15.3 Convexity

15.24   Definition (Convexity) Let $f$ be a differentiable function on an interval $(a,b)$. We say that $f$ is convex upward over $(a,b)$ or that $f$ holds water over $(a,b)$ if and only if for each point $t$ in $(a,b)$, the tangent line to graph($f$) at $(t,f(t))$ lies below the graph of $f$.

Since the equation of the tangent line to graph($f$) at $(t,f(t))$ is

$$y = f(t) + f'(t)(x-t),$$

the condition for $f$ to be convex upward over $(a,b)$ is that for all $x$ and $t$ in $(a,b)$

$$f(t) + f'(t)(x-t)\leq f(x).$$ (15.25)

Condition (15.25) is equivalent to the two conditions:

$$f'(t) \leq \frac{f(x) -f(t)}{x-t} \mbox{ whenever } t<x,$$

and

$$\frac{f(t) - f(x)}{t-x} \leq f'(t) \mbox{ whenever } x<t.$$

These last two conditions can be written as the single condition

$$f'(p) \leq \frac{f(q) - f(p)}{q-p} \leq f'(q) \mbox{ whenever } p < q.$$ (15.26)

We say that $f$ is convex downward over $(a,b)$, or that $f$ spills water over $(a,b)$ if and only if for each point $t$ in $(a,b)$, the tangent line to graph($f$) at $(t,f(t))$ lies above the graph of $f$.

This condition is equivalent to the condition that for all points $p,q \in (a,b)$

$$f'(p) \geq \frac{f(q) - f(p)}{q-p} \geq f'(q) \mbox{ whenever } p < q.$$

15.27   Theorem. Let $f$ be a differentiable function over the interval $(a,b)$. Then $f$ is convex upward over $(a,b)$ if and only if $f'$ is increasing over $(a,b)$. (and similarly $f$ is convex downward over $(a,b)$ if and only if $f'$ is decreasing over $(a,b)$.)

Proof:     If $f$ is convex upward over $(a,b)$, then it follows from (15.26) that $f'$ is increasing over $(a,b)$.

Now suppose that $f'$ is increasing over $(a,b)$. Let $p,q$ be distinct points in $(a,b)$. By the mean value theorem there is a point $c$ between $p$ and $q$ such that

$$f'(c) = \frac{f(p)-f(q)}{p-q}.$$

If $p<q$ then $p<c<q$ so since $f'$ is increasing over $(a,b)$

$$f'(p) \leq f'(c) \leq f'(q),$$

i.e.

$$f'(p) \leq \frac{f(p)-f(q)}{p-q} \leq f'(q).$$

Thus condition (15.26) is satisfied, and $f$ is convex upward over $(a,b)$.

15.28   Corollary. Let $f$ be a function such that $f''(x)$ exists for all $x$ in the interval $(a,b)$. If $f''(x) \geq 0$ for all $x\in (a,b)$ then $f$ is convex upward over $(a,b)$. If $f''(x) \leq 0$ for all $x\in (a,b)$ then $f$ is convex downward over $(a,b)$.

15.29   Exercise. A Prove one of the two statements in corollary 15.28.

15.30   Lemma (Converse of corollary 12.26) Let $f$ be a real function such that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f$ is increasing on $[a,b]$, then $f'(x) \geq 0$ for all $x\in (a,b)$.

Proof:     let $p \in (a,b)$. Choose $\delta > 0$ such that $(p-\delta,p +\delta) \subset (a,b)$. Then $\{p+\frac{\delta}{2n}\}$ is a sequence such that

$$\{p+\frac{\delta}{2n}\} \to p,$$

and hence

$$\{ \frac{ f(p+\frac{\delta}{2n})-f(p)}{(p+\frac{\delta}{2n})-p}\} \to f'(p).$$

Since $f$ is increasing on $(a,b)$, we have

$$\frac{ f(p+\frac{\delta}{2n})-f(p)}{(p+\frac{\delta}{2n})-p} \geq 0$$

for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, and it follows that

$$f'(p) \geq 0 \mbox{ for all }p \in (a,b).\mbox{ $\diamondsuit$}$$

15.31   Definition (Inflection point) Let $f$ be a real function, and let
$a \in \mbox{{\rm dom}}{f}$. We say that $a$ is a point of inflection for $f$ if there is some $\epsilon >0$ such that $(a-\epsilon,a+\epsilon) \subset \mbox{{\rm dom}}{f}$, and $f$ is convex upward on one of the intervals $(a-\epsilon ,a)$, $(a,a+\epsilon)$, and is convex downward on the other.

15.32   Theorem (Second derivative test for inflection points) Let $f$ be a real function, and let $a$ be a point of inflection for $f$. If $f''$ is defined and continuous in some interval $(a-\delta,a+\delta)$ then $f''(a)=0.$

Proof:     We will suppose that $f$ is convex upward on the interval $(a-\delta,a)$ and is convex downward on $(a,a+\delta)$. (The proof in the case where these conditions are reversed is essentially the same). Then $f'$ is increasing on $(a-\delta,a)$, and $f'$ is decreasing on $(a,a+\delta)$. By (15.30), $f''(x) \geq 0$ for all $x \in (a-\delta,a)$, and $f''(x) \leq 0$ for all $x \in (a,a+\delta)$. We have

$$f''(a) = \lim \{f''(a+\frac{\delta}{2n})\} \leq 0,$$

and

$$f''(a) = \lim\{f''(a -\frac{\delta}{2n})\} \geq 0.$$

It follows that $f''(a)=0.$ $\diamondsuit$

15.33   Example. When you look at the graph of a function, you can usually ``see'' the points where the second derivative changes sign. However, most people cannot ``see'' points where the second derivative is undefined.

By inspecting graph$(f)$, you can see that $f$ has a discontinuity at $p$.

By inspecting graph$(g)$, you can see that $g$ is continuous everywhere, but $g'$ is not defined at $q$.

By inspecting graph$(h)$ in figure a below, you can see that $h'$ is continuous, but you may have a hard time seeing the point where $h''$ is not defined.

The function $h$ is defined by

$$h(x) = \cases{ x^2 - {5\over2}x+2 & if $0 \leq x \leq {3\ov... ...r {1\over 2}x^2 - x +{7\over 8} & if ${3\over 2} < x \leq 2$.}$$ (15.34)

so $h''(x) = 2$ for $0 < x < {3\over 2}$, and $h''(x) = 1$ for ${3\over 2} < x < 2$, and $h''({3\over 2})$ is not defined. We constructed $h$ by pasting together two parabolas. Figure b shows the two parabolas, one having a second derivative equal to 1, and the other having second derivative equal to 2.

15.35   Exercise. Let $h$ be the function described in formula (15.34). Draw graphs of $h'$ and $h''$.

15.36   Entertainment (Discontinuous derivative problem.) There exists a function $f$ such that $f$ is differentiable everywhere on $\mbox{{\bf R}}$, but $f'$ is discontinuous somewhere. Find such a function.

15.37   Exercise. Let $f(x) = x^4$. Show that $f''(0)=0$, but $0$ is not a point of inflection for $f$. Explain why this result does not contradict theorem 15.32

15.38   Example. Let

$$f(x) = \frac{1}{1+x^2}.$$

Then

$$f'(x) = \frac{-2x}{(1+x^2)^2},$$

and

$$f''(x) = \frac{(1+x^2)^2(-2)-(-2x)(2(1+x^2)(2x))}{(1+x^2)^4} =\frac{2(3x^2-1)}{(1+x^2)^3}.$$

Thus the only critical point for $f$ is $0$. Also,

$$(f'(x) > 0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}x <0) \... ...and }(f'(x)<0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}x>0),$$

so $f$ is increasing on $(-\infty,0)$ and is decreasing on $(0,\infty)$. Thus $f$ has a maximum at $0$, and $f$ has no minima.

We see that $f''(x) = 0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}x^2= \frac{1}{3}$, and moreover

$$(f''(x) <0) \hspace{1ex}\Longleftrightarrow\hspace{1ex}x \in\left(-\sqrt{\frac{1}{3}},\sqrt{\frac{1}{3}}\right),$$

so $f$ spills water over the interval ${\Big(-\sqrt{\frac{1}{3}},\sqrt{\frac{1}{3}}\Big)}$, and $f$ holds water over each of the intervals ${\Big(-\infty,-\sqrt{{1\over 3}}\Big)}$ and ${\Big(\sqrt{{1\over 3}},\infty\Big)}$. Thus $f$ has points of inflection at $\pm \sqrt{1\over 3}$. We can use all of this information to make a reasonable sketch of the graph of $f$. Note that $f(x)>0$ for all $x$, $f(0) = 1$, and $f\Big(\pm\sqrt{{1\over 3}} \Big) =\frac{3}{4}$, and $\sqrt{\frac{1}{3}}$ is approximately 0.58.

15.39   Exercise. Discuss the graphs of the following functions. Make use of all the information that you can get by looking at the functions and their first two derivatives.

a) $f(x) = 5x^4 - 4x^5$.

b) $G(x) = 5x^3 - 3x^5$.

c) $${ H(x) = e^{- {1 \over x^2}}}$$.