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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: $$ \langle 1, f\rangle = 1, $$ $$ \langle x, f\rangle = 0, $$ $$ \langle x^2, f\rangle = 1. $$

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    $\begingroup$ 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. $\endgroup$ – Mindlack Jun 30 '19 at 13:15
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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 $f:=\mathscr{F}(\phi)$.

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  • $\begingroup$ StarBug, so for the same price, you get a function which is orthogonal to the entire space of polynomials ? $\endgroup$ – Teddy Jul 3 '19 at 8:23
  • $\begingroup$ @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. $\endgroup$ – StarBug Jul 3 '19 at 13:20

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