Bourbaki

Exercise 25, § 5

Integration, V, p. 105

Bourbaki marks this one as harder than the rest.

a) If uu and vv are any two real numbers, prove, for 1<p<+∞1 < p < +\infty, the inequality

∣u+v∣p≤∣u∣p+pvu∣u∣p−2+a∑r=2[p]∣v∣r∣u∣p−r+b∣v∣p, |u + v|^p \leq |u|^p + pvu|u|^{p-2} + a \sum_{r=2}^{[p]} |v|^r |u|^{p-r} + b|v|^p,
where aa and bb are two constants depending only on pp, and [p][p] is the integral part of pp (FRV, III, §2, Exer. 6).

b) Let (f~n)(\tilde{f}_n) be a sequence that converges to 0 in LpL^p for the weakened topology (1<p<+∞1 < p < +\infty). Show that there exists a subsequence (fnk)(f_{n_k}) of (fn)(f_n) such that, on setting sm=∑k=1mfnks_m = \sum_{k=1}^m f_{n_k}, one has

∣∫sm−1∣sm−1∣p−2fnm dμ∣≤1 \left| \int s_{m-1}|s_{m-1}|^{p-2} f_{n_m} \, d\mu \right| \leq 1
(define the sequence (nk)(n_k) recursively). Deduce from this, using a) and Hölder’s inequality, that if p>2p > 2 there exist two constants a,ba, b such that
N1(∣sn∣p)≤N1(∣sn−1∣p)+a+bN1(∣sn−1∣p−2), N_1(|s_n|^p) \leq N_1(|s_{n-1}|^p) + a + bN_1(|s_{n-1}|^{p-2}),
and that if 1<p≤21 < p \leq 2 there exists a constant cc such that
N1(∣sn∣p)≤N1(∣sn−1∣p)+c. N_1(|s_n|^p) \leq N_1(|s_{n-1}|^p) + c.
Conclude that
Np(sn)=O(n1/2)for p>2Np(sn)=O(n1/p)for 1<p≤2. N_p(s_n) = O(n^{1/2}) \quad \text{for } p > 2 \\ N_p(s_n) = O(n^{1/p}) \quad \text{for } 1 < p \leq 2.

c) Show that the preceding results cannot in general be improved upon (cf. Exers. 19 and 20 b)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.