Lemma 4

Tag 02X2, permanent. Cite it as [Bourbaki, Tag 02X2].

Théories spectrales, chapter IV, § 1. Opérateurs compacts sur un espace hilbertien

Let A be a ring. Let $n\in \mathbf{N}$ and let $(a_i)_{0\leqslant i\leqslant n}$ and $(b_i)_{0\leqslant i\leqslant n}$ be families of elements of A. For $0\leqslant j\leqslant n$ , put

A_j=\sum_{i=0}^ja_i

Then

\sum_{i=0}^na_ib_i= A_nb_n-\sum^{n-1}_{i=0}A_i(b_{i+1}-b_i)

Put $A_{-1}= 0$ . We have $a_i= A_i-A_{i-1}$ for $0\leqslant i\leqslant n$ , hence

\sum_{i=0}^na_ib_i=\sum_{i=0}^n(A_i-A_{i-1})b_i=\sum^{n-1}_{i=0}A_i(b_i-b_{i+1}) + A_nb_n

as desired.

Let I be an interval of $\overline{\mathbf{R}}$ not containing $+\infty$ . Recall (FVR, I, p. 38, remark) that a continuous function $f: I\rightarrow [-\infty ,+\infty [$ is said to be convex if its restriction to the interior of I is a convex function with values in $\mathbf{R}$ ; it then has a limit (finite or infinite) on the right at inf I. In the statements below, the product of 0 and an element of $\{-\infty ,+\infty \}$ is by convention equal to 0.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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