Is $\cos(x^2)$ globally Lipschitz in x?

Question

I know that $\cos\left(x^2\right)$ is continuous and differentiable. I wonder whether I can conclude that $\cos\left(x^2\right)$ is also globally Lipschitz on $\mathbb{R}$?

Answer 1

For a differentiable function, the global Lipschitz constant is given by $\sup|f'(x)|$. In this case, $$\cos(x^2)' = -2x\sin(x^2).$$ But this expression is is unbounded as $|x|\to\infty$, take the sequence $(x_n)_{n\in\mathbb N}$ with $x_n = \sqrt{2\pi n + \frac \pi 2}$: $$\left|-2x_n\sin(x_n^2)\right| = \sqrt{2\pi n +\frac\pi 2}\cdot 1 \ge \sqrt n.$$ Hence the first derivative becomes arbitrarily large, and so does the lower bound for any Lipschitz constant. Since we are not able to find a Lipschitz constant, $f$ is not globally Lipschitz.

Answer 2

Let $f = \cos x^2$. For any $x, y ∈ ℝ$ with $x < y$, the mean value theorem for differential maps assures the existence of some intermediate value $ξ ∈ (x..y)$ with $$f'(ξ) = \frac{f(y) - f(x)}{x - y}, \quad\text{so}\quad |f(x) - f(y)| = |f'(ξ)||x - y|.$$ But $f' = -2x \sin x^2$, so …

Answer 3

Consider the sequence $a_n=\sqrt{\pi n/2}$.

Then $a_{n+1}-a_n\to 0$, but $|f(a_{n+1})-f(a_n)|=1$.

Intuitively, the slope between peaks and valleys grow arbitrarily when $x \to \infty$.