Convexity and Derivatives

From Chord Slopes to Derivatives

In “Convex Functions,” the three-point secant-slope inequality showed that the slopes of successive chords of a convex graph are ordered. When the function is differentiable, derivatives are limits of these chord slopes. This connects the geometric definition of convexity to a useful test involving the derivative. The test also gives a stronger geometric conclusion: at each point, the tangent line lies below the entire graph.

Throughout this tutorial, let \(I\) be an open interval, and let \(f:I\to\mathbb{R}\) be differentiable at every point of \(I\). We use “nondecreasing” to mean that \(f'(x)\leq f'(y)\) whenever \(x<y\). The condition is not that the derivative must be strictly increasing.

The Derivative of a Convex Function Is Nondecreasing

Fix two points \(x<y\) in \(I\). For a convex function, a secant slope from \(x\) to a point just to its right cannot exceed the slope from \(x\) to \(y\). Letting that nearby point approach \(x\) compares the derivative at \(x\) with the slope across the whole interval. A corresponding comparison near \(y\) bounds that same slope by the derivative at \(y\).

Proof. Fix \(x<y\) in \(I\). For any \(t\) with \(x<t<y\), the Three-Point Secant-Slope Inequality gives

As \(t\to x\) from the right, the leftmost quotient tends to \(f'(x)\), by the definition of the derivative. The middle quotient is fixed, so this limit gives

As \(t\to y\) from the left, the rightmost quotient tends to \(f'(y)\). Indeed, it is the difference quotient at \(y\) with increment \(t-y\), since \((f(t)-f(y))/(t-y)=(f(y)-f(t))/(y-t)\). Thus

Combining the two inequalities proves the claimed secant-slope bounds. In particular, \(f'(x)\leq f'(y)\) whenever \(x<y\), so \(f'\) is nondecreasing. \(\square\)

This argument does not require \(f'\) to be continuous. It uses only differentiability at the two endpoints of each secant interval and the ordering of slopes that convexity provides.

A Derivative Characterization of Convexity

The preceding theorem proves that convexity forces the derivative to be nondecreasing. Conversely, the Mean Value Theorem shows that a nondecreasing derivative implies convexity, as proved below. Together, these results give an exact criterion for differentiable functions on open intervals.

Proof. If \(f\) is convex, the Theorem on Derivative Monotonicity for Convex Functions proves that \(f'\) is nondecreasing. Conversely, suppose \(f'\) is nondecreasing on \(I\). For \(x<z<y\) in \(I\), the Mean Value Theorem gives \(c_1\in(x,z)\) and \(c_2\in(z,y)\) such that \(\frac{f(z)-f(x)}{z-x}=f'(c_1)\leq f'(c_2)=\frac{f(y)-f(z)}{y-z}\). Since \(c_1<c_2\), the inequality follows from the assumed monotonicity. Rearranging gives \(f(z)\leq \frac{y-z}{y-x}f(x)+\frac{z-x}{y-x}f(y)\), the convexity inequality for interior points \(z\) between \(x\) and \(y\). The endpoint cases are immediate, so \(f\) is convex. These two implications prove the equivalence. \(\square\)

The open-interval hypothesis ensures that every point of \(I\) is an interior point where the derivative is defined, and that the derivative criterion applies throughout the domain. This is a criterion for differentiable functions; it does not replace the chord definition for functions that may fail to be differentiable.

Tangent Lines as Global Lower Bounds

For a differentiable convex function, the slope of the chord from \(x\) to \(y\) is at least \(f'(x)\) when \(x<y\). Rearranging this slope inequality says that \(f(y)\) lies above the tangent line drawn at \(x\). The same conclusion holds when \(y<x\), using the other side of the secant-slope bounds.

Proof. First suppose \(a<y\). The secant-slope bounds in the Theorem on Derivative Monotonicity for Convex Functions give

Since \(y-a>0\), multiplying preserves the inequality and gives \(f(y)\geq f(a)+f'(a)(y-a)\).

Now suppose \(y<a\). Applying the same secant-slope bounds to the ordered pair \(y<a\) gives

Multiplying by \(a-y>0\) gives \(f(a)-f(y)\leq f'(a)(a-y)\). Rearranging yields \(f(y)\geq f(a)+f'(a)(y-a)\).

If \(y=a\), both sides of the claimed inequality equal \(f(a)\). All cases are covered, so the supporting-line inequality holds for every \(y\in I\). \(\square\)

There is also a converse that explains why the tangent-line condition captures convexity. Suppose \(f\) is differentiable on \(I\) and that for every \(a,y\in I\), \(f(y)\geq f(a)+f'(a)(y-a)\). Take \(x,y\in I\), \(0\leq\lambda\leq1\), and set \(z=(1-\lambda)x+\lambda y\). Since \(I\) is an interval, \(z\in I\). Apply the assumed inequality at \(a=z\), first with the point \(x\), and then with the point \(y\). Multiplying the resulting inequalities by \(1-\lambda\) and \(\lambda\), respectively, and adding gives

The final equality holds because \(z=(1-\lambda)x+\lambda y\). This is the chord inequality, so \(f\) is convex. The tangent-line condition and convexity are therefore equivalent for differentiable functions on an open interval.

What the Derivative Test Does—and Does Not—Say

The derivative criterion is efficient because it replaces comparisons across all chords with an order condition on \(f'\). The supporting-line inequality adds a practical bound that can be used without calculating a secant slope. Both results depend on convexity or a nondecreasing derivative over the whole interval; checking the derivative at only a few points is not enough to establish either property.

A common mistake is to demand that the derivative be strictly increasing. Convexity requires only that it be nondecreasing, so intervals on which \(f'\) is constant are allowed. On such an interval, the Mean Value Theorem implies that \(f\) is affine there. Another mistake is to use a derivative test when differentiability has not been established. The chord definition remains the appropriate definition in that case.

The next tutorial will develop a second-derivative criterion for convexity. The first-derivative viewpoint remains useful: it explains why that later test works and provides a direct route from increasing slopes to supporting tangent lines.