Exercise 36, § 1

The notations are the same as in exerc. 35. a) Let $a$ be a vector of norm 1 in E, $K_\varepsilon$ the set of all $x \in S$ such that $|\langle x|a \rangle| \leq \sin \varepsilon$ and $L_\varepsilon$ the set of all $x \in S$ such that $d(x, S_a) \geq \varepsilon$ (where $S_a$ is the set of all points of S orthogonal to $a$). Show that for $\varepsilon > 0$ small enough, we have $m(K_\varepsilon) \geq 4e^{-n\varepsilon^2/2}$ and $m(L_\varepsilon) \leq 4e^{-n\varepsilon^2/2}$ (we observe that the image of the measure $m$ under the mapping $x \mapsto \langle x|a \rangle$ from S into the interval $(-1, 1)$ of $\mathbf{R}$ is of the form $c_n(1 - t^2)^{(n-3)/2}dt$ with a suitable constant $c_n > 0$. b) Let $f$ be a continuous mapping from S into $\mathbf{R}$ and $M(f)$ a real number such that the set of all $x \in S$ for which $f(x) \leq M(f)$ (resp. $f(x) \geq M(f)$) has a measure $\geq \frac{1}{2}$ for $m$. Let B be the set of all $x \in S$ such that $f(x) = M(f)$. Deduce from a) that, for every $\varepsilon > 0$ small enough, the set of all $x \in S$ such that $d(x, B) \geq \varepsilon$, has a measure for $m$ at most equal to $4e^{-n\varepsilon^2/2}$. c) For every $\varepsilon > 0$, let $h(n, \varepsilon)$ be the smallest integer $h \geq 1$ for which there exist points $x_1, ..., x_h$ in S such that $S = \bigcup_{i=1}^h B(x_i, \varepsilon)$ show that $\lim_{\varepsilon \to 0} (\log h(n, \varepsilon))/|\log \varepsilon| = n$. d) Recall that E is a real hilbertian space of dimension $n$. Let k be a positive integer and $\varepsilon, \varepsilon'$ two strictly positive numbers such that $4h(k, \varepsilon) < e^{n\varepsilon'/2}$. Let $f$ be a mapping from S into $\mathbf{R}$ such that $|f(x) - f(y)| \leq \|x - y\|$ for all $x, y$ in S, and $M(f)$ a real number satisfying the relation stated in b). Show that there exists a vector subspace F of E, of dimension k, satisfying the following condition : for every $x \in F \cap S$, there exists a point $y$ in $F \cap S$ such that $\|x - y\| < \varepsilon$ and $|f(y) - M(f)| \leq \varepsilon'$.