prove that the function, $f(x)=x^2$ for $x \in [0,1]$, is uniformly continuous

Question

Example 3.4.3: $f: [0, 1] \to \mathbb{R}$ defined by $f(x) := x^2$ is uniformly continuous.

Proof: Note that $0 \le x, c \le 1$. Then $$|x^2 - c^2| = |x + c||x - c| \le (|x| + |c|)|x - c| \le (1 + 1)|x - c|.$$ Therefore given $\varepsilon \gt 0$, let $\delta := \varepsilon/2$. Take $x \gt 0$ and let $c := x + \delta/2$. Write $$\varepsilon \gt |x^2 - c^2| = |x + c||x - c| = (2x + \delta/2)\delta/2 \ge \delta x.$$ Therefore $x \lt \varepsilon/\delta$ for all $x \gt 0$, which is a contradiction.

1. In this proof, I don't understand why the author chose $c=x + \frac {\delta}{2}$. Could you explain to me how $c$ is decided when proving this kind of statement?

2. Why is $x \lt \varepsilon / \delta$ a contradiction?

Thank you in advance!

Answer 1

Because we are trying to prove this is true for all of $\Bbb R$ we can choose any $c$ we want, there is no method to decide on which $c$ to take(although this is a classic example, so you probably won't find other choices online)

We have a contradiction because it is not possible for a constant to be greater than all numbers in $\Bbb R$, after all $\Bbb R$ is closed under addition, for example $x=\frac{\epsilon}{\delta}$ implies that $x\not< \epsilon/\delta$(how can it be equal and strictly less at the same time?)

So to prove those something is not forall you need to prove that at least one exists

So they proved that exists $\epsilon$ such that for all $\delta$ there exists $x$ and $c$ such that the requirement for the claim are false, in other words:

Not for all $\epsilon$ exists $\delta$ such that for all $x$ and $c$ the requirements for the claim hold

Answer 2

1) If $f$ were uniformly continuous, such a $\delta>0$ would meet the epsilon challenge for any $x\in \mathbb{R}$. $x=c+\frac{\delta}{2}$ is a choice of a particular $x$ which shows this is false (remember, not "for all" is "there exists one").

2) This is a contradiction as $\epsilon>0$ was fixed, and $\delta$ is supposedly a function of only $\epsilon$ (perhaps it would help to explicitly write $\delta(\epsilon)$). But $x$ can be anything on the real line, which is of course unbounded.

Answer 3

$x<\varepsilon/\delta$ for all $x>0$ is a contradiction because $\varepsilon/\delta$ is a constant. It's obviously false that all positive real numbers are less than some constant. As to why he chose $c=\delta/2$ all I can say is that it was a big enough difference to deduce a contradiction.

Uniform continuity says that the difference between function values is arbitrarily will be arbitrarily small, provided the points at which the function is evaluated are sufficiently close, and "sufficiently close" is the same over the entire domain. But that isn't true for the squaring function. For a given difference between the $x$ values, I can make the $y$ values as large as I like. $(x+\delta)^2-x^2=2\delta x +\delta^2$ Clearly I can make that bigger than any $\varepsilon$ you give me by taking $x$ large enough. The proof just translates this observation into the language of uniform continuity.