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?

  • $\begingroup$ 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

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 $|f(x)-L| <\delta$.

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


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 $\epsilon$

  • $\begingroup$ When OP says 'affirmative', does this mean 'the converse'? $\endgroup$ – fosho Jul 1 '16 at 18:51
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    $\begingroup$ 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

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