Proposition 10

— Under the hypothesis $\Phi(0) - \Re \Phi \leq \varepsilon + q$, we have

for every $r > 0$.

One writes $\operatorname{Tr}(q/h)$ for the trace of q with respect to h (cf. Annex, No. 1). The formula (55) is trivial when $\operatorname{Tr}(q/h)$ is infinite. We assume henceforth that $\operatorname{Tr}(q/h)$ is finite, hence that $h(x) = 0$ implies $q(x) = 0$ for $x \in T$.

Let $a_1, \ldots, a_n$ be elements of $T$, and $D$ the set of $x' \in T'$ such that $\sum_{j=1}^n \langle a_j, x' \rangle^2 > 1$. For every real $t \geq 0$ we have $3(1 - e^{-t/2}) \geq 0$, and we even have

for $t > 1$, because $e > \frac{9}{4}$. Applying these inequalities to $t = \sum_{j=1}^n \langle a_j, x' \rangle^2$, we obtain

Let $\gamma$ be the measure on $\mathbf{R}$ having density $t \mapsto (2\pi)^{-1/2} e^{-t^2/2}$ with respect to Lebesgue measure. By Lemma 2 of No. 4,

for all real $u$. Consequently,

for all $x' \in T'$. The function of $x', t_1, \ldots, t_n$ to be integrated in the second member is continuous and is bounded above in absolute value by 2, and the measures $\mu$ and $\gamma$ are bounded; one can therefore integrate the two members of (57) with respect to $d\mu(x')$ and interchange the integrations with respect to $\mu$ and $\gamma$; one obtains

Since $q$ is a quadratic form on $T$, there exist real numbers $q_{jk}$ such that

for $t_1, \ldots, t_n$ real; in particular, $q_{jj} = q(a_j)$ for $1 \leq j \leq n$. Moreover, the integral $\int_{\mathbf{R}} t^n d\gamma(t)$ has the values 1, 0, 1 for $n = 0, 1, 2$, respectively (No. 4, Lemma 1). From this, one deduces immediately Now, the first member of (58) and $\Phi(0)$ are real numbers; one can therefore replace $\Phi$ by $\Re \Phi$ in the second member of (58). The inequality $\Phi(0) - \Re \Phi \leq \varepsilon + q$ and the formulas (56), (58) and (59) then imply Let us fix the number $r > 0$. Since the quadratic form $h$ is positive, there exist a basis $(e_1, \ldots, e_n)$ of $T$ and an integer $m$ between 0 and $n$ such that for $t_1, \ldots, t_n$ real (Annex, No. 1, Prop. 2). It is then immediate that $C_r$ consists of the $x' \in T'$ such that For every integer $l \geq 1$, let $D_l$ be the set of $x' \in T'$ satisfying the inequality One sees easily that the sequence $(D_l)_{l \geq 1}$ is increasing with union $T' - C_r$, whence

But by (60),

for $j = m+1, \ldots, n$ we have $h(e_j) = 0$, therefore $q(e_j) = 0$. Moreover, $\operatorname{Tr}(q/h) = \sum_{j=1}^m q(e_j)$ (Annex, No. 1, Prop. 2). The relation (55) then follows from (61) and (62).

Q.E.D.