Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 23

Problem

A real-valued function f defined in (a,b) is said to be convex if

f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y)

whenever a<x<b,a<y<b,0<\lambda<1. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. (For example, if f is convex, so is e^f.)

If f is convex in (a,b) and if a<s<t<u<b, show that

\frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}.

Answer

(1) Let f be a convex function in (a,b) and a<s<t<u<b. Then note that

t = \frac{t-s}{u-s}u + (1-\frac{t-s}{u-s})s.

Then

f(t) \leq \frac{t-s}{u-s}f(u) + (1-\frac{t-s}{u-s})f(s).

From the above inequality, we can deduce the following two inequalities.

\frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s},

\frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}.

(2) Fix x \in (a,b) and \epsilon >0. For y \in (a,b), choose a_0,a_1,b_0,b_1 \in (a,b) satisfying a_0<a_1<x<b_0<b_1 and a_0<a_1<y<b_0<b_1. If x>y, we have

\frac{f(a_1) - f(a_0)}{a_1 - a_0} \leq \frac{f(x) - f(y)}{x-y} \leq \frac{f(b_1) - f(b_0)}{b_1-b_0},

and if x<y, we have

\frac{f(a_1) - f(a_0)}{a_1 - a_0} \leq \frac{f(y) - f(x)}{y-x} \leq \frac{f(b_1) - f(b_0)}{b_1-b_0},

Then

\frac{|f(x) - f(y)|}{|x-y|} \leq \max(\frac{|f(a_1) - f(a_0)|}{|a_1 - a_0|},\frac{|f(b_1) - f(b_0)|}{|b_1 - b_0|}).

Choose \delta = \epsilon\min(1, \frac{|a_1-a_0|}{|f(a_1)-f(a_0)|},  \frac{|b_1-b_0|}{|f(b_1)-f(b_0)|}). Then we have |f(x) - f(y)| < \epsilon for y \in (a,b) satisfying |x-y|<\delta.

(3) Let f be a convex function and g be a convex increasing function on (a,b). We will show that h = g \circ f is also a convex function. For x,y \in (a,b) and \lambda \in (0,1),

h(\lambda x + (1-\lambda)y)  = g(f(\lambda x + (1-\lambda)y)) \leq g(\lambda f(x) + (1-\lambda)f(y)) \leq \lambda g(f(x)) + (1-\lambda)g(f(y)) = \lambda h(x) + (1-\lambda)h(y).

 


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