Strict Convexity

When Convexity Is Strict

In “Second Derivative and Convexity,” the second-derivative criterion connected $f''\geq0$ with ordinary convexity. Strict convexity sharpens the comparison with a chord: between any two distinct points, the function must lie strictly below the line segment joining its values. This stronger inequality rules out straight portions of the graph, even though a strictly convex function may still have a zero second derivative at some points.

Strict convexity is defined for functions on any interval; differentiability is not part of the definition. Derivative tests will be useful when differentiability is available, but the underlying condition is a comparison of function values.

Definition (Strict Convexity): Let $I$ be an interval and $f:I\to\mathbb{R}$. The function $f$ is strictly convex on $I$ if, for every pair of distinct points $x,y\in I$ and every $t\in(0,1)$, $$ f((1-t)x+ty)<(1-t)f(x)+tf(y). $$

The point $(1-t)x+ty$ lies strictly between $x$ and $y$ when $x\ne y$ and $0<t<1$. The right-hand side is the value at that point of the line segment joining $(x,f(x))$ and $(y,f(y))$. Thus strict convexity says that every such interior point of the graph lies below the chord. If $x=y$, or if $t$ is an endpoint of $[0,1]$, the corresponding equality is expected; strictness is required only for distinct inputs and interior weights.

Every strictly convex function is convex: its strict inequality in particular implies the non-strict inequality in the definition of convexity. The converse is false. Convexity permits a chord to coincide with part of the graph, whereas strict convexity does not.

The Derivative Characterization

For a differentiable function on an open interval, strict convexity can be detected by whether the first derivative is strictly increasing. This is stronger than the derivative characterization of ordinary convexity, which requires only that the derivative be nondecreasing. The distinction is important: a nondecreasing derivative may remain constant over an interval, producing a straight segment in the graph.

Theorem (Derivative Characterization of Strict Convexity): Let $I$ be an open interval, and let $f:I\to\mathbb{R}$ be differentiable. Then $f$ is strictly convex on $I$ if and only if $f'$ is strictly increasing on $I$.

Proof. First suppose that $f'$ is strictly increasing. Choose $x,z,y\in I$ with $x<z<y$. The Mean Value Theorem applies to $f$ on both $[x,z]$ and $[z,y]$, since $f$ is continuous on these closed intervals and differentiable on their interiors. Therefore there are $c_1\in(x,z)$ and $c_2\in(z,y)$ such that

\frac{f(z)-f(x)}{z-x}=f'(c_1), \qquad \frac{f(y)-f(z)}{y-z}=f'(c_2).

Since $c_1<c_2$ and $f'$ is strictly increasing, the first secant slope is strictly less than the second:

\frac{f(z)-f(x)}{z-x} < \frac{f(y)-f(z)}{y-z}.

Both denominators are positive. Multiplying by $(z-x)(y-z)$, then collecting the terms involving $f(z)$, gives

(y-z)\bigl(f(z)-f(x)\bigr) < (z-x)\bigl(f(y)-f(z)\bigr), $$ $$ (y-x)f(z)<(y-z)f(x)+(z-x)f(y).

Divide by $y-x>0$. With $t=(z-x)/(y-x)$, we have $0<t<1$, $z=(1-t)x+ty$, and $(y-z)/(y-x)=1-t$. Hence

$$ f((1-t)x+ty)<(1-t)f(x)+tf(y). $$

This proves strict convexity. Any two distinct points can be ordered as $x<y$, so considering ordered points loses no cases.

Conversely, suppose $f$ is strictly convex. It is then convex, so the Theorem on Derivative Monotonicity for Convex Functions from “Convexity and Derivatives” implies that $f'$ is nondecreasing. We show it is strictly increasing. Take any $x<y$ in $I$. Nondecrease gives $f'(x)\leq f'(y)$. If equality held, then for every $t\in[x,y]$,

f'(x)\leq f'(t)\leq f'(y)=f'(x),

so $f'(t)=f'(x)$ throughout $[x,y]$. For any $u<v$ in $[x,y]$, the Mean Value Theorem gives a point $c\in(u,v)$ with

\frac{f(v)-f(u)}{v-u}=f'(c)=f'(x).

Thus $f(v)-f(u)=f'(x)(v-u)$ for every such pair, and $f$ is affine on $[x,y]$. In particular, its value at the midpoint is the average of its endpoint values:

f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}.

This contradicts strict convexity, applied to $x$, $y$, and $t=1/2$. Therefore equality $f'(x)=f'(y)$ is impossible. Since $f'(x)\leq f'(y)$, it follows that $f'(x)<f'(y)$. This holds for every $x<y$, so $f'$ is strictly increasing. $\square$

Applying the Characterization

Worked Example: A Strictly Convex Function with a Zero Second Derivative

Let $f(x)=x^4$ on $\mathbb{R}$. Its derivatives are $f'(x)=4x^3$ and $f''(x)=12x^2$. The second derivative vanishes at $0$, so the condition $f''(x)>0$ everywhere does not apply. Instead, we check directly that $f'$ is strictly increasing. If $x<y$, then

$$ f'(y)-f'(x)=4(y^3-x^3) =4(y-x)(y^2+xy+x^2). $$

The factor $y-x$ is positive. Also,

y^2+xy+x^2=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}.

This sum is nonnegative, and it can be zero only if $y=0$ and $x+y/2=0$, which would imply $x=y=0$, contrary to $x<y$. Thus $y^2+xy+x^2>0$, so $f'(y)-f'(x)>0$. The derivative characterization proves that $x^4$ is strictly convex on $\mathbb{R}$, despite $f''(0)=0$.

Theorem (Positive Second Derivative Implies Strict Convexity): Let $I$ be an open interval, and let $f:I\to\mathbb{R}$ be twice differentiable. If $f''(x)>0$ for every $x\in I$, then $f$ is strictly convex on $I$.

Proof. Take $x,y\in I$ with $x<y$. The function $f'$ is continuous on $[x,y]$ and differentiable on $(x,y)$, so the Mean Value Theorem applied to $f'$ gives a $c\in(x,y)$ such that

$$ f'(y)-f'(x)=f''(c)(y-x). $$

By hypothesis $f''(c)>0$, and $y-x>0$, so $f'(y)-f'(x)>0$. Therefore $f'$ is strictly increasing on $I$. The Derivative Characterization of Strict Convexity now shows that $f$ is strictly convex on $I$. $\square$

Worked Example: A Positive Second Derivative on a Restricted Domain

Let $f(x)=x^3$ on $I=(0,\infty)$. Differentiating gives

f'(x)=3x^2, \qquad f''(x)=6x.

For every $x\in(0,\infty)$, $6x>0$. The Positive Second Derivative Implies Strict Convexity theorem therefore proves that $f$ is strictly convex on $(0,\infty)$. The domain matters: on all of $\mathbb{R}$, the second derivative is negative for $x<0$, and the derivative $3x^2$ is not increasing throughout that larger interval.

Worked Example: Convexity Without Strict Convexity

Consider the affine function $g(x)=5x-3$ on $\mathbb{R}$. For any $x,y\in\mathbb{R}$ and $t\in(0,1)$,

$$ g((1-t)x+ty) =5((1-t)x+ty)-3 =(1-t)(5x-3)+t(5y-3) =(1-t)g(x)+tg(y). $$

The convexity inequality holds with equality for every pair of inputs. In particular, it is not strict when $x\ne y$, so $g$ is convex but not strictly convex. The derivative test gives the same conclusion: $g'(x)=5$ is nondecreasing but not strictly increasing. This example also shows why a nondecreasing derivative alone cannot establish strict convexity.

What the Strict Test Tells You

The two derivative tests have different hypotheses and conclusions. The Second-Derivative Criterion from the previous tutorial says that $f''\geq0$ everywhere is equivalent to convexity for a twice-differentiable function on an open interval. The strict test proved here gives a sufficient condition: $f''>0$ everywhere implies strict convexity. It is not a necessary condition, as $x^4$ demonstrates.

A zero value of $f''$ at one point therefore does not decide whether a function is strictly convex. What matters in the derivative characterization is whether $f'$ is strictly increasing across every pair of distinct points. For $x^4$, the derivative increases across every such pair even though its rate of change is zero at the origin. By contrast, if $f'$ is constant on a nontrivial interval, the Mean Value Theorem shows that $f$ is affine there, which is incompatible with strict convexity.

Strict convexity is a global condition on the interval under consideration. A calculation on a smaller domain proves a conclusion only on that domain, as the $x^3$ example illustrates. When using a second-derivative test, first identify the interval, then check the sign throughout it. If strict positivity fails, do not conclude that strict convexity fails: examine whether the first derivative is nevertheless strictly increasing.

The chord definition remains available even when derivatives do not exist. The derivative characterization is a convenient equivalent test under its differentiability hypothesis, not a replacement for the definition in every setting. Keeping these roles separate prevents a common error: treating a derivative test as though differentiability were part of strict convexity itself.

Check Your Understanding

Use the definition and the derivative results to answer the following questions.

State the strict convexity inequality and specify which choices of points and weights it applies to.
Why does a strictly increasing derivative imply a strict inequality between the secant slopes on $[x,z]$ and $[z,y]$ when $x<z<y$?
In the converse direction of the derivative characterization, why would $f'(x)=f'(y)$ contradict strict convexity?
Does $f''(a)=0$ at one point rule out strict convexity? Explain using an example from this tutorial.
Why is the affine function $g(x)=5x-3$ convex but not strictly convex?

What You'll Learn