Bourbaki

Exercise 29, § 1

Tag 02L5, permanent. Cite it as [Bourbaki, Tag 02L5].

Spectral Theories, II, p. 274

Let f∈L2(R)f\in L^2(\mathbf{R}) such that

∫R∣f(x)∣2ex2dx⩽1,∫R∣f^(y)∣2ey2dy⩽1 \int_{\mathbf{R}}|f(x)|^2e^{x^2}dx\leqslant 1,\int_{\mathbf{R}}|\widehat{f}(y)|^2e^{y^2}dy\leqslant 1

a) For every integer n∈Nn\in \mathbf{N}, the function gng_n defined by

gn(z)=(∫Rf(x)xng_n(z) =(\int_{\mathbf{R}}f(x)x^n exp(−zx2/2-zx^2/2)dx)2dx)^2

is holomorphic in the open set U={z∈C∣R(z)>−1}U =\{z\in \mathbf{C}|\mathscr{R}(z)>-1\} of C\mathbf{C}.

b) For every integer n∈Nn\in \mathbf{N}, the function hnh_n defined by

hn(z)=(−1)nn(∫f^(y)ynh_n(z) =\frac{(-1)^n}{n}(\int\widehat{f}(y)y^n exp(−y2/(2z)-y^2/(2z))dy)2dy)^2

zR2+1 z^{2+1}_{\mathbf{R}}

is holomorphic in the open set V={z∈C∣R(z−1)>−1}={z∈C∣V =\{z\in \mathbf{C}|\mathscr{R}(z^{-1})>-1\}=\{z\in \mathbf{C}| ∣z+1/2∣>1/2}|z+ 1/2|>1/2\} of C\mathbf{C}.

c) There exists a holomorphic function k0k_0 in U∪V=C−{−1}U\cup V =\mathbf{C}-\{-1\} such that k0∣U=g0k_0|U =g_0 and k0∣V=h0k_0|V =h_0.

d) The function k0k_0 is zero. (Exercise prove that there exists c∈Cc\in \mathbf{C} such that ∣k0(z)∣⩽|k_0(z)|\leqslant c/∣z+1∣c/|z+ 1| for all z∈C−{−1}z\in \mathbf{C}-\{-1\} and apply Liouville's theorem to z↦(z+1)k0(z)z\mapsto (z+ 1)k_0(z).)

e) Now prove by induction on n∈Nn\in \mathbf{N} that there exists a holomorphic function knk_n in U∪V=C−{−1}U\cup V =\mathbf{C}-\{-1\} such that kn∣U=gnk_n|U =g_n and kn∣V=hnk_n|V =h_n, then that kn=0k_n= 0. f) We have f=0f= 0.

g) Let a,b>0a,b >0 such that ab⩾1ab\geqslant 1. If f∈L2(R)f\in L^2(\mathbf{R}) satisfies

∫R∣f(x)∣2eax2dx<+∞,∫R∣f^(y)∣2eby2dy<+∞ \int_{\mathbf{R}}|f(x)|^2e^{ax^2}dx <+\infty ,\int_{\mathbf{R}}|\widehat{f}(y)|^2e^{by^2}dy <+\infty

then f=0f= 0 (“Hardy's uncertainty principle”).

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Machine translation of the English, by gpt-5-6. Not checked by a person.