## Geometria analitica con elementi di algebra lineare by Marco Abate By Marco Abate

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Suppose that f admits a right derivative at all points of the complement with respect to [a, b[[ of a countable subset A of this interval. Show that there exists a point x ∈ ]a, b[[ ∩ A such that f(b) − f(a) fd (x) (b − a). h, and use th. 2 of I, p. ) § 3. 1) With the same hypotheses as in prop. 2 of I, p. g( p) ]. p 2) With the notation of prop. 2 of I, p. y] 0 for all y ∈ F implies that a 0 in E. Under these conditions, if gi (0 i n) are n + 1 vector 40 Ch. f (n) ] 0 then the functions gi are identically zero.

Sufﬁcient that it be convex on ]a, b[[ and that one has f (a) 5) Let f be a convex function on an open interval ]a, +∞[[; if there exists a point c > a such that f is strictly increasing on ]c, +∞[[ then lim f (x) +∞. x→+∞ 6) Let f be a convex function on an interval ]a, +∞[[; show that f (x)/x has a limit (ﬁnite or equal to +∞) as x tends to +∞; this limit is also that of f d (x) and of f g (x); it is > 0 if f (x) tends to +∞ as x tends to +∞. 7) Let f be a convex function on the interval ]a, b[[ where a 0; show that on this interval the function x → f (x) − x f (x) (the “ordinate at the origin” of the right semi-tangent at the point x to the graph of f ) is decreasing (strictly decreasing if f is strictly convex).

23, th. 2). rn (y) M COROLLARY. If f is a ﬁnite real function with a derivative of order n + 1 on I, and if m f (n+1) (x) M on I, then for all x a in I one has m (x − a)n+1 (n + 1)! rn (x) M (x − a)n+1 (n + 1)! (10) and the second term cannot be equal to the ﬁrst (resp. to the third) unless f (n+1) is constant and equal to m (resp. M) on the interval [a, x]]. The proof proceeds in the same way, but applying th. 1 of I, p. 14. Remarks. 1) We have already noticed in the proof of th. 1 that if f has a derivative of order n on I, and if f(x) a0 + a1 (x − a) + a2 (x − a)2 + · · · + an (x − a)n + rn (x) (11) is its Taylor expansion of order n at the point a, then the Taylor expansion of order n − 1 for f at the point a is f (x) a1 + 2a2 (x − a) + · · · + nan (x − a)n−1 + rn (x).