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Let $\mathcal S(\mathbb R)$ be the Schwartz-space. Look at the linear operator $T:\mathcal S(\mathbb R)\rightarrow \mathcal S(\mathbb R)$
$(Tf)(x)=\sqrt{2\pi}xf(x)-\frac{1}{\sqrt{2\pi}}f'(x)$ and define $\phi_0(x)=e^{-\pi x^2}$, $\phi_n(x)=T^n\phi_0$ for alle $n\in \mathbb N$. Show $\phi_n\neq 0$ for all $n\in\mathbb N$. I think I have to show it with induction, but that does not lead to the goal. Do you have a hint?

Leon
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  • When you say $\phi_n \ne 0$, you mean $\phi_n$ is not the constant function equal to $0$, or do you mean $\phi_n$ has no zeroes? In either case, $T^nf(x) = m_n(x)f(x)$ for some very explicit function $m_n(x)$, so hopefully starting from this observation is helpful for you. – Alex Ortiz Jul 01 '22 at 15:20
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    Is there a derivation missing in the definition of $T$ ? If $T$ is just the multiplication by $x \mapsto cx-c^{-1}$ with $c=\sqrt{2\pi}$, the question is trivial. – Christophe Leuridan Jul 01 '22 at 15:35
  • I corrected the definition of $T$ – Leon Jul 01 '22 at 17:09
  • $f' = ax f$ implies $ f = C e^{ax^2/2}$ – reuns Jul 01 '22 at 19:35
  • Yes, this means $T(\phi_0)\neq 0$. But why is $T^n(\phi_0)\neq 0$? – Leon Jul 01 '22 at 21:08
  • Make a recursion shows that $T^nf = P_nf$, where $P_n$ is a polynomial with leading term $c_nX^n$ (look at the first terms to guess a formula for the coefficient $c_n$). – Christophe Leuridan Jul 05 '22 at 21:16

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