A function f is convex on an interval I if every secant line
is above the graph on I.
Algebraically, convexity is expressed by: If X1 < X2, then for
X1 < X < X2,
f(X) £ f(X1) +
f(X2) - f(X1)
X2- X1
(X - X1).
1.
Show that if X1 < X2 < X3, then
f(X2) - f(X1)
X2 - X1
£
f(X3) -f(X1)
X3 - X1
,
f(X3) -f(X1)
X3 - X1
£
f(X3) - f(X2)
X3 - X2
.
This is an algebraic verification of the geometric observation
that slope[`(P1 P2)] £ slope[`(P1 P3)] and
slope[`(P1 P3)] £ slope[`(P2 P3)].
In particular, as X3decreases to X2, the difference quotient
f(X3) - f(X2)
X3 - X2
decreases and is bounded below by
f(X2) - f(X1)
X2 - X1
.
Thus
lim
X ® X2+
f(X) - f(X2)
X - X2
= D+f(X2)
exists. The right hand limit of the difference quotient,
D+f(x) =
lim
Dx® 0+
f(x + Dx) - f(x)
Dx
is called the right derivative or right derivate of
f at x.
Similarly the left derivative
lim
X ® X2-
f(X) - f(X2)
X - X2
= D-f(X2)
exists (and is £ D+f(X2)).
Thus we have shown: If f is convex on an open interval I = (a,b), then for each x Î I, the left derivative D-f(x)
and the right derivative D+f(x) both exist. If they are the same,
then f¢(x0) exists. Note that if a < x1 < x2 < b, then
D-f(x1) £ D+f(x1) £ D-f(x2) £ D+f(x2).
It
follows that D-f(x) and D-f(x) are nondecreasing
functions on I.
2.
Is the following result true? If not, give a counterexample.
Theorem. If f is convex on an open interval I = (a,b), then f is continuous on I.
File translated from
TEX
by
TTH,
version 3.79. On 03 Dec 2007, 15:48.
