# Uniqueness of a Schwartz function orthogonal to a set of polynomials

Let us define the following set of polynomials: $$p_k(x) := x^k - (k-1) (x^{k-1} + x^{k-2}) + \binom{k-1}{2}x^{k-3},$$ for $$k\geq 3$$.

Let $$S$$ denote the space of Real Schwartz functions, with domain $$\mathbb{R}$$.

Consider the functions $$f\in S$$, such that for all $$k\geq 3$$, $$\langle p_k, f\rangle := \int_{\mathbb{R}} p_k(x) f(x)\, dx =0.$$

Is $$f$$ unique (up to a scalar multiple) ?

Added after the comment and answer below: What if we add the requirements: $$\langle 1, f\rangle = 1,$$ $$\langle x, f\rangle = 0,$$ $$\langle x^2, f\rangle = 1.$$

• Consider any smooth compactly supported function $\chi$, any $a > 0$, and take $f=\mathscr{F}\left(\chi(x)\cdot \exp{-\frac{a}{x^2}}\right)$. So, no. – Mindlack Jun 30 '19 at 13:15

No, $$f$$ is not unique. To see this, we can employ the Fourier transform: \begin{align*} \langle p_k,f \rangle = 0 \iff \langle \mathscr{F}(p_k),\mathscr{F}^{-1}f \rangle = 0. \end{align*} Recalling the Fourier transform of a polynomial $$\mathscr{F}\big( \sum_{j=0}^k a_j x^j \big) = \sum_{j=0}^k a_j \partial_x^j\delta\in S',$$ we see that any $$f\in S$$ with $$\partial_x^j\mathscr{F}^{-1}(f)(0)=0$$ for all $$j$$ is "orthogonal" to all $$p_k$$. Clearly, a very large class of Schwartz functions satisfy this condition. Take for example any $$\phi\in S$$ with $$\phi=0$$ on $$[-\epsilon,\epsilon]$$ and put $$f:=\mathscr{F}(\phi)$$.
• @Teddy: Yes. In fact, in the space of Schwartz functions $f$ with the property the $f$ is orthogonal to all polynomials has a name. In the books of Triebel, it is called $Z(\mathbb{R}^n)$. In the books and papers of Peetre, it has another name (I forget). Anyway, the space is useful to define homogeneous Bessel-potential spaces. – StarBug Jul 3 '19 at 13:20