Exercise 8, § 6

Let T be a compact space and L a vector subspace of $\mathscr{C}(T; \mathbf{R})$, that is of finite dimension r; give to L the norm induced by that of $\mathscr{C}(T; \mathbf{R})$ and to its dual $L^*$ the norm $\|x'\| = \sup_{\|x\| \leq 1} \langle x, x' \rangle$, so that if B is the ball $\|x\| \leq 1$ in L, then $B^\circ$ is the ball $\|x'\| \leq 1$ in $L^*$.

a) For all $t \in T$, write $e'_t$ for the linear form $x \mapsto x(t)$ on L. Show that $B^\circ$ is the convex envelope of the set of the $\pm e'_t$, where t varies in T (obtain a contradiction using prop. 4 of II, p. 38).

Deduce that every linear form $x' \in L^*$ such that $\|x'\| = 1$ we can write $x' = \sum_{i=1}^r \lambda_i e'_{t_i}$, where the $t_i$ are r points of T and the $\lambda_i$ are real numbers such that $\sum_{i=1}^r |\lambda_i| = 1$ (cf. II, p. 66, exerc. 9, a)).

b) For every $y \in \mathscr{C}(T; \mathbf{R})$, there exists a unique $x \in L$ such that $\|y - x\| = d(y, L)$, if and only if for every non null $z \in L$, there exist at most $r - 1$ distinct points $t_i \in T$ such that $z(t_i) = 0$ (Haar’s th.). (To show that this condition is sufficient, observe first that it is equivalent to saying that for r distinct points $t_i \in T$ ($1 \leq i \leq r$) the $e'_{t_i}$ are linearly independent in $L^*$. Now argue by assuming the conclusion is false and obtaining a contradiction. If there exist two distinct points $x', x''$ of L such that $\|y - x'\| = \|y - x''\| = d(y, L)$ then there exists $x_0 \in L$ and $z \in L$ such that, for all sufficiently small real $\lambda$, we have $\|y - (x_0 + \lambda z)\| = d(y, L)$. Apply the last part of a) to the subspace $L \oplus \mathbf{R}y$ of $\mathscr{C}(T; \mathbf{R})$ and to a suitable linear form on this space which vanishes in L. To see that the condition is necessary, note that if it is not true, then there exist r distinct points $t_i \in T$ ($1 \leq i \leq r$) such that the $e'_{t_i}$ are linearly dependent and there exists a function $z \in L$ that is not null and vanishes at the points $t_i$. If $\alpha_i$ ($1 \leq i \leq r$) are numbers not all zero such that $\sum_{i=1}^r \alpha_i e'_{t_i} = 0$, consider a function $w \in \mathscr{C}(T; \mathbf{R})$ such that $\|w\| = 1$, $w(t_i) = \operatorname{sgn}(\alpha_i)$ for $1 \leq i \leq r$, and the function $y = w(1 - |\beta z|)$ with $|\beta|$ sufficiently small and $\neq 0$.)

c) Suppose that T is a compact interval in $\mathbf{R}$ and that L satisfies the condition of Haar’s th.; let $(t_i)_{1 \leq i \leq r+1}$ be a strictly increasing sequence of $r + 1$ points of T; then there exist $r + 1$ real non zero numbers $\lambda_i$ such that $\sum_{i=1}^{r+1} \lambda_i e'_{t_i} = 0$. Show that then $\operatorname{sgn}(\lambda_i) \operatorname{sgn}(\lambda_{i+1}) = -1$ for $1 \leq i \leq r$. (Consider separately the case $r = 1$ and the case $r > 1$. In the second case, suppose on the contrary, that for some index $i \leq r - 1$, the number $\lambda_i$ is of the same sign as $\lambda_{i-1}$ or as $\lambda_{i+1}$ and that $\lambda_{i-1}$ and $\lambda_{i+1}$ are of opposite signs. If, for example, $\lambda_i > 0$, take $\alpha_{i-1} > 0$, $\alpha_{i+1} > 0$ such that $\alpha_{i-1} \lambda_{i-1} + \alpha_{i+1} \lambda_{i+1} = 0$, then $z \in L$ such that $z(t_{i-1}) = \alpha_{i-1}$, $z(t_{i+1}) = \alpha_{i+1}$ and $z(t_j) = 0$ for j distinct from $i - 1$, i and $i + 1$. Deduce that $z(t_i) < 0$ and show that this contradicts the given hypotheses.)

d) With the same hypotheses and notations as those of c), suppose that there exists $y \in \mathscr{C}(T; \mathbf{R})$ and $z \in L$ such that $y(t_i) - z(t_i) = (-1)^i \alpha_i$ with $\alpha_i > 0$ for $1 \leq i \leq r + 1$. Show that we then have $d(y, L) \geq \inf \alpha_i$. (Use a) applied to $L \oplus \mathbf{R}y$, and c).)

e) With the hypotheses of c) let $y \in \mathscr{C}(T; \mathbf{R})$, and z be the unique point of L such that $\|y - z\| = d(y, L)$. Show that there exists a strictly increasing sequence $(t_i)_{1 \leq i \leq r+1}$ in T such that

with $\varepsilon = \pm 1$. Conversely if $z$ has this property, then $z$ is the unique point of $L$ such that $\|y - z\| = d(y, L)$. (Use c) and d)). Consider the case when T is an interval of $\mathbf{R}$ and when L is the set of the restrictions to T of polynomials of degree < r (Tchebycheffs th.).