Functions of a Real Variable, II, p. 32
. b) Prove that there exists a convex function h in (0, 1) such that h≥f and that
∫01h(t)dt≤2∫01f(t)dt.
(For every a such that 0<a≤1, let fa be the function equal to f for 0≤t≤a and to f(a) for a≤t≤1; let A be the set of the a∈]0,1[ for which there exists a convex function ha in (0,1) such that ha≥fa and ∫01ha(t)dt≤2∫01fa(t)dt. Prove that the upper bound b of A still belongs to A, using exerc. 1 of I, p. 52. Proving subsequently that fb=f by reasoning by contradiction. For this, one reduces to proving the following result: if φ is continuous and increasing in (0,1], nonconstant and such that φ(0)=0, there is a point c∈]0,1[ such that φ(c)>0 and that the linear function ψ such that ψ(c)=φ(c) and
ψ(2c−1)=0
verifies the relation ψ(t)≥φ(t) for max(0,2c−1)≤t≤c.
23) Let f be a continuously differentiable vector function in an interval (a,b)⊂R, with values in a complete normed space. a) Prove that for a≤t≤b, one has
f(t)=b−a1∫abf(x)dx+∫abb−ax−af′(x)dx+∫tbb−ax−bf′(x)dx.
b) Deduce the inequalities
∥f(t)∥≤b−a1∫ab∥f(x)∥dx+∫ab∥f′(x)∥dx
and
∥f(21(a+b))∥≤b−a1∫ab∥f(x)∥dx+21∫ab∥f′(x)∥dx.