ dx=0$ for every $n$, then $f=0$. — Winther, Nov 21 '18 at 15:47

score 3 · Accepted Answer · answered Nov 21 '18 at 15:20

3

The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.

Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.

answered Nov 21 '18 at 15:20

dallonsi

312

Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice – Richard Martin Nov 21 '18 at 15:21

hmakholm left over Monica · Answer 2 · 2018-11-21T15:34:56.073

$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and $$ f \mapsto \left (\int_a^b f(x)dx, \int_a^b x f(x)dx, \int_a^b x^2 f(x)dx\right) $$ is a linear transformation $C^0[a,b] \to \mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.

(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).

score 1 · Answer 3 · answered Nov 21 '18 at 15:19

1

No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $\ge 3$.

answered Nov 21 '18 at 15:19

Richard Martin

1,671

TonyK · Answer 4 · 2018-11-21T15:49:50.020

1

To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial $$f(x)=x^3+cx^2+dx+e$$

Then: $$\int_0^1 f(x)dx=\frac14+\frac{c}{3}+\frac{d}{2}+e=0$$ $$\int_0^1 xf(x)dx=\frac15+\frac{c}{4}+\frac{d}{3}+\frac{e}{2}=0$$ $$\int_0^1 x^2f(x)dx=\frac16+\frac{c}{5}+\frac{d}{4}+\frac{e}{3}=0$$ We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-\frac32,\frac35,-\frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives $$f(x)=20x^3-30x^2+12x-1$$

edited Nov 21 '18 at 15:49

answered Nov 21 '18 at 15:35

TonyK

64,559

Thank you, by the way I think the last term is $-1$ – Chiquita Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now. – TonyK Nov 21 '18 at 15:59

Is $f$ that satisfies $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$ a constant?

4 Answers4