1
$\begingroup$

I'm asked to analyze what happens when I have $\delta$ exchanged with $\epsilon$ in the limit definition like this:

$\forall \epsilon>0, \exists \delta>0 / |x-a|<\epsilon\implies |f(x)-L|<\delta$

I need to show that $f$ matches this condition $\iff$ it's bounded in any bounded interval of center $a$. In affirmative case, $L$ is real.

The limit definition is already difficult to me, now I can't understand this one.

Any ideas on how to prove it?

$\endgroup$
  • $\begingroup$ Shouldn't it be $\exists \delta <\infty$ instead of $>0$? That $\delta$ must be positive is implicit, as the RHS cannot hold if $\delta$ is negative $\endgroup$ – b00n heT Jul 1 '16 at 18:24
  • $\begingroup$ @b00nheT no, it should be $>0$. The fact that $\delta$ exists is to say it exists in $\mathbb{R}$ i.e. is finite. $\endgroup$ – fosho Jul 1 '16 at 18:26
  • $\begingroup$ By affirmative, do you mean converse? $\endgroup$ – fosho Jul 1 '16 at 18:50
1
$\begingroup$

We first prove the forward ($\Rightarrow$) direction.Suppose $f$ meets your definition. Fix an $\epsilon>0$. Then for any interval of length $\epsilon$ centered at $a$ i.e. $(a-\epsilon, a+\epsilon)$ you know that $|f(x) - L|<\delta$ where $\delta$ is some fixed positive number now that we have chosen an epsilon.

Now note that for $x\in(a-\epsilon, a+\epsilon)$ $$|f(x)| = |f(x)+L-L|\leq |f(x)-L|+|L|\leq \delta +L$$

Conversely, let $\epsilon$ be given. Now $(a-\epsilon, a+\epsilon)$ is a bounded interval centered at $a$. So by hypothesis, $f$ is bounded on this interval. This means that $\exists \delta>0$ such that $|f(x)|<\delta$. Since $L$ is any real number, we can let $L = 0$ so this is just the same thing as $|f(x)-L| <\delta$.

$\color{blue}{\text{Does this help?}}$

$\endgroup$
1
$\begingroup$

Remark: To me, if you allow $\infty$ then the above is a void statement, as it holds for any well defined function $f$, so I say $\delta<\infty$.

That said, the proof is quite easy.

Let $A$ be any bounded interval centered at $a$. Then there exists some $\epsilon_0$ such that $A\subset \{x\mid |x-a|<\epsilon_0\}$ but then there exists some $\delta(\epsilon_0)$ such that $$|f(x)-L|<\delta(\epsilon_0),$$ i.e. the function $f\big|_A$ is bounded by $L+\delta(\epsilon_0)<\infty.$

This same direction can also be proven as follows: Assume $f$ is unbounded in some bounded $\epsilon$ neighbourhood of $a$, call it $A$. Then $$|f(x)-L|\geq|f(x)|-|L|$$ is also unbounded, so there cannot exist any $\delta<\infty$ satisfying the condition for the given $\epsilon$.

The other direction is also not too difficult.

Let $\epsilon_0>0$ and let $B:=\{x\mid |x-a|<\epsilon_0\}$. Note that $B$ is bounded.Then as $f$ is bounded on $B$, by very definition there exists some $\delta(\epsilon_0)$ such that $$f(B)\subset (-\delta(\epsilon_0),\delta(\epsilon_0)).$$ Now choose (for example) $L=0$ and you are done, as you have found, for any given $\epsilon$

$\endgroup$
  • $\begingroup$ When OP says 'affirmative', does this mean 'the converse'? $\endgroup$ – fosho Jul 1 '16 at 18:51
  • 1
    $\begingroup$ Also '$\exists \delta$' disallows $\infty$ in any case. $\endgroup$ – fosho Jul 1 '16 at 18:57
  • $\begingroup$ To be honest, I really am not sure what exactly was meant by that... In the end you may pick any real $L$ and it holds... $\endgroup$ – b00n heT Jul 1 '16 at 19:01

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.