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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