7/31/2019 Conv Esse
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C.10
Convex functions
We begin with a lemma, which tells that local convexity (i.e.,
convexity on aneighbourhood of every point of I ) is in fact a
global condition (valid on all of I ).
Lemma C.10.1 Let f be differentiable on the interval I . Then f
is convex on I if and only if for every x 0 I
f (x ) f (x 0 ) + f (x 0 )(x x 0 ) x I . (C.10.1)
Proof. Obviously, it is enough to show that if f is convex
according to Denition6.32 on I , then also (C.10.1) holds. To this
end, one usefully notes that f is convexon I if and only if the map
g(x ) = f (x )+ ax + b, a, b R is convex; in fact, requiringf (x )
f (x 0 ) + f (x 0 )(x x 0 ) is equivalent to g(x ) g(x 0 ) + g (x 0
)(x x 0 ).
Let then x 0 I be xed arbitrarily and consider the convex map
g(x ) =f (x ) f (x 0 ) f (x 0 )(x x 0 ), which satises g(x 0 ) = g
(x 0 ) = 0. We have to provethat g(x ) 0, x I . Suppose x 0 is not
the right end-point of I and let us showg(x ) 0, x I, x > x 0 ;
a symmetry argument will complete the proof.
Being g convex at x 0 , we have g(x ) 0 on a (right)
neighbourhood of x 0 . Itmakes then sense to dene
P = {x > x 0 : g(s ) 0, s [x 0 , x ]}
and x 1 = sup P .If x 1 coincides with the right end-point of I
, the assertion follows. Let us
assume, by contradiction, x 1 lies inside I ; By denition g(x )
0, x [x 0 , x 1 ),while in each (right) neighbourhood of x 1 there
exist points x I at which g(x ) 0, andlet x (x 0 , x 1 ) be a
pre-image of g(x ) = M . By Fermats Theorem 6.21 g (x ) = 0,so the
convexity at x yields a neighbourhood of x on which g(x ) g(x ) = M
;but M is the maximum of g on [x 0 , x 1 ], so g(x ) = M on said
neighbourhood. Nowdene
Q = {x > x : g(s ) = M, s [x, x ]}
and x 2 = sup Q . The map g is continuous, hence x 2 = max Q ,
and moreoverx 2 < x 1 because g(x 1 ) = 0. As before, the
hypothesis of convexity at x 2 leads toa contradiction. 2
Pag. 191 Proof of Theorem 6.36
Teorema 6.36 Given a differentiable map f on the interval I
,
a) if f is convex on I , then f is increasing on I .b1) If f is
increasing on I , then f is convex on I ;b2) if f is strictly
increasing on I , then f is strictly convex on I .
Proof.a) Take x 1 < x 2 two points in I . From (C.10.1) with
x 0 = x 1 and x = x 2 weobtain
f (x 1 ) f (x 2 ) f (x 1 )
x 2 x 1,
while putting x 0 = x 2 , x = x 1 gives
f (x 2 ) f (x 1 )x 2 x 1
f (x 2 ) .
Combining the two inequalities yields the result.b1) Let x >
x 0 be chosen in I . The second formula of the nite increment of f
on[x 0 , x ] prescribes the existence of a point x (x 0 , x ) such
that
f (x ) = f (x 0 ) + f (x )(x x 0 ) .
The map f is monotone, so f (x ) f (x 0 ) hence (C.10.1). When x
< x 0 theargument is analogous.b2) In the proof for b1) we now
have f (x ) > f (x 0 ), whence (C.10.1) is strict (forx = x 0 ).
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