Uniform Continuity of $x^2$.

Question

Assume all the standards, such as $f:\mathbb R→\mathbb R$ and all that other jargon. I guess it doesn't really matter, but my main question is why isn't $f(x)=x^2$ uniformly continuous. I know if we restrict the domain to some subset $D\subset \mathbb R$, then $f$ is uniformly continuous on $D$, but not on the entire domain of $\mathbb R$.

Answer 1

There are a lot of interpretations of what uniform continuity is. That's what makes it so useful, similar to how a matrix is useful due to the many interpretations (e.g. linear transformation, 2D array of information, vector of vectors, etc.).

One of the best analytical interpretations of uniform continuity can be found within the definition of continuity. For a function $f$ to be continuous at a point $x_0$, you obviously need to fix $\epsilon > 0$ and find (or prove there exists) a specific $\delta>0$ where $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$. In other words, you must find (or prove there exists) a special function $N:(0,\infty) \times D \rightarrow (0,\infty) $, where $N(\epsilon,x_0)=\delta$. This function (often called the modulus) maps any $\epsilon$ and $x_0$ to some $\delta$ that satisfies that continuity predicate. For example, in order to prove $x^2$ is continuous at $x_0$, you are (whether you like it or not) forced to find (or prove there exists) such a function. I think for $x^2$, we can say $N(\epsilon,x_0)=\min(1,\frac{\epsilon}{1+2|_0|})$. The modulus exists, so $x^2$ is continuous. Yay.

But what about uniform continuity? Note that $N$ is a function that depends on $\epsilon$ and $x_0$. If you vary $x_0$, the value of $N$ will change. If you can find a modulus that does not depend on $x_0$, as in if you vary $x_0$, then the value of $N$ doesn't change, then your function is uniformly continuous. You cannot do this for $x^2$. Any modulus you find, you will see that $x_0$ crops up somewhere in the definition (like the example I wrote). Showing this is true is kinda difficult obviously. You usually have to use a proof by contradiction.

I like to interpret moduli as a sort of quantization of continuity. Some functions are more "continuous" than others. How can you measure continuity? The modulus serves as a good way of doing that.

For a more geometric interpretation of uniform continuity means, I suggest checking out the wikipedia page on it.

Answer 2

Basically because the slope of your curve is unbounded as $|x|\to\infty$. Given, say, $\epsilon=1$, at $x=0$ the corresponding $\delta$ in the limit definition is just 1 but at $x=2$ such $\delta$ is not good enough and you need say $\delta=0.2$ which in turn is not enough at $x=10$, etc. This lack of a universal $\delta$ is the reason why your function is not uniformly continuous.

Answer 3

How do you show that a function $f:\mathbb R→\mathbb R$ is not uniformly continuous?

Since you have such a 'clean' domain, there is an easy (necessary) test that must be passed in order for $f$ to be uniformly continuous.

Define a sequence $a_n = f(n) - f(n-\frac{1}{n})$.
If the function $f$ is uniformly continuous then the sequence $a_n$ converges to $0$.

For $f(x) = x^2$,

$\tag 1 a_n = n^2 - (n - \frac{1}{n})^2 = 2 - \frac{1}{n^2} $

and therefore

$\tag 2 {\displaystyle \lim _{n\to +\infty}} a_n = 2$

We conclude that $f(x) = x^2$ is not uniformly continuous.