[Math] f(x)=cos(x^2) is not uniformly continuous

uniform-continuity

Prove that $f(x)=cos(x^2)$ is not uniformly continuous on the set R.

I started out by solving $cos(x_k^2)$=1⟹cosX=1 ⟹ $X=2_k\pi$
$x_k^2$=$2k\pi$=>$x_k$⟹$\sqrt(2k\pi)$ -> goes to infinity

$cost_k^2$=-1=>$t_k^2$=$k\pi$⟹$t_k$=$\sqrt(k\pi)$ goes to 0 as k goes to infinity

  1. show difference of $\|x_k-t_k|$->0 but $|f(x_k)-f(t_k)|$=2≠0.

$\sqrt(2k\pi)$–$\sqrt(k\pi)$->0 when k->infinity.

Then we are told to conjugate (a+b)(a-b)=$a^2-b^2$.

So I got $\frac{k\pi}{\sqrt(2k\pi)+\sqrt(k\pi)}$

This is where I am confused on how to relate this back to showing it is not uniformly continuous.

Best Answer

Think about the definition of uniform continuity:

$f$ is uniformly continuous if for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x$ and $y$, if $|x-y| < \delta$, then $|f(x) - f(y)| < \epsilon$.

When you negate such a statement, the quantifiers flip. So:

$f$ is not uniformly continuous if there exists an $\epsilon > 0$ such that for all $\delta > 0$ there exist $x$ and $y$ such that $|x-y| < \delta$, but $|f(x) - f(y)| \geq \epsilon$.

Now look at a graph of the function:

graph of y=cos(^2)

You see that the function oscillates between $1$ and $-1$, but at an increasing frequency the further away from zero you go. That's because $f'(x) = 2x \sin(x^2)$.

Thinking about it another way, the horizontal distance between each consecutive peak and valley shrinks the larger $x$ gets, and in fact tends to $0$ as $x \to \infty$. So we can find points $x$ and $y$ arbitrarily close together such that $f(x) = 1$ and $f(y) = -1$. That is what you're trying to find with all your $x_k$'s and $t_k$'s.

Related Question