Wikipedia's proof that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous

Question

Prove that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous

Now I actually ask this question, as Wikipedia's proof of this seems wrong to me.

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Wikipedia's Proof:

$f$ is uniformly continuous on $\mathbb{R}$ if $\forall \epsilon > 0$, there exists a $\delta > 0$ for all $x_1, x_0 \in \mathbb{R}$ such that

$$|x_1 - x_0| < \delta \implies |f(x_1) -f(x_0)| < \epsilon$$

But $$\begin{aligned} f(x+ \delta) - f(x) &= x^2 + 2x\delta +\delta^2 - x^2\\ &= 2x\delta + \delta^2\\ &= \delta(2x + \delta)\\ &> \epsilon \ \ \ \text{for suffiently large $x$} \ \ \ \square \end{aligned}$$

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Now what I don't get is that if we set $x_1 = x+ \delta$ and $x_0 = x$, then $|x_1 -x_0| = |x+ \delta - x| = |\delta| \not< \delta$, so everything above should not apply as $|x+ \delta - x| \not < \delta$ in the first place.

Is Wikipedia's proof correct? If so, then why would my argument above be incorrect?

Answer 1

You can replace the $|x - y| < \delta$ with $|x - y | \leq \delta$ in the definition of uniform continuity and not fundamentally change anything. However, if it makes you feel comfier, you can fix it up by choosing a $\rho < \delta$ and proceed exactly as shown, but with $f(x+ \rho) - f(x)$.

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We can establish the equivalence of the two slightly different definitions as follows:

First, if there exists a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$, then there exists a $\rho$ such that $|x - y| \leq \rho \implies |f(x) - f(y)| < \varepsilon$. For example, $\rho = \delta / 2$.

On the other hand, if there exists a $\delta$ such that $|x - y| \leq \delta \implies |f(x) - f(y)| < \varepsilon$, then of course it's also true that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

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One can also change the $< \varepsilon$ condition to $\leq \varepsilon$. To establish this equivalence, first suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)|< \varepsilon$. It follows immediately that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$.

On the other hand, suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$. By choosing a $\delta$ that satisfies this condition for some $\varepsilon' < \varepsilon$, we thus have a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

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I suspect the reason why the strict less-than condition for both $\varepsilon$ and $\delta$ is prefered in analysis texts is because this emphasizes the more general topological definition of continuity. Namely, the preimage of an open set is open.

Answer 2

You miss an $x$ in $\delta(2+\delta)$. It should be, as the Wikipedia article shows:

But $$ {\displaystyle f(x+\delta )-f(x)=2x\delta +\delta ^{2}=\delta (2\color{blue}{x}+\delta )\ ,}\tag{1} $$ and for all sufficiently large $x$ this quantity is greater than $\epsilon$ .

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[Added after OP put the $x$ back.]

The function $f$ being not uniformly continuous is the same as the following:

there exist $\epsilon>0$ such that for any $\delta>0$ there exist $x_1,x_2\in\mathbb{R}$ with $|x_1-x_2|<2\delta$ and $$ |f(x_1)-f(x_2)|\geq \epsilon $$

Answer 3

If we put

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

$$x_n=\sqrt{n+1},$$

and $$y_n=\sqrt{n}.$$

then we have

$$\lim_{n\to+\infty}(x_n-y_n)=0$$

but

$$|f(x_n)-f(y_n)|=1$$

So, $\;\;f\;\;$is not uniformly continuous at $[0,+\infty)$.

If we take $\epsilon=0.5$ for example,

then $\forall \eta>0$,

we will always find $x_n$ and $ y_n$ such that

$|x_n-y_n|<\eta$ and

$|f(x_n)-f(y_n)|=1>\epsilon$.