Convexity And Supporting Lines For Jensen's Inequality

Convexity and the Supporting-Line Inequality

Convexity turns a global comparison between values of a function into a useful pointwise lower bound. At an interior point of its interval, a finite convex function always lies above at least one line passing through its graph there. This supporting-line inequality is the elementary geometric fact behind Jensen's inequality: one can apply the same line to each value being averaged, then average the resulting inequalities. The interior-point qualification is essential; at an endpoint, a finite convex function may have no supporting line with finite slope.

Definition: Let $I$ be an interval in $\mathbb{R}$, and let $f:I\to\mathbb{R}$ be real-valued and finite. The function $f$ is convex if, for all $x,y\in I$ and $t\in[0,1]$, $$ f((1-t)x+ty)\le (1-t)f(x)+tf(y). $$ If $x_0$ is an interior point of $I$, a supporting line from below at $x_0$ is a line $y\mapsto f(x_0)+c(y-x_0)$, for some finite $c\in\mathbb{R}$, such that $$ f(y)\ge f(x_0)+c(y-x_0)\qquad\text{for every }y\in I. $$

Convexity does not require differentiability or continuity as an additional hypothesis. The supporting-line result below follows directly from convexity and the finiteness of $f$. Its proof uses slopes of chords, not derivatives.

How Convexity Orders Secant Slopes

For $a<b$ in $I$, write the secant slope between $a$ and $b$ as $(f(b)-f(a))/(b-a)$. Convexity orders these slopes when their intervals are arranged from left to right.

Lemma: If $f:I\to\mathbb{R}$ is convex and $a<b<c$ are in $I$, then $$ \frac{f(b)-f(a)}{b-a}\le \frac{f(c)-f(b)}{c-b}. $$

Proof. Since $b=(1-t)a+tc$, where $t=(b-a)/(c-a)\in(0,1)$, convexity gives $$ f(b)\le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c). $$ Multiplying by $c-a>0$ and rearranging yields $(c-b)(f(b)-f(a))\le(b-a)(f(c)-f(b))$. Dividing by the positive number $(b-a)(c-b)$ proves the claimed inequality. $\square$

The lemma compares adjacent secants. It also implies the familiar ordering for nested secants: slopes cannot decrease as their intervals move to the right. This order will make it possible to choose one slope that works simultaneously for every point of $I$.

Supporting Lines at Interior Points

Theorem: If $I$ is an interval, $f:I\to\mathbb{R}$ is finite and convex, and $x_0$ is an interior point of $I$, then $f$ has a supporting line from below at $x_0$. More precisely, define $$ L=\sup_{y\in I,\ y<x_0}\frac{f(x_0)-f(y)}{x_0-y}, \qquad R=\inf_{y\in I,\ y>x_0}\frac{f(y)-f(x_0)}{y-x_0}. $$ Both $L$ and $R$ are finite real numbers, $L\le R$, and every $c\in[L,R]$ is a supporting slope. Conversely, every supporting slope belongs to $[L,R]$.

Proof. For each $u<x_0<v$ in $I$, the lemma applied to $u,x_0,v$ gives $$ \frac{f(x_0)-f(u)}{x_0-u}\le \frac{f(v)-f(x_0)}{v-x_0}. $$ There are points on both sides of $x_0$, because it is an interior point. Fix one such $u_0<x_0$ and one such $v_0>x_0$. Every left slope is at most the fixed right slope at $v_0$; hence the supremum $L$ is bounded above. It is bounded below by the finite left slope at $u_0$. Thus $L$ is finite. Every right slope is at least the fixed left slope at $u_0$, so $R$ is bounded below; it is bounded above by the finite right slope at $v_0$. Thus $R$ is finite. The inequality between every left and every right slope implies $L\le R$.

Choose $c\in[L,R]$. If $y<x_0$, its left slope is at most $L\le c$, so \[ \frac{f(x_0)-f(y)}{x_0-y}\le c. \] Multiplying by $x_0-y>0$ and rearranging gives $f(y)\ge f(x_0)+c(y-x_0)$. If $y>x_0$, its right slope is at least $R\ge c$, which directly gives $f(y)\ge f(x_0)+c(y-x_0)$. At $y=x_0$, equality holds. Thus $c$ is a supporting slope. Conversely, suppose $c$ is a supporting slope. Applying its defining inequality first to $y<x_0$ and then to $y>x_0$, and dividing by the positive distances, shows that every left slope is at most $c$ and every right slope is at least $c$. Consequently $L\le c\le R$. This proves both the existence and the stated characterization. $\square$

The interval $[L,R]$ describes every possible supporting slope, not just one choice. If $f$ is differentiable at $x_0$, this interval reduces to the derivative. Indeed, for $y<x_0<z$, the secant-slope lemma gives \[ \frac{f(x_0)-f(y)}{x_0-y}\le\frac{f(z)-f(x_0)}{z-x_0}. \] Letting $z$ approach $x_0$ from the right shows that each left slope is at most $f'(x_0)$; letting $y$ approach $x_0$ from the left shows that each right slope is at least $f'(x_0)$. The same limits show that the supremum of the left slopes and the infimum of the right slopes both equal $f'(x_0)$. Thus the supporting line is the tangent line when the derivative exists.

Worked Example: Supporting Lines for the Square Function

Take $I=\mathbb{R}$, $f(y)=y^2$, and $x_0=1$. For $y<1$, the left slope is \[ \frac{1-y^2}{1-y}=\frac{(1-y)(1+y)}{1-y}=1+y. \] Its supremum over $y<1$ is $2$: all these slopes are less than $2$, and values $y$ approaching $1$ make $1+y$ approach $2$. For $y>1$, the right slope is \[ \frac{y^2-1}{y-1}=\frac{(y-1)(y+1)}{y-1}=y+1, \] whose infimum is $2$. Therefore $L=R=2$, and the supporting line is $1+2(y-1)=2y-1$. Check it for every real $y$: $$ y^2-(2y-1)=(y-1)^2\ge0. $$ Equality holds at $y=1$, as a supporting line must pass through $(1,f(1))=(1,1)$.

Worked Example: Many Supporting Lines for the Absolute-Value Function

Let $I=\mathbb{R}$, $f(y)=|y|$, and $x_0=0$. For $y<0$, $f(y)=-y$, so the left slope is \[ \frac{|0|-|y|}{0-y}=\frac{0-(-y)}{-y}=-1. \] For $y>0$, the right slope is \[ \frac{|y|-|0|}{y-0}=\frac{y}{y}=1. \] Thus $L=-1$ and $R=1$. Every $c\in[-1,1]$ supports the function at zero. To check this directly, if $y>0$, then $cy\le y=|y|$ because $c\le1$. If $y<0$, then $cy\le -y=|y|$ because $c\ge-1$. At $y=0$, both sides are zero. In particular $c=0$ gives the supporting line $0$, and the calculations verify the full range of allowed slopes.

Why the Endpoint Condition Cannot Be Dropped

Worked Example: A Convex Function Without a Supporting Line at an Endpoint

On $I=[0,1]$, let $f(y)=-\sqrt{y}$. First, this function is convex: for $x,y\in[0,1]$ and $t\in[0,1]$, $$ \big((1-t)\sqrt{x}+t\sqrt{y}\big)^2 \le (1-t)x+ty, $$ because the right side minus the left side is $t(1-t)(\sqrt{x}-\sqrt{y})^2\ge0$. Both sides before squaring are nonnegative, so taking square roots gives the concavity inequality for $\sqrt{\cdot}$; multiplying by $-1$ gives convexity of $-\sqrt{\cdot}$.

At the endpoint $x_0=0$, a supporting line with finite slope $c$ would require $-\sqrt{y}\ge cy$ for every $y\in[0,1]$. For $y>0$, division by $y$ would require $c\le-1/\sqrt{y}$. Given any finite $c$, choose $y=1/(|c|+2)^2$, which lies in $(0,1]$. Then $-1/\sqrt{y}=-(|c|+2)<c$, contradicting the required inequality. So no finite supporting slope exists at zero.

At the interior point $x_0=1/4$, however, the line with slope $-1$ is supporting. Its equation is \[ f(1/4)-\big(y-1/4\big)=-1/2-y+1/4=-y-1/4. \] For every $y\in[0,1]$, \[ -\sqrt{y}-(-y-1/4)=y+1/4-\sqrt{y}=(\sqrt{y}-1/2)^2\ge0. \] Thus the line lies below the function on the whole interval and meets it at $y=1/4$. This checks the inequality directly, including both endpoints.

This endpoint issue matters when the point of support is an average. In a finite weighted average with positive weights summing to one, if all inputs belong to $[a,b]$ and their average is $a$, then every input equals $a$: each input is at least $a$, and any input strictly larger than $a$ would make the weighted average strictly larger than $a$. The corresponding statement holds at $b$. Thus an argument using a supporting line at the average must either establish that the average is interior or treat an endpoint average separately. The theorem above makes no claim of a supporting line at an endpoint.

Using the Pointwise Inequality Under an Integral

Here is the precise way the supporting-line fact enters an integral argument. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $X:\Omega\to\mathbb{R}$ be a measurable, integrable real-valued function with $X\in I$ almost everywhere, and suppose $f(X)$ is integrable. If $a=\mathbb{E}[X]$ is an interior point of $I$, choose a supporting slope $c$ at $a$. The theorem gives, almost everywhere, \[ f(X)\ge f(a)+c(X-a). \] The right side is integrable because $X$ is integrable and $f(a),c,a$ are finite. Monotonicity and linearity of the Lebesgue integral therefore give \[ \mathbb{E}[f(X)]\ge f(a)+c\big(\mathbb{E}[X]-a\big)=f(a). \] This calculation illustrates the pointwise mechanism; the conditions on the mean and integrability are stated because the supporting-line theorem alone does not supply them.

If the mean is an endpoint of a closed interval and $X$ takes values in that interval almost everywhere, the supporting-line step may be unavailable. Instead, for example, if $X\ge a$ almost everywhere and $\mathbb{E}[X]=a$, then the nonnegative function $X-a$ has integral zero, so $X=a$ almost everywhere. In that case $f(X)=f(a)$ almost everywhere. The finite weighted version has the same endpoint phenomenon. The safe rule is to use an interior supporting line only when the point of support is actually interior.

A Common Mistake: Using a Chord as a Supporting Line

A chord joining two points of a convex graph is generally not a line lying below the whole graph. For $f(y)=y^2$ on $[0,2]$, the chord through $(0,0)$ and $(2,4)$ is $2y$. At $y=1$, it gives $2$, while $f(1)=1$, so the purported supporting inequality $y^2\ge2y$ fails: $1\ge2$ is false. The supporting line at $y=1$ instead has slope $2$ and equation $2y-1$; substitution gives $y^2-(2y-1)=(y-1)^2\ge0$ for every $y$. The chord is above the graph between its endpoints, while a supporting line from below must stay below the graph throughout the interval.

Check Your Understanding

For each question, identify which hypotheses are relevant before applying the supporting-line result.

For $f(y)=y^2+2y$ on $\mathbb{R}$, compute the supporting slope at $x_0=3$ and verify the inequality for every $y$.
For $f(y)=|y|$ at $x_0=0$, determine the full interval of supporting slopes directly from the defining inequality.
Explain why a finite convex function on $[a,b]$ need not have a supporting line at $a$, using $f(y)=-\sqrt{y}$ on $[0,1]$.
If a finite weighted average of points in $[a,b]$, with every weight positive, equals $a$, what must each point be? Justify the conclusion.
On a probability space, state the interior-point and integrability assumptions needed to integrate a supporting-line inequality at $a=\mathbb{E}[X]$.