2018 Putnam B Problems/Problem 5

Problem

Let $f = (f_1, f_2)$ be a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ with continuous partial derivatives $\tfrac{\partial f_i}{\partial x_j}$ that are positive everywhere. Suppose that $\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left(\frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0$ everywhere. Prove that $f$ is one-to-one.

Solution 1

Let $ f = (f_1, f_2) : \mathbb{R}^2 \rightarrow \mathbb{R}^2 $ be a function with continuous partial derivatives, and suppose all the partials $ \frac{\partial f_i}{\partial x_j} $ are positive. The given inequality is $\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0$ everywhere on $ \mathbb{R}^2 $.

Let $ a = \frac{\partial f_1}{\partial x_1} $, $ b = \frac{\partial f_1}{\partial x_2} $, $ c = \frac{\partial f_2}{\partial x_1} $, $ d = \frac{\partial f_2}{\partial x_2} $. All of these are positive functions. The inequality becomes $ad - \frac{1}{4} (b+c)^2 > 0.$ This implies $ ad > \frac{1}{4} (b+c)^2 \geq bc $, using the AM–GM inequality: $ (b+c)^2 \geq 4bc $. So $ ad > bc $, and the Jacobian determinant $ \det Df = ad - bc > 0 $ everywhere. This shows that $ Df $ is invertible at all points, and by the inverse function theorem, $ f $ is a local diffeomorphism everywhere.

To show that $ f $ is globally one-to-one, apply the global inverse function theorem (Hadamard's theorem). This requires that $ f $ is $ C^1 $, its Jacobian is everywhere invertible (already established), and that $ f $ is proper, meaning the preimage of any compact set is compact.

Because all partial derivatives of $ f_1 $ and $ f_2 $ are positive, each component $ f_i $ is strictly increasing in each variable. If we fix $ x_2 $, then increasing $ x_1 $ increases both $ f_1 $ and $ f_2 $; similarly for increasing $ x_2 $. So if $ x_n \rightarrow \infty $ in any direction, then $ f(x_n) \rightarrow \infty $. This means that the image of unbounded sets is unbounded, and hence, the preimage of any bounded set must be bounded. Since $ f $ is continuous, preimages of compact sets are closed, so they are compact. This proves that $ f $ is proper.

Therefore, all conditions of Hadamard’s theorem are satisfied: $ f $ is $ C^1 $, has positive Jacobian determinant everywhere, and is proper.

So $ f $ is a global diffeomorphism, and in particular, one-to-one.

~Pinotation

Solution 2 (Easier)

We are given a function $ f = (f_1, f_2) : \mathbb{R}^2 \rightarrow \mathbb{R}^2 $ with continuous and positive partial derivatives, and the inequality $\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0$ everywhere.

We want to show that $ f $ is one-to-one, and we’ll do it by showing that if $ f(x) = f(y) $, then $ x = y $. We'll use an integral-based approach.

Let $ x = (x_1, x_2) $, $ y = (y_1, y_2) $, and suppose $ f(x) = f(y) $. Define a path $ \gamma(t) = (1 - t)x + ty $, for $ t \in [0, 1] $, which is a straight line from $ x $ to $ y $.

Then the difference $ f(y) - f(x) $ is $f(y) - f(x) = \int_0^1 \frac{d}{dt} f(\gamma(t)) \, dt.$

By the chain rule: $\frac{d}{dt} f(\gamma(t)) = Df(\gamma(t)) \cdot (y - x),$ so $f(y) - f(x) = \int_0^1 Df(\gamma(t)) \cdot (y - x) \, dt.$

Now suppose $ f(y) = f(x) $. Then the integral of a vector-valued function is zero: $\int_0^1 Df(\gamma(t)) \cdot (y - x) \, dt = 0.$

Let $ v = y - x $. Then we have: $\int_0^1 Df(\gamma(t)) \cdot v \, dt = 0.$

This means that the average of the Jacobian matrices along the path, acting on the vector $ v $, gives zero: $\left( \int_0^1 Df(\gamma(t)) \, dt \right) \cdot v = 0.$

Let’s define the matrix $ A = \int_0^1 Df(\gamma(t)) \, dt $. Since each entry of $ Df $ is continuous and positive everywhere, each entry of $ A $ is also positive. So $ A $ is a real $ 2 \times 2 $ matrix with positive entries.

Moreover, the inequality $\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0$ is preserved under averaging because the determinant is strictly positive at all $ t $, and the inequality is strict. So the matrix $ A $ satisfies the same inequality: $A_{11} A_{22} - \frac{1}{4} (A_{12} + A_{21})^2 > 0.$

This implies that the symmetric part of $ A $, $\frac{1}{2} (A + A^T),$ is positive definite. That means for all nonzero $ v $, $v^T A v > 0.$

But earlier we assumed that $ f(y) = f(x) $, so $ Av = 0 $, which implies $v^T A v = 0.$

This contradicts the fact that $ v^T A v > 0 $ for all nonzero $ v $. Therefore, $ v = 0 $, so $ y = x $.

Hence, $ f $ is one-to-one.

~Pinotation

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2018 Putnam B Problems/Problem 5

Problem

Solution 1

Solution 2 (Easier)