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

1,240

Solution 1

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.

Solution 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.

Share:
1,240

Related videos on Youtube

student_t
Author by

student_t

Updated on August 01, 2022

Comments

  • student_t
    student_t over 1 year

    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.

  • student_t
    student_t over 6 years
    Is this the same idea as finding a convergent sequence $x_n \rightarrow x_o$ so that $f(x_n) \not\rightarrow f(x_o)$?
  • Shashi
    Shashi over 6 years
    @danny no, that only works when you prove discontinuous. You might have a function that is continuous but not uniform continuous. So what you have said won't work.
  • Shashi
    Shashi over 6 years
    @danny I must add that it works when the function is discontinuous somewhere, because uniform continuity implies continuity. What I wanted to say is that it won't give a conclusion when the function is continuous.
  • student_t
    student_t over 6 years
    This makes a lot of sense!
  • student_t
    student_t over 6 years
    yes that makes sense. Thank you.