Tips for proving a $f(x)$ is not uniformly continuous

Question

I've recently been introduced to uniform continuity and I will be asked to disprove that several functions are not uniformly continuous. It seems the trick is to play around with $|x-y| < \delta$ and $|f(x) - f(y)| < \varepsilon$ to arrive at a contradiction.

I was wondering if there were any other general tips/strategies. For instance, one that I have seen is to choose $y = x + \frac{\delta}{2}$ and then pick $x$ accordingly to arrive at a contradiction. And that it is usually convenient to restrict $\delta$ to be less than one because for any $\delta' \geq 1, |x - y| < \delta < \delta' $ the inequality will still hold.

Any other useful tricks or intuition that people have picked up would be greatly appreciated.

Answer 1

Find some particular $\epsilon >0$ and construct sequences $x_n$ and $y_n$ in the domain of the function $f$ that satisfy $|x_n-y_n|\leq \frac 1 n$ and $|f(x_n) - f(y_n) |\geq \epsilon$

Answer 2

The process of finding a proof typically goes in the reverse direction from the proof you eventually write down. You start with the conclusion you want to reach (in this case, that you have some $\varepsilon > 0$ and for all $\delta > 0$ there are $x$ and $y$ with $|x - y| < \delta$ but $|f(x) - f(y)| > \varepsilon$), and you try to satisfy this. Informally, you ask yourself what points on the graph of $f$ will be separated by at least some fixed positive distance in the vertical direction but very close in the horizontal direction.

Answer 3

You can use the following equivalences:

1) $f$ is not uniformly continuous on $\Omega$.

2) There is an $\epsilon > 0$ such that for all $\delta > 0$ exits points $\mathbf{x}, \mathbf{y} \in \Omega$ sucht that $||\mathbf{x} - \mathbf{y}|| < \delta$ but $||f(\mathbf{x}) - f(\mathbf{y})|| > \epsilon$.

3) There is $\epsilon > 0$ and two sequences $(\mathbf{x}_k)$ and $(\mathbf{y}_k)$ in $\Omega$ such that $\lim_{k \to \infty} (\mathbf{x}_k - \mathbf{y}_k) = 0$ and $||f(\mathbf{x}_k) - f(\mathbf{y}_k)|| > \epsilon$ for all $k \in \mathbb{N}$.

Example:

$f(x) = 1/x$ is not uniformly continuous on $\mathbb{R}^+$. Let $0 < \epsilon < 1$ and consider the sequences $(1/k)$ and $(1/(k+1))$. Then $$\lim_{k \to \infty} (1/k) - (1/(k+1)) = 0$$ but $(1/(k+1)) - (1/k) = k + 1 - k = 1 > \epsilon$ for all $k \in \mathbb{N}$.

Answer 4

Let me show one of my favourite ways with an example. Let's show that $f(x)=x^2$ is not uniformly continuous on $[0,+\infty)$. We know that it is uniformly continuous in any compact set in $[0,+\infty)$, so the ''problem'' arises ''at infinity''.

That's the intuition. To actually prove that (using the intuition), fix $\varepsilon = 1$ and show there is no $\delta$ such that $|x^2-y^2| \le \varepsilon$ whenever $|x-y| \le \delta$.

Let's then fix $\delta >0$ and $x,y \ge 0$ such that $x-y = \frac{\delta}{2}$. We want to ''catch'' the problem ''at infinity'', so let's ''send'' $x$ and $y$ to infinity by taking $y= \frac{2}{\delta}$. Then we have

$$|x^2-y^2| = (x+y)(x-y) =1+ \frac{\delta^2}{4} > 1 = \varepsilon.$$

Of course we may have made different choices, such as $y=\frac{1}{\delta^2}$ or something more complicated. That really depends on the form of the function you are studying.