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I'm trying to prove the continuity of $f(x)=x^2$ using my book's definition of continuity and the distance function $d(x,a)=|x-a|$.

In order to prove that, I must prove that given $\epsilon \gt 0$, there exists $\delta$ defined in terms of $\epsilon$ such that $$|x-a|\lt\delta$$ implies $$|x^2-a^2|\lt\epsilon$$ I got so far as to "unsimplify" the second inequality to get $$|x+a||x-a|\lt \epsilon$$ But I can't figure out how to define $\delta$ in order to complete the proof. I can't use $\delta=\sqrt\epsilon$, because $a$ could be arbitrarily large, meaning that $|x+a|$ could be much larger than $|x-a|$. And I'm not allowed to define $\delta$ in terms of $x$ or $a$, either.

Any ideas?

Ivo Terek
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Franklin Pezzuti Dyer
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    $\delta$ is a function of $\epsilon$ and $a$, except if we are talking of uniform continuity. – Masacroso Jul 11 '17 at 15:09
  • @Masacroso Perhaps I should have specified that, then... my book doesn't differentiate between the two anywhere, and it has never used $\delta$ as a function of $a$. – Franklin Pezzuti Dyer Jul 11 '17 at 15:10
  • This https://math.stackexchange.com/a/2124026/135775 will hep – Naive Jul 11 '17 at 15:11
  • @Nilknarf I know. Some books of analysis are not enough explicit about the definition of continuity when it shows for the first time. I only read one book that was completely explicit about $\delta$ as a function of $\epsilon$ and the point around we want to study the continuity of $f$. In many books this relation is assumed to be understood implicitly, what IMO is a mistake, especially for someone trying to understand the topic for the first time. – Masacroso Jul 11 '17 at 15:20

4 Answers4

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Just force a bound on $\delta$. If $\delta < 1$, then $|x|-|a| \leq |x-a| < 1$ implies $|x|<1+|a|$. So $$|x^2-a^2| = |x+a||x-a| < (|x|+|a|)\delta \color{red}{<} (2|a|+1)\delta$$tells us that $\delta \doteq \min\{ 1, \epsilon/(2|a|+1) \}$ fits the bill. My answer here might be useful too.

Ivo Terek
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  • Why is the $min$ required? Why not just $$\delta=\frac{\epsilon}{2|a|+1}$$? – Franklin Pezzuti Dyer Jul 11 '17 at 15:53
  • Because if this happens to be greater than $1$, the step pointed in red (right now) might fail. – Ivo Terek Jul 11 '17 at 15:55
  • So if I wanted to prove the continuity of $x^3+x^2$, I could use $$\delta=\operatorname{min}\bigg(1,\sqrt{\frac{\epsilon}{2|a|+1}}\bigg)$$ ? – Franklin Pezzuti Dyer Jul 11 '17 at 17:16
  • I think it is more complicated than that. $$|x^3+x^2 - a^3-a^2| \leq |x^3-a^3| + |x^2-a^2| \leq (|x^2|+|a||x|+|a|^2 + |x|+|a|)|x-a|,$$and the bound $\delta < 1$ would give something of the form $\delta = \min{1, \epsilon/p(|a|)}$, where $p$ is some polynomial. – Ivo Terek Jul 11 '17 at 17:19
  • When dealing with polynomials you can always avoid taking roots of $\epsilon$. – Ivo Terek Jul 11 '17 at 17:20
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As a first step, take $\Delta=1$. Then$$|x-a|<\Delta\iff|x-a|<1\Longrightarrow|x+a|=|x-a+2a|\leqslant1+2|a|.$$Now, take $\delta=\min\left\{1,\frac\epsilon{1+2|a|}\right\}$. Then$$|x-a|<\delta\Longrightarrow|x^2-a^2|=|x+a|.|x-a|<(1+2|a|)\frac\epsilon{1+2|a|}=\epsilon.$$

José Carlos Santos
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From Stephen Abbott's Understanding Analysis,

Let's show $$ \lim_{x \to 2} g(x) = 4 $$ where $g(x) = x^2$. ... a little algebra reveals $$ |g(x) -4| = |x^2 -4| = |x+2||x-2|. $$ ...we need an upper bound on $|x+2|$ in order to know how small to choose $\delta$. If we agree that our $\delta$-neighborhood around $c = 2$ must have radius no bigger than $\delta = 1$, then we get the upper bound $|x+2| \leq |3 + 2| = 5$ for all [$x \in (c - \delta, c+\delta)$].

Now choose $\delta = \min\left( 1, \epsilon/5 \right)$.

This choice baffled me at first, but the idea is to choose a way to bound $|x + 2|$ so that it won't get too large. The proof finishes with that $|x-2| < \delta$, we obtain $|x^2 - 4| < \epsilon$.

You can use a similar idea by saying, perhaps, that $|x + a| \leq |2a + 1|$ and choosing $\delta = \min \left(1, \frac{\epsilon}{|2a+1|} \right)$. There may be some tweaks needed here.

(I'm aware that the above passage asks for a limit, but it applies to continuity as well.)

Sean Roberson
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f(x) = x2 is only a uniformly continuous function if it is restricted to a bounded interval- by which I mean, it is impossible to find δ in terms of ϵ such that for all x, d(x, a) < δ implies d(x2, a2) < ϵ. This is because the slope of the x2 function increases boundlessly as x tends towards plus or minus infinity. If you fix ϵ and δ dependent on ϵ, for large (or small) enough x, the rate of change of x2 will be high enough that the implication d(x, a) < δ ==> d(x2, a2) < ϵ does not hold.

It is possible to do this if your domain is bounded, however. Say we are examining the function on the closed interval [c,d] (if your domain is bounded but not a closed interval, take some closed interval [c,d] that contains your domain). Let ϵ > 0. The absolute value of the derivative |f'(x)| = 2|x| is continuous on [c,d] and thus attains its maximum at some point x' in [c,d]. Take δ < min{1,ϵ/(1+2|x'|)}. As noted by previous answers, this δ works at x'. Moreover, because 2|x'| > 2|a| for all other a in the interval [c,d] (by our choice of x'), we have that δ < min{1,ϵ/(1+2|x'|)} < min{1,ϵ/(1+2|a|)} for all a in [c,d]. Thus the choice of δ works everywhere on our domain, and is dependent only on ϵ and a constant x'.

Olivia Strahan
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