6 Proof of Friedman’s theorem
6.1 The free group and its reduced algebra
We denote as \(\mathbb {F}_r= \langle g_1, \ldots , g_r \rangle \) the free group with \(r\) generators.
6.2 Support of a real distribution
6.2.1 Stieltjes transform
In this section we introduce a useful tool which will be essential to the next section. Let \(\hat{\mathbb {C}}\) denote the Riemann sphere.
Let \(z \in \hat{\mathbb {C}}\) such that \(z \notin \mathbb {R}\). The function \(x \in \mathbb {R}\mapsto \frac{1}{2i\pi } \frac{1}{x-z} \in \mathbb {C}\) is an element of \(\mathcal{E}(\mathbb {R}, \mathbb {C})\).
Usual functions.
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). The Stieltjes transform of \(\nu \) is defined as the function \(F_\nu : \hat{\mathbb {C}}- \mathbb {R}\rightarrow \mathbb {C}\) by \(F(z) = \nu (x \mapsto \frac{1}{2 i \pi }\frac{1}{x-z})\).
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). For any \(h \in \mathcal{E}(\mathbb {R}, \mathbb {C})\),
This can be done by computing explicitely the values of \(F_\nu (x \pm i \epsilon )\) and seeing the resulting integral as an approximation of \(\delta _x\) as \(\epsilon \rightarrow 0^+\), [todo write the details].
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). The function \(F_\nu : \hat{\mathbb {C}}- \mathbb {R}\rightarrow \mathbb {C}\) is analytic. It can be analytically extended to a \(x \in \mathbb {R}\) if and only if \(x \notin \mathrm{supp} \, \nu \).
This proof is not complete yet as it requires a bit of work on distributions.
[todo: find a general result describing when \(\nu (h(\cdot , z))\) is holomorphic, there is information on this in Mathematical Methods in Physics: Distributions, Hilbert Space Operators, Variational Methods, and Applications in Quantum Physics]
[todo: write the details]
For \(z \in \mathbb {R}\), if \(z \notin \mathrm{supp} \, (\nu )\), then we take a \(\phi \in \mathcal{D}(\mathbb {R}, \mathbb {C})\) such that \(\phi \) is identically equal to \(1\) on \(\mathrm{supp} \, (\nu )\) and identically equal to \(0\) on a neighbourhood of \(z\). We then observe that \(x \in \mathbb {R}\mapsto \frac{1}{2 i \pi } \frac{\phi (x)}{x-z}\) is an element of \(\mathcal{E}(\mathbb {R}, \mathbb {C})\) and hence we can define \(\tilde{F}_\nu (z) = \nu (\frac{1}{2 i \pi } \frac{\phi (x)}{x-z})\). We check it is an extension of \(F_\nu \) and analytic on a neighbourhood of \(z\).
Let \(z \in \mathbb {R}\). We assume that \(F_\nu \) admits an analytic extension on the ball of radius \(r\) centered at \(z\). Let \(r' \in (0,r)\) and \(h \in \mathcal{E}(\mathbb {R}, \mathbb {C})\) such that \(\mathrm{supp} \, (h) \subset [z-r',z+r']\). By Theorem 6.2.3,
For small enough \(\epsilon {\gt}0\), and any \(x \in [z-r',z+r']\), \(F_\nu (x \pm i \epsilon )\) lies in the ball of radius \(r\) around \(z\) where \(F_\nu \) is analytically extended. We use this information to prove the that limit above (and hence \(\nu (h)\)) is equal to \(0\). Since this holds for any \(h\) supported on \([z-r',z+r']\), \(z \notin \mathrm{supp} \, (h)\).
6.2.2 Spectral radius and support
We shall now show how one can access the support of a distribution from its moments.
Let \(p \geq 0\). The monomial function \(x \mapsto x^p\) belongs in \(\mathcal{E}(\mathbb {R}, \mathbb {C})\). Furthermore, for any integer \(m \geq 0\), for any compact set \(K\) of \(\mathbb {R}\), if \(K+ := \sup _{x \in K} |x|\), then
The regularity is well-known. Let \(x \in K\) and \(0 \leq k \leq m\) an integer. Then, the \(k\)-th derivative of the monomial at \(x\) is given by \(p(p-1) \ldots (p-k+1) x^{p-k}\) if \(p\geq k\) and \(0\) otherwise. In both cases, this is smaller in absolute value than \(\max (p,1)^k \max (K_+,1)^p\) which is smaller than \(\max (p,1)^m \max (K_+,1)^p\).
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). For \(p \in \mathbb {N}\), we define the \(p\)-th moment \(\nu _p := \nu (x \mapsto x^p)\).
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). We define
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). Then \(\rho (\nu ) {\lt} \infty \).
By Lemma 5.2.6, there exists a compact set \(K\), a constant \(C\) and an integer \(m\) such that, for all \(h \in \mathcal{E}(\mathbb {R}, \mathbb {C})\), \(|\nu (h)| \leq C \| h\| _{C^m(K)}\). Let \(p \geq 0\) be an integer. Then by Lemma 6.2.5, if \(K_+ := \sup _{x \in K} |x|\),
and hence \((|\nu _p|^{1/p})_{p \geq 0}\) is bounded.
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). There exists \(M {\gt} 0\) such that \(z \mapsto -\frac{1}{2 i \pi } \sum _{p=0}^\infty \frac{\nu _p}{z^{p+1}}\) is an analytic extension of \(F_\nu \) in \(\{ |z| {\gt} M\} \).
Let \(K\), \(c\), \(m\) be such that \(|\nu (h)| \leq c \| h\| _{C^m([-K,K])}\) for all \(h \in \mathcal{E}(\mathbb {R}, \mathbb {C})\). Let \(N \geq 0\), \(z \neq 0\). By linearity,
We need to prove that for \(z \notin \mathbb {R}\) large enough, \(\sum _{p=0}^N \left(\frac{x}{z}\right)^p\) converges to \(\frac{1}{1-x/z}\) w.r.t. \(\| \cdot \| _{C^m([-K,K])}\), which then implies that the above quantity converges to \(F_\nu (z)\). [todo write the details]
The main result of this section is as follows.
Let \(\nu \in \mathcal{E}'(\mathbb {R}, \mathbb {C})\). Then \(\mathrm{supp} \, \nu \subseteq [-\rho (\nu ), \rho (\nu )]\).
As in [ .