4
$\begingroup$

I realize this may be a simple question, but I am having trouble proving proving this, mostly with selecting a suitable delta.

My attempt so far:

$x^2$ is continuous on $[0,1]$ if $\forall\epsilon>0,\exists\delta>0$ such that $$0<\vert x-a\vert<\delta\,x\in [0,1]\Rightarrow\vert x^2-a^2\vert<\epsilon$$ for all $a\in [0,1]$.

Since $a$ and $x$ are in $[0,1]$, the maximum value of $a+x$ is 2. Take $\delta=\min\{1,\frac{\epsilon}{2}\}$. Then $$0<\vert x-a\vert<\delta\Rightarrow\vert x^2-a^2\vert=\vert x-a\vert\vert x+a\vert<2\vert x-a\vert=2\delta\leq\epsilon$$

I think my logic makes sense but I am unsure about my step involving $\delta$. Is this a correct proof? If not, how would I improve it?

$\endgroup$
  • 1
    $\begingroup$ Looks good to me! Now I have a question for you: why take $\delta$ to be the minimum of $1$ and $\epsilon/2$? Why not just let it be $\epsilon/2$? $\endgroup$ – layman Oct 23 '15 at 4:57
  • 1
    $\begingroup$ In the end you need to get a strict inequality according to the definition So you might want to make $\delta = \epsilon /4 $ $\endgroup$ – happymath Oct 23 '15 at 4:58
  • $\begingroup$ This proof is perfect and is exactly how a continuity proof is supposed to go! My only thought is as $x^2$ is continuous on unbound R, can you do this without using the max value of a + x? $\endgroup$ – fleablood Oct 23 '15 at 5:04
  • $\begingroup$ Oooh, the "less than 2*...." should be a less than or equal. Very minor but as it need to be a strict inequality.... $\endgroup$ – fleablood Oct 23 '15 at 5:11
0
$\begingroup$

The only improvement I can see here (as mentioned by happymath in the comments), is that the implication requires strict inequality. So, choosing something like $\delta=\frac{\epsilon}{3}$ yields a better result. This gives the implication:

$$ 2\delta=\frac{2\epsilon}{3}<\epsilon$$

This is the required condition. Of course, as you probably know, $\delta=\frac{\epsilon}{3}$ is a somewhat arbitrary choice. You can choose anything less than $\frac{\epsilon}{2}$ (i.e. $\delta=\frac{\epsilon}{2.0000000001}$, or $\delta=\frac{\epsilon}{123456789}$ ). Or, more generally, as proposed in the comments, consider $\delta<\min(1,\frac{\epsilon}{2})$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Or he could simply say, "let $\delta <\min(1 , \epsilon/2)$" $\endgroup$ – fleablood Oct 23 '15 at 5:07
  • $\begingroup$ It doesn't hurt that $\delta$ is chosen as the minimum of $1$ and $\epsilon/2$, but how does it help in this case? Why not just take $\delta = \epsilon/2$? $\endgroup$ – layman Oct 23 '15 at 5:13
  • $\begingroup$ @fleablood Also valid. I'm just trying to underscore the somewhat arbitrary nature of the choice here. $\endgroup$ – Alekos Robotis Oct 23 '15 at 5:14
  • $\begingroup$ It is arbitrary. But as it's arbitrary it needn't actually be specified. As you point out is be be anything less (very true and a good point; we're basically saying the same thing) and as anything less can always be found, we only have to say "let it be less". $\endgroup$ – fleablood Oct 23 '15 at 5:23
  • 1
    $\begingroup$ I took $\delta=\min\{1,\epsilon/2\}$ because when I was initially working out the problem, I set $\delta=1$ and noted that $1<x-a<1\Rightarrow -(x+a)<x^2-a^2<x+a\leq2$, and then I modified my definition of $\delta$ from that. But I didnt have a firm idea of what I was doing $\endgroup$ – Matt G Oct 23 '15 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.