Convexity and the Second Derivative

We break the proof into steps, each a self-contained lemma, using the Mean Value Theorem as the sole analytic engine and the established equivalence between sign of a derivative and monotonicity from Monotonicity and the Sign of the Derivative for part (C).

Step 1: An equivalent three-point form of convexity. Claim: $f$ is convex on $I$ if and only if for all $a, b, c \in I$ with $a < b < c$,

Proof of claim. Write $b = ta + (1-t)c$ where $t = \frac{c-b}{c-a} \in (0,1)$ (so $1-t = \frac{b-a}{c-a}$). Convexity applied to $x_1=a, x_2=c$ gives

Subtracting $f(a)$ and dividing by $b - a > 0$:

This is the left inequality of $(\ast)$. For the right inequality, rewrite the same convexity inequality as

and divide by $c-b>0$ to get $\frac{f(c)-f(b)}{c-b} \geq \frac{f(c)-f(a)}{c-a}$, the right inequality. Conversely, if $(\ast)$ holds for all $a<b<c$ in $I$, reversing the algebra above (for any $x_1 \neq x_2$ in $I$ and $t\in(0,1)$, set $a = \min(x_1,x_2)$, $c=\max(x_1,x_2)$, $b = tx_1+(1-t)x_2$, which lies strictly between $a$ and $c$) recovers the defining inequality of convexity. This proves the claim, with strict inequalities throughout corresponding to strict convexity by the identical argument with $\leq$ replaced by $<$. $\blacksquare$ (end of Step 1)

Step 2 (A, $\Rightarrow$): Convex implies $f'$ nondecreasing. Suppose $f$ is convex on $I$. Let $u < v$ in $I$; we must show $f'(u) \leq f'(v)$. Pick any $w$ with $u < w < v$ (possible since $I$ is an interval, hence contains $(u,v)$). By $(\ast)$ with $a=u, b=w, c=v$:

Now let $w \to u^+$ in the leftmost expression: since $f$ is differentiable at $u$, $\frac{f(w)-f(u)}{w-u} \to f'(u)$. This gives

Similarly, let $w \to v^-$ in the rightmost expression: $\frac{f(v)-f(w)}{v-w} \to f'(v)$, giving

Chaining these: $f'(u) \leq \frac{f(v)-f(u)}{v-u} \leq f'(v)$, so $f'(u) \leq f'(v)$. Since $u<v$ were arbitrary, $f'$ is nondecreasing on $I$.

Step 3 (A, $\Leftarrow$): $f'$ nondecreasing implies convex. Suppose $f'$ is nondecreasing on $I$. We verify $(\ast)$ for arbitrary $a<b<c$ in $I$. By the Mean Value Theorem applied to $f$ on $[a,b]$ and on $[b,c]$ (valid since $f$ is differentiable, hence continuous, on these closed sub-intervals contained in $I$), there exist $\xi_1 \in (a,b)$ and $\xi_2 \in (b,c)$ with

Since $\xi_1 < b < \xi_2$ and $f'$ is nondecreasing, $f'(\xi_1) \leq f'(\xi_2)$, i.e.

This gives the outer inequality of a rearranged form of $(\ast)$; we still must derive the two inequalities in $(\ast)$ involving the chord over $[a,c]$. Note that

is a weighted average of $\frac{f(b)-f(a)}{b-a}$ and $\frac{f(c)-f(b)}{c-b}$: explicitly, with $\lambda = \frac{b-a}{c-a} \in (0,1)$,

which is verified directly: $f(c)-f(a) = (f(b)-f(a)) + (f(c)-f(b))$, and dividing through appropriately recovers this weighted-average identity (a short algebraic check: $\lambda(b-a) + (1-\lambda)(c-b) \cdot \frac{c-a}{c-a}$ reduces correctly since $\lambda = \frac{b-a}{c-a}$ makes the two terms combine to $\frac{(f(b)-f(a))+(f(c)-f(b))}{c-a} = \frac{f(c)-f(a)}{c-a}$). Since a weighted average of two numbers lies between them, and by $(\dagger)$ the first is $\leq$ the second, we get

which is exactly $(\ast)$. By Step 1, $f$ is convex on $I$.

Step 4 (B): Strict version. If $f'$ is strictly increasing, the inequality $f'(\xi_1) \leq f'(\xi_2)$ in Step 3 becomes strict (since $\xi_1 < \xi_2$), making $(\dagger)$ strict, and the weighted-average argument then gives strict inequalities in $(\ast)$; by the strict case of Step 1's claim, $f$ is strictly convex. Conversely, if $f$ is strictly convex, then $(\ast)$ holds strictly for all $a<b<c$; running the limiting argument of Step 2 with $u<v$ fixed and inserting a strict inequality point $w$ shows $f'(u) < \frac{f(v)-f(u)}{v-u} < f'(v)$ whenever such a $w$ exists strictly between $u,v$ — which it always does since $I$ is an interval — giving $f'(u) < f'(v)$ strictly. (The key point: the two limiting inequalities from Step 2 remain non-strict in the limit even when the pre-limit inequality is strict, but here we get the sharper strict middle term $\frac{f(v)-f(u)}{v-u}$ sandwiched strictly, which forces $f'(u) \neq f'(v)$ combined with the already-established $f'(u)\le f'(v)$ from part (A); hence $f'(u)<f'(v)$.) This establishes (B).

Step 5 (C): The second-derivative criterion. Now suppose $f$ is twice differentiable on $I$, so $f''=(f')'$ exists on $I$. By the result Monotonicity and the Sign of the Derivative applied to the function $g := f'$ (which is differentiable with $g' = f''$):

i.e. $f'$ is nondecreasing on $I$ if and only if $f''(x) \geq 0$ for all $x \in I$. Combined with part (A), this gives:

For the strict statement: the same cited result gives that $g'(x) > 0$ for all $x$ implies $g$ strictly increasing (this is the direction of that theorem that does not require a converse), i.e. $f''>0$ everywhere implies $f'$ strictly increasing, which by part (B) implies $f$ strictly convex. The converse fails — a strictly convex $f$ need not have $f''>0$ everywhere — as exhibited in the counterexample below, so we assert only the one-directional implication. $\blacksquare$

Using the technique of direct estimation from the definition of the second derivative as a limit (avoiding any circular appeal to Theorem (C) restricted to a one-sided neighborhood, so this is proved independently and can even be used as an alternate route to local statements of Theorem (C)): Suppose $f''(c) > 0$. By definition,

using $f'(c) = 0$. Since this limit is a positive number, there exists $\delta > 0$ such that for all $h$ with $0 < |h| < \delta$,

(This is the standard "limit positive $\Rightarrow$ eventually positive" fact: taking $\varepsilon = f''(c)/2 > 0$ in the definition of the limit produces such a $\delta$.) This means: for $0<h<\delta$, $f'(c+h) > 0$; and for $-\delta < h < 0$, $f'(c+h) < 0$ (dividing by a negative $h$ flips the sign). So $f'$ is negative immediately to the left of $c$ and positive immediately to the right of $c$, on $(c-\delta,c+\delta)$. By the theorem Monotonicity and the Sign of the Derivative, $f$ is strictly decreasing on $(c-\delta, c]$ and strictly increasing on $[c, c+\delta)$. Hence for any $x \in (c-\delta,c+\delta)$ with $x \neq c$: if $x<c$, $f(x) > f(c)$ (strict decrease means values to the left are larger); if $x > c$, $f(x) > f(c)$ (strict increase means values to the right are larger). In both cases $f(x) > f(c)$, so $c$ is a strict local minimum. The case $f''(c) < 0$ is identical with all inequalities reversed (or apply the above to $-f$, noting $(-f)''(c) = -f''(c) > 0$ and a local min of $-f$ at $c$ is a local max of $f$ at $c$). For $f''(c)=0$ inconclusiveness: exhibit $f_1(x) = x^4$, $f_2(x)=-x^4$, $f_3(x) = x^3$, all with $c=0$, $f_i'(0)=0$, $f_i''(0)=0$. Direct check: $x^4$ has a strict local (indeed global) min at $0$; $-x^4$ a strict local max; $x^3$ is strictly increasing everywhere (since $(x^3)' = 3x^2 \geq 0$ with equality only at an isolated point, which by the strict-monotonicity criterion still gives strict increase) hence has neither a local max nor min at $0$. This exhibits all three possible outcomes, proving the inconclusiveness claim. $\blacksquare$