Exercise 5.14

Question

Exercise 14: Let $f$ be a differentiable real function defined in $(a,b)$. Prove that $f$ is convex if and only if $f'$ is monotonically increasing. Assume next that
    $f''(x)$ exists for every $x\in(a,b)$, and prove that $f$ is convex if and only if $f''(x)\ge0$ for all $x\in(a,b)$.

analambanomenos · Accepted Answer

\def\ints{\mathbb{Z}}

If $f''(x)$ exists for every $x\in(a,b)$, then $f''(x)\ge0$ for all $x\in(a,b)$ if and only if $f'$ is monotonically increasing, so the second part of the Exercise
    follows from the first part.

Suppose that $f$ is convex and let $a<x<y<b$. By Exercise 4.23 $$\frac{f(t)-f(x)}{t-x}\le\frac{f(u)-f(y)}{u-y}$$ for $t$ and $u$ close to $x$ and $y$, respectively. Taking
    limits as $tightarrow x$ and $uightarrow y$, we get $f'(x)\le f'(y)$.

Conversely, suppose is $f'$ is monotonically increasing, let $a<x<y<b$, and let $z=\lambda x+(1-\lambda)y$ for $0<\lambda<1$. By Theorem 5.10 there are points $w_1,w_2$
    such that $x<w_1<z<w_2<y$ such that 
    \begin{align*}
        \frac{f(z)-f(x)}{z-x}=f'(w_1) &\le f'(w_2)=\frac{f(y)-f(z)}{y-z} \
        \frac{f(z)-f(x)}{(\lambda-1)x+(1-\lambda)y} &\le \frac{f(y)-f(z)}{\lambda(y-x)} \
        f(z)(y-x) &\le \bigl(\lambda(y-x)\bigr)f(x)+\bigl((1-\lambda)(y-x)\bigr)f(y) \
        f\bigl(\lambda x+(1-\lambda)y\bigr) &\le \lambda f(x)+(1-\lambda)f(y)
    \end{align*}
    which shows that $f$ is convex on $(a,b)$. (The algebra above is much easier if you just take $\lambda=1/2$, then apply Exercise 4.24.)

Exercise 5.14

Answers

Comments

Exercise 5.14

Answers

Comments

Add answer