Is a function that integrates to zero against all polynomials constant?

Question

Suppose $f := [0,\infty) \rightarrow \mathbb{R}$ satisfies $|f(x)| \leq e^{-x}$ for all $x \in (0,\infty)$, and also has the property that

$$ \int_0^{\infty} f(x) x^n dx = 0 \qquad \forall n \in \{0,1,2,3,...\}. $$

Does it follow that $f$ is constant? If so, is this a standard theorem?

Many thanks for your help.

$f(x) \equiv 0$ is the only constant function that meets your criteria. — Fly by Night, Oct 11 '16 at 17:18
@HagenvonEitzen Woops, yes, $f$ is continuous. Though if it weren't I suppose the same result would hold with the statement changed to: $f$ is constant almost everywhere (wrt Lebesgue measure) — Frank, Oct 11 '16 at 17:21
@saz : Actually, it's not a duplicate. Dealing with the issue of an infinite interval is non-trivial. For example, how would you justify interchanging a limit $\lim_n \int_{0}^{\infty}p_n(x)f(x)dx$ with the integral if you have a sequence of polynomials $p_n$ converging to $f$, for example. — Disintegrating By Parts, Oct 11 '16 at 17:55
@TrialAndError you are right; it's more complicated than I thought; sorry for that. So let's call it a related question, okay?http://math.stackexchange.com/q/858441/ — saz, Oct 11 '16 at 18:32
@saz : :) It's a related question that's more difficult. I don't see a nice solution at the moment, unless I can use Complex Analysis. — Disintegrating By Parts, Oct 11 '16 at 18:42

Disintegrating By Parts · Accepted Answer · 2016-11-13T19:24:58.943

For any real $s > 0$, the series for $e^{-sx}$ is an alternating series for $x > 0$, which gives an error bound in terms of the first neglected term: $$ e^{-x}\left|\sum_{n=0}^{N}(-1)^n\frac{(sx)^n}{n!}-e^{-sx}\right| \le e^{-x}\frac{(sx)^{N+1}}{(N+1)!}. $$ Integrating the right side in $x$ over $[0,\infty)$ and applying integration by parts repeatedly gives $$ \int_{0}^{\infty}e^{-x}\frac{(sx)^{N+1}}{(N+1)!}dx = s^{N+1}\int_{0}^{\infty}e^{-x}dx \rightarrow 0 \mbox{ as } N\rightarrow\infty \mbox{ for } 0 < s < 1. $$ Therefore, by your assumptions, $$ F(s)=\int_0^{\infty}f(x)e^{-sx}dx = \lim_{N\rightarrow\infty}\int_{0}^{\infty}f(x)\sum_{n=0}^{N}\frac{(-sx)^{n}}{n!}dx=0,\;\;\; 0 < s < 1. $$ At this point you can invoke uniqueness theorems about the Laplace transform in order to conclude that $f$ is $0$ a.e... Or you can use Morera's Theorem to show that $F$ is holomorphic for $\Re s > -1$ because (a) it is continuous and (b) integrals over triangles in this right half plane give $0$. Then, because $F$ is holomorphic and vanishes on a non-zero interval of the positive real axis, the identity theorem implies that $F$ is identically $0$ for $\Re s > -1$. Using that, $F(ir)$ is $0$, which is the Fourier transform of $f$, and that proves $f$ is $0$ a.e. by the Plancherel identity for Fourier transforms.

score 1 · Answer 2 · answered Oct 11 '16 at 17:38

Hints:

$f$ is integrable (in fact $L^p$ for any $p\ge 1$).
The set of compactly supported smooth functions $C_c^\infty$ is dense in $L^p$ for $1\le p < \infty$.
Polynomials are dense in $C(K)$ for any compact set $K$.

Could you approximate $\int |f|^2$?

Is a function that integrates to zero against all polynomials constant?

2 Answers2