Convex Functions

This scene studies functions whose chords float above the graph. Convexity turns out to be the derivative increasing.

Picture a bowl shape opening upward. A curve whose slope keeps growing is convex. The middle height never exceeds the average of the end heights.

For example, stretch a 🧵 string taut between two points on a bowl-shaped curve and it always floats above the curve. The string's middle sits higher than the curve's middle.

f(tx+(1−t)y)≤tf(x)+(1−t)f(y)f(tx+(1-t)y) \le tf(x)+(1-t)f(y)
f convex   ⟺  f′ increasing,f′′≥0  ⟹  convexf \text{ convex } \iff f' \text{ increasing}, \quad f'' \ge 0 \implies \text{convex}

Part of these learning paths

Examples & Exercises

Prove directly that f(x)=x2f(x) = x^2 is convex.

(tx+(1−t)y)2≤tx2+(1−t)y2  ⟺  t(1−t)(x−y)2≥0(tx+(1-t)y)^2 \le tx^2+(1-t)y^2 \iff t(1-t)(x-y)^2 \ge 0
a=−1,b=1a=-1, b=1 — the chord floats above the parabola
Show proof

  1. Expanding the difference gives tx2+(1−t)y2−(tx+(1−t)y)2=t(1−t)(x−y)2tx^2+(1-t)y^2-(tx+(1-t)y)^2 = t(1-t)(x-y)^2. (Squaring (tx+(1−t)y)(tx+(1-t)y) and collecting terms factors t(1−t)t(1-t) out of the x2x^2 and y2y^2 coefficients.)

  2. Since t∈[0,1]t \in [0,1], we have t(1−t)≥0t(1-t) \ge 0, and (x−y)2≥0(x-y)^2 \ge 0 as a square. So the right-hand side is nonnegative.

  3. That is, (tx+(1−t)y)2≤tx2+(1−t)y2(tx+(1-t)y)^2 \le tx^2+(1-t)y^2 — the definition of convexity holds, so x2x^2 is convex. This is also why the max gap is 11 for a=−1,b=1a=-1, b=1 in the picture above.

Prove that if ff is differentiable and f′f' is increasing, then ff is convex.

f(x2)−f(x1)x2−x1=f′(c1)≤f′(c2)=f(x3)−f(x2)x3−x2\frac{f(x_2)-f(x_1)}{x_2-x_1} = f'(c_1) \le f'(c_2) = \frac{f(x_3)-f(x_2)}{x_3-x_2}
exe^x — an example with f′′>0f'' > 0
Show proof

  1. Take x1<x2<x3x_1 < x_2 < x_3.

    The mean value theorem on [x1,x2][x_1, x_2] and [x2,x3][x_2, x_3] gives c1∈(x1,x2)c_1 \in (x_1, x_2) and c2∈(x2,x3)c_2 \in (x_2, x_3) with f′(c1)=f(x2)−f(x1)x2−x1f'(c_1) = \frac{f(x_2)-f(x_1)}{x_2-x_1} and f′(c2)=f(x3)−f(x2)x3−x2f'(c_2) = \frac{f(x_3)-f(x_2)}{x_3-x_2}.

    Since c1<c2c_1 < c_2 and f′f' is increasing, f′(c1)≤f′(c2)f'(c_1) \le f'(c_2).

  2. Rearranging the slope inequality f(x2)−f(x1)x2−x1≤f(x3)−f(x2)x3−x2\frac{f(x_2)-f(x_1)}{x_2-x_1} \le \frac{f(x_3)-f(x_2)}{x_3-x_2} gives f(x2)≤(x3−x2)f(x1)+(x2−x1)f(x3)x3−x1f(x_2) \le \frac{(x_3-x_2)f(x_1)+(x_2-x_1)f(x_3)}{x_3-x_1} — the middle value lies below the chord joining the endpoints.

  3. With x2=tx1+(1−t)x3x_2 = tx_1+(1-t)x_3 we have x3−x2x3−x1=t\frac{x_3-x_2}{x_3-x_1} = t and x2−x1x3−x1=1−t\frac{x_2-x_1}{x_3-x_1} = 1-t, so the inequality becomes f(tx1+(1−t)x3)≤tf(x1)+(1−t)f(x3)f(tx_1+(1-t)x_3) \le tf(x_1)+(1-t)f(x_3) — the definition of convexity.

    The degenerate cases are trivial: t=0t = 0 or t=1t = 1 gives equality on both sides, and equal points (x1=x3x_1 = x_3) likewise leave only equality.

    Note that f′′≥0f'' \ge 0 makes f′f' increasing — under an extra hypothesis: if ff is twice differentiable (so f′f' is continuous and differentiable on the interior with derivative f′′f''), the mean value theorem applied to f′f' gives this.

Derive the AM–GM inequality for two numbers a,b>0a, b > 0 from the concavity of log.

log⁡a+b2≥log⁡a+log⁡b2=log⁡ab\log\frac{a+b}{2} \ge \frac{\log a + \log b}{2} = \log\sqrt{ab}
the chord picture applies to log just the same
Show proof

  1. log⁡\log is C2C^2 on (0,∞)(0, \infty), and (log⁡)′(x)=1x(\log)'(x) = \frac{1}{x} is decreasing (0<x1<x20 < x_1 < x_2 gives 1x1>1x2\frac{1}{x_1} > \frac{1}{x_2}). Problem 2's proof, with every inequality reversed, shows a function with decreasing derivative is concave — applying it to log⁡\log makes log⁡\log concave: the chord lies below the graph. (In fact C1C^1 with decreasing derivative suffices.)

  2. Jensen's inequality (concave version) with t=12t = \frac12 gives log⁡a+b2≥log⁡a+log⁡b2\log\frac{a+b}{2} \ge \frac{\log a + \log b}{2}.

  3. The right-hand side is 12(log⁡a+log⁡b)=log⁡ab\frac12(\log a + \log b) = \log\sqrt{ab}, so exponentiating gives a+b2≥ab\frac{a+b}{2} \ge \sqrt{ab} — the AM–GM inequality. (Exponentiation is increasing, so the inequality is preserved.)

10. Differentiation