Compactly supported smooth function with zero moments

Question

I want to determine whether it is possible to construct a (non-zero) function $\psi: \mathbb{R}\to \mathbb{R}$ such that

1. $\psi$ vanishes outside $[-1, 1]$,
2. $\psi$ is smooth, i.e. infinitely differentiable,
3. $\int_{-1}^1 \psi(x) \, \mathrm{d}x = 1$,
4. $\forall k \in \mathbb{N}$ it holds that $\int_{-1}^1 x^k \psi(x) \, \mathrm{d}x = 0$.

I know it is possible to construct a $C^m$ function satisfying 1., 3. and 4. for $k$ up to $m$ using orthogonal polynomials.

My intuition says that it should not be possible to construct such a function but I have no clue how to show it. Any thoughts?

Answer 1

Hints: $\int_{-1}^{1}[p(x)-p(0)]\psi (x)dx=0$ for every polynomial $p$. This gives $\int_{-1}^{1}p(x)\psi (x)dx=\int p(x)d\mu(x)$ where $\mu=\delta_0$. by Weierstrass Theorem, this implies that the same equation holds for any continuous function $p$. But then $\mu$ would be absolutely continuous with density $\psi$, a contardiction.

Answer 2

It suffices that $\psi $ is continuous on $[-1,1]$ and the moments vanish for $k\ge k_0$ for a fixed $k_0,$ not necessarily equal $1.$ Moreover the condition $3.$ is not essential.

Assume $\psi\neq 0$ and $$\int\limits_{-1}^1x^k\psi(x)\,dx=0,\quad k\ge k_0$$ Thus $$\int\limits_{-1}^1x^kx^{k_0}\psi(x)\,dx=0,\quad k\ge 0$$ By the Weierstrass theorem $$\int\limits_{-1}^1f(x)x^{k_0}\psi(x)\,dx=0$$ for any $f\in C[-1,1].$ Hence $x^{k_0}\psi(x)=0,$ which implies $\psi=0,$ a contradiction.

Answer 3

There are posts already about variations of this problem on this website: post 1 and post 2 for instance. The answer is no, and it can be deduced immediately from the arguments in the first post and its answers (the function $x\mapsto x\psi(x)$ in your post is continuous on $[-1,1]$ and satisfies 4 for $k\geq 0$, so by the result in the first post $\psi$ must be equal to $0$ on $[-1,1]$, in particular $\psi$ has to be zero on $(0,1]$, which contradicts assumption 3).

My personal comments on this problem.

1. The regularity assumptions on $\psi$ do not really matter if 4 is stated for $k\geq 0$, in a very general sense: the only integrable function satisfying 1 and 4 (for $k\geq 0$) is the zero function (considering functions to be equal when coinciding pointwise almost everywhere). More generally, the only distribution (that is, an arbitrarily irregular “function”) that is supported on $[-1,1]$ and that satisfies 4 for all $k\geq 0$ is the zero distribution (4 has to be stated with the integral taken on the whole $\mathbb R$ in this case). This can be proved passing through the Fourier transform of $\psi$: see my answer to the second post cited above.

   • Notes:
     • If the Fourier transform is too advanced to think of, you can extend the function/distribution by periodicity on the interval $[-2,2]$ and use Fourier series instead: in that case, 4 with $k\geq 0$ implies that all the Fourier coefficients of the periodic distribution are zero, so $\psi$ has to be zero by the injectivity of the Fourier series (the injectivity of Fourier series/transform is true in the space of tempered distributions; note that a compactly supported distribution is tempered, so the Fourier transform makes sense, and a periodic distribution is also tempered, so the Fourier series makes sense as well).
     • If distributions are too advanced for you, just think of continuous, piecewise-continuous, or integrable functions. The argument with the Fourier series applies without too many troubles (however, I will stick to the distributional setting for the next comment).

2. However, going back and considering assumption 4 to be valid only for $k\geq 1$, the argument in the first paragraph implies that the distribution $x\psi$ is identically zero: this, however, only implies that $\psi$ is a scalar multiple of the Dirac delta. In fact, if in your problem you replace assumption $2$ with “$\psi$ is a distribution”, the answer is yes, and the unique distribution satisfying 1-4 is $$ \psi=\frac 12 \delta_0, $$ where $\delta_0$ is the Dirac delta. This means that your problem is sensitive on the regularity of $\psi$: if the regularity of $\psi$ is constrained to be higher than the one of the Dirac delta (for instance, if you assume that $\psi$ is a continuous or an integrable function), no such $\psi$ exists); on the other hand, if your regularity assumption is weaker (e.g., if you assume that $\psi$ is a distribution, or a Radon measure), there is a solution to the problem being $\delta_0/2$, and such solution is unique.