Exercise 36, § 1

The notations are those of Exercise 35.

(a) Let $a$ be a vector of norm 1 in $E$, $K_\varepsilon$ the set of the $x \in S$ such that $|\langle x|a \rangle| \leqslant \sin \varepsilon$ and $L_\varepsilon$ the set of the $x \in S$ such that $d(x, S_a) \geqslant \varepsilon$ (where $S_a$ is the set of the points of $S$ orthogonal to $a$). Prove that for $\varepsilon > 0$ sufficiently small, one has $m(K_\varepsilon) \geqslant 4e^{-n\varepsilon^2/2}$ and $m(L_\varepsilon) \leqslant 4e^{-n\varepsilon^2/2}$ (one may remark that the image of the measure $m$ by 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 the $x \in S$ such that $f(x) \leqslant M(f)$ (resp. $f(x) \geqslant M(f)$) has measure $\geqslant \frac{1}{2}$ for $m$. Let $B$ be the set of the $x \in S$ such that $f(x) = M(f)$. Deduce from a) that, for every $\varepsilon > 0$ sufficiently small, the set of the $x \in S$ such that $d(x, B) \geqslant \varepsilon$ has 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 $h \geqslant 1$ for which there exist points $x_1, \ldots, x_h$ of $S$ such that $S = \bigcup_{i=1}^h B(x_i, \varepsilon)$. Prove that one has $\lim_{\varepsilon \to 0} (\log h(n, \varepsilon))/|\log \varepsilon| = n$.

d) Recall that $E$ is a real Hilbert space of dimension $n$. Let $k$ be a positive integer and $\varepsilon, \varepsilon'$ two strictly positive real numbers such that $4h(k, \varepsilon) < e^{n\varepsilon'/2}$. Let $f$ be a mapping from $S$ into $\mathbf{R}$ such that $|f(x) - f(y)| \leqslant \|x - y\|$ for $x, y$ in $S$, and $M(f)$ a real number satisfying the relation stated in b). Now prove 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 $y$ of $F \cap S$ such that $\|x - y\| < \varepsilon$ and $|f(y) - M(f)| \leqslant \varepsilon'$.