2000 USAMO Problems/Problem 1

Problem

Call a real-valued function $f$ very convex if

$\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|$

holds for all real numbers $x$ and $y$ . Prove that no very convex function exists.

Solution

Solution 1

Let $y \ge x$ , and substitute $a = x, 2b = y-x$ . Then a function is very convex if $\frac{f(a) + f(a+2b)}{2} \ge f(a + b) + 2b$ , or rearranging,

$\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4$

Let $g(a) = \frac{f(a+b) - f(a)}{b}$ , which is the slope of the secant between $(a,f(a))(a+b,f(a+b))$ . Let $b$ be arbitrarily small; then it follows that $g(a+b) - g(a) > 4$ , $g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4$ . Summing these inequalities yields $g(a+kb)-g(a) > 4k$ . As $k \rightarrow \infty$ (but $b << \frac{1}{k}$ , so $bk < \epsilon$ is still arbitrarily small), we have $\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty$ . This implies that in the vicinity of any $a$ , the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.

Solution 2

Suppose, for the sake of contradiction, that there exists a very convex function $f.$ Notice that $f(x)$ is convex if and only if $f(x) + C$ is convex, where $C$ is a constant. Thus, we may set $f(0) = 0$ for convenience.

Suppose that $f(1) = A$ and $f(-1) = B.$ By the very convex condition, $\frac{f(0) + f\left(2^{-n}\right)}{2} \ge f\left(2^{-(n+1)}\right) + \frac{1}{2^n}.$ A straightforward induction shows that: $f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}$ for all nonnegative integers $n.$ Using a similar line of reasoning as above, $f\left(-2^{-n}\right) \le \frac{B - 2n}{2^n}.$ Therefore, for every nonnegative integer $n,$ $f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.$ Now, we choose $n$ large enough such that $n > \frac{A+B}{4} - 1.$ This is always possible because $A$ and $B$ are fixed for any particular function $f.$ It follows that: $f\left(2^{-n}\right) + f\left(-2^{-n}\right) < \frac{1}{2^{n-2}}.$ However, by the very convex condition, $f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.$ This is a contradiction. It follows that no very convex function exists.

Solution 3

We consider the function $f(x, y) = \frac{x + y}{2} + |x - y|,$ defined on $\mathbb{R}^2$ . Define the graph of $f$ as the manifold $\mathcal{M} = \{ (x, y, z) \in \mathbb{R}^3 : z = f(x, y) \}.$

Claim: The manifold $\mathcal{M}$ is not the graph of a function $\mathbb{R} \to \mathbb{R}^2$ , i.e., it cannot be inverted to express $(x, y)$ as a function of $z$ .

Proof: We first observe that $f$ is symmetric: $f(x, y) = f(y, x),$ and convex, as it is the sum of a linear function and the convex function $|x - y|$ .

Consider the behavior of $f$ along the line $x = y$ . For $x > y$ , we have $f(x, y) = \frac{x + y}{2} + (x - y) = \frac{3x - y}{2},$ while for $x < y$ , we have $f(x, y) = \frac{x + y}{2} + (y - x) = \frac{3y - x}{2}.$ The partial derivatives with respect to $x$ and $y$ do not match along $x = y$ , creating a non-differentiable crease. Specifically, $\nabla f(x, y) = \begin{cases} \left( \frac{3}{2}, -\frac{1}{2} \right), & x > y, \\ \left( -\frac{1}{2}, \frac{3}{2} \right), & x < y. \end{cases}$ At $x = y$ , the gradient does not exist; the directional derivatives have a jump discontinuity, forming a kink along the line $x = y$ .

Suppose for contradiction that $\mathcal{M}$ is the graph of a function $g: \mathbb{R} \to \mathbb{R}^2$ such that $g(z) = (x, y)$ with $f(x, y) = z$ . Then each level set $L_z = \{ (x, y) \in \mathbb{R}^2 : f(x, y) = z \}$ must be the image of a unique point $g(z) \in \mathbb{R}^2$ .

However, since $f(x, y) = f(y, x)$ , the level sets are symmetric across the line $x = y$ . Furthermore, near any $z$ for which the level set intersects the kink $x = y$ , the gradient directions from either side differ, and the level set contains multiple points mapping to the same height $z$ . Therefore, the preimage $f^{-1}(z)$ contains at least two distinct points, violating the definition of a function $z \mapsto (x, y)$ .

Hence, $\mathcal{M}$ cannot be the graph of a function $\mathbb{R} \to \mathbb{R}^2$ . The surface folds along the kink, failing the vertical line test in the direction of $z$ .

$\boxed{\text{Therefore, } \mathcal{M} \text{ is a non-functional manifold.}} \quad \blacksquare$

~Lopkiloinm

2000 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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