Using the chord test, why is f(x) = x^2 convex but f(x) = x^3 not?

## The chord (secant) definition Let `D` be a convex set. A function `f : D -> R` is **convex** if for all `x, y` in `D` and all `t` in `[0, 1]`, ``` f(t*x + (1 - t)*y) <= t*f(x) + (1 - t)*f(y) ``` The left side is the function evaluated at a point on the segment between `x` and `y`. The right side is the height of the straight **chord** joining `(x, f(x))` to `(y, f(y))`, evaluated at that same point. So the condition reads: **the chord lies on or above the graph**. A function is **concave** when the reverse inequality holds — its chords lie on or below the graph — and `f` is concave exactly when `-f` is convex. Note what the definition does *not* require: no derivative, no continuity assumption baked in, no particular shape. The chord test is the primitive; everything else is a consequence or a convenient sufficient condition. ## Strict convexity `f` is **strictly convex** if the inequality is strict whenever `x != y` and `t` is in the open interval `(0, 1)`. Then the chord touches the graph only at its two endpoints. `f(x) = x^2` is strictly convex. An affine function `f(x) = a*x + b` is convex *and* concave, since the chord coincides with the graph — it is convex but not strictly so. `f(x) = |x|` is convex but not strictly convex, because it is affine on each side of the origin, so a chord from `1` to `2` lies exactly on the graph. ## Working the two examples **Squaring.** For `f(x) = x^2`, expand the difference between the chord and the graph: ``` t*x^2 + (1-t)*y^2 - (t*x + (1-t)*y)^2 = t*(1-t)*(x - y)^2 ``` Since `t*(1-t) >= 0` on `[0, 1]` and `(x - y)^2 >= 0`, the chord minus the graph is non-negative for every pair, and strictly positive when `x != y` and `0 < t < 1`. That is a complete proof of strict convexity with no calculus at all. **Cubing.** For `f(x) = x^3` on the whole real line, one counterexample settles it. Take `x = -2`, `y = 0`, `t = 0.5`. The convex combination is the midpoint `-1`, and ``` f(-1) = -1 chord value at -1 = 0.5*f(-2) + 0.5*f(0) = 0.5*(-8) + 0 = -4 ``` Convexity demands `f(midpoint) <= chord value`, i.e. `-1 <= -4`, which is false. The graph rises above its own chord on the negative half-line — cubing is concave there. Restricted to `x >= 0`, `x^3` is convex, which is a reminder that convexity is a statement about a function **on a specified domain**, not an intrinsic label on a formula. A cautionary note about the midpoint you might be tempted to try first: with `x = -1`, `y = 1`, `t = 0.5`, the midpoint value `f(0) = 0` exactly equals the chord value `0`. That single pair does not violate anything, so a hasty check with a symmetric pair proves nothing. You need a witness that actually breaks the inequality. ## Midpoint convexity Checking only `t = 1/2` — `f((x + y)/2) <= (f(x) + f(y))/2` — is called **midpoint convexity**. It is weaker in general, but for a continuous function (indeed for any measurable one) midpoint convexity implies full convexity, because dyadic values of `t` are dense in `[0, 1]` and continuity fills in the rest. Pathological counterexamples exist only among non-measurable functions built with the axiom of choice, so in practice the midpoint check is a sound quick test — provided you actually verify continuity. ## Why the chord form is the one to memorise Second-derivative tests are more convenient when a function happens to be twice differentiable, but they say nothing about functions with corners, and they can only be applied pointwise. The chord inequality applies to every convex function, differentiable or not, and it is the form used to prove that convexity is preserved by non-negative weighted sums, by composition with an affine map, and by taking pointwise maxima. If you remember one thing about convexity, remember that the chord sits on top.