3
$\begingroup$

True of false: $|f(x)-g(x)|<\epsilon$ $\forall x\in I$ $\Rightarrow$ $|\sup(f(x))-\sup(g(x))|\leq \epsilon$

I actually have an idea for a proof in case it is correct, but just to make sure what I'll write isn't rubbish I would like to get a confirmation from you all.

Intuitively, it seems logical to me.

If anyone has a counter example I would love to see it.

Thanks a lot.

$\endgroup$
  • 4
    $\begingroup$ As long as the supremums are finite, then the statement is OK. Perhaps you should include a sketch of your argument so that this question is more useful to others. $\endgroup$ – Michael Burr May 25 '19 at 18:02
  • $\begingroup$ I just dont know if the first part is about the same x for f and g , and the second part isnt $\endgroup$ – Milan May 25 '19 at 18:23
  • $\begingroup$ What is $I$? I assume an interval. Bounded, unbounded, closed? $\endgroup$ – max_zorn May 26 '19 at 0:26
1
$\begingroup$

Let $a=\sup (f(x))$ and $b=\sup(g(x))$. Then, for all $\delta>0$, we have that there is some $x_0\in I$ such that $ f(x_0)\geq a-\delta$. Since $|f(x_0)-g(x_0)|<\epsilon$. Then, $$a-\delta-\epsilon\leq f(x_0)-\epsilon< g(x_0)\leq b$$ So, $a-\delta-\epsilon< b$. Since this inequality works for any $\delta>0$, by taking $\delta\to 0^+$ we conclude $a-\epsilon\leq b$.

Similarly, $b-\epsilon\leq a$. So, we conclude $|b-a|\leq \epsilon$.

$\endgroup$
  • $\begingroup$ Sorry for the unintended downvote, it was accidental. I can & will undo it when the answer will get an edit. $\endgroup$ – Hanno May 25 '19 at 21:35
  • $\begingroup$ @Hanno I see, I wondered why I got+8. $\endgroup$ – Julian Mejia May 26 '19 at 14:46
1
$\begingroup$

A simple explanation :

If the distance between any two given values of $f(x)$ and $g(x)$ is smaller than $\epsilon$, then the distance between their supremums can follow as such as well. Of course, that only fails in the case of either one (or both) of the supremums being infinite.

So, yes, what you write isn't wrong. You may continue with your desired statement/proof.

$\endgroup$
  • $\begingroup$ I'm not sure that the OP intends for $\epsilon$ to be universally quantified, so I would be careful about the phrase "arbitrarily small" $\endgroup$ – Michael Burr May 25 '19 at 18:07
  • 1
    $\begingroup$ @MichaelBurr Smaller than would do then, thought the opposite. $\endgroup$ – Rebellos May 25 '19 at 18:09

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