For a Schwartz function $f$, if $\int_{\mathbb{R}} f(x) x^n dx = 0$ for all nonnegative integers $n$, is $f$ identically 0?

This is an old exam question I'm practicing with. The associated hint is to use the Fourier transform. I'm pretty lost, but here are my thoughts so far.

First, in this old stack exchange question a user referenced Classical Fourier Analysis by Loukas Grafakos, which in Prop 2.3.25 defines $$S_\infty (\mathbb{R}^n)$$ to be a subspace of $$S(\mathbb{R}^n)$$ such that for $$f \in S(\mathbb{R}^n)$$ $$\int_{\mathbb{R}^n} x^\alpha f(x) = 0$$ which leads me to think there are non-trivial $$f$$ in this space.

Moreover, as user Jonas Teuwen wrote in the question linked above, the Fourier transform maps the Schwartz function to itself, so the question is equivalently asking whether the Fourier tranform evaluated at $$0$$ of $$\hat{x^\alpha} f(x) = 0$$ implies that $$f$$ is identically $$0$$. Since $$x^\alpha \mapsto i^\alpha d/dx^\alpha$$, we want $$f$$ so that $$\frac{d}{dx^{\alpha}} \hat{f(x)} = 0$$ when evaluated at $$x = 0$$.

But I'm at a loss as to how to construct such a function.

Any help or hints would be very much appreciated!

• it means that $(\hat f)^{(n)}(0)=0$ for all $n$. You'll easily find a non-zero Schwartz function $\hat f$ with this property (you can have $\hat f$ with a compact support if you wish) – user8268 Dec 30 '18 at 21:06
• Now if $f \in C^\infty_c(\mathbb{R}^n)$ then $\hat{f}$ is analytic so $\forall \alpha,\int_{\mathbb{R}^n} x^\alpha f(x)dx = i^{|\alpha|} D^\alpha\hat{f}(0) = 0$ implies $\hat{f} = 0$ and $f=0$. – reuns Dec 30 '18 at 21:14

The answer is unfortunately, no. To see this, note that $$\int_{\mathbb{R}} f(x) x^n dx = 0 \Leftrightarrow \widehat{f}^{(n)}(0) = 0,$$ where $$\widehat{f}$$ denotes the Fourier transform of $$f$$. Since the fourier transform is a bijection on the Schwartz space $$\mathcal{S}$$, the problem comes down to the existence of $$\phi \in \mathcal{S}\setminus \{0\}$$ such that $$\phi^{(n)}(0) =0,\quad \forall n\ge 0.$$ Let $$\varphi(x) = e^{-\frac{1}{x}}1_{(0,\infty)}(x),$$ and $$\eta(x)$$ be a $$C^\infty$$ function such that $$\eta(x) = 1$$ on $$|x|\leq 1$$ and $$\eta(x) = 0$$ on $$|x|\ge 2$$. It is well-known that $$\varphi\in C^\infty$$ and $$\varphi^{(n)}(0)=0$$ for all $$n\ge 0$$. Thus, if we let $$\phi(x) = \varphi(x)\eta(x)$$, then it follows that $$\phi \in C_c^\infty(\mathbb{R})\subset \mathcal{S}$$ and $$\phi^{(n)}(0) = 0$$ for all $$n\ge 0$$ as desired.
• One question: why do we take the product of $\phi$ and $\eta$ when $\phi$ is already smooth? Do we just want to get a smooth function with compact support so we can be sure that it's in the Schwartz space without checking the derivative is rapidly decreasing? – kkc Dec 31 '18 at 2:58
• You are right. One can show in fact $\varphi\in \mathcal{S}$, but to avoid tedious arguments to prove this, I chose a mollifier $\eta$. – Song Dec 31 '18 at 3:07