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I'm working on the following problem:

A real valued function $f$ defined on $\mathbb{R}$ has the property that $(\forall \epsilon>0)(\exists\delta>0)$ s.t. $|x-1| \geq \delta \implies |f(x)-f(1)| \geq \epsilon$

The choices are either: $f$ is unbounded, $\lim_{|x|\rightarrow \infty}|f(x)|=\infty$, or $\int^{\infty}_0|f(x)|dx=\infty$.

The key says the answer is $\lim_{|x|\rightarrow \infty}|f(x)|=\infty$. I'm having trouble seeing this immediately though. In fact I'm also unable to distinguish it from $f$ is unbounded - doesn't the second choice imply the first?

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    $\begingroup$ The second choice does not imply the first (take $f (x)=1$). The first statement does imply $f$ is unbounded. The $\epsilon$ does not seem to have a role in what you have written. Are your statements as intended? $\endgroup$ – AnyAD Sep 30 '18 at 22:48
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    $\begingroup$ 1. How does $f = 1$ work? The second choice is $\lim_{|x|\rightarrow \infty}|f(x)|=\infty$, but $\lim_{|x|\rightarrow \infty}|1|=1$. 2. Indeed, the statements are intended. It is problem 60 on this practice GRE - rambotutoring.com/GR1268.pdf $\endgroup$ – yoshi Sep 30 '18 at 22:56
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    $\begingroup$ I guess the second delta should actually be an epsilon in your statement. $\endgroup$ – Alonso Delfín Sep 30 '18 at 23:14
  • $\begingroup$ YES! Sorry, someone just edited $\endgroup$ – yoshi Sep 30 '18 at 23:23
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    $\begingroup$ @yoshi I think there was a misunderstanding (on my part) of which statement you meant for 'second choice' $\endgroup$ – AnyAD Sep 30 '18 at 23:37
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Given $\epsilon >0$ there exists $\delta >0$ such that

$$|x-1| \geq \delta \implies |f(x)-f(1)| \geq \epsilon.$$

In other words

$$x\in (-\infty,1-\delta)\cup (1+\delta,\infty)\implies f(x)\in (-\infty,f(1)-\epsilon)\cup (f(1)+\epsilon,\infty).$$

As a consequence we get

$$|x|\ge 1+\delta \implies |f(x)|\ge \min\{|f(1)-\epsilon|,|f(1)+\epsilon|\}.$$

This shows that $$\lim_{|x|\to \infty} |f(x)|=\infty.$$

I'm having trouble seeing this immediately though. In fact I'm also unable to distinguish it from $f$ is unbounded - doesn't the second choice imply the first?

Note that $f(x)=e^x$ is unbounded but $\lim_{|x|\to \infty} |f(x)|$ doesn't exist. So both statements are not the same. Of course $\lim_{|x|\to \infty} |f(x)|=\infty$ implies that $f$ is unbounded.

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    $\begingroup$ I don't know if you read the comments, but you agree also that the second choice implies the first, right? $\endgroup$ – Ovi Sep 30 '18 at 23:34
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    $\begingroup$ @Ovi $\lim_{|x|\to \infty} |f(x)|=\infty$ implies that $f$ is unbounded. But the first stamement gives much more information about the behaviour of $f.$ $\endgroup$ – mfl Sep 30 '18 at 23:37
  • $\begingroup$ okay so $\lim_{|x|\rightarrow \infty}|f(x)| = \infty$ is specifying a type of way to be unbounded. Namely, both ends need to go to positive infinity? Where as unbounded could mean both directions? $\endgroup$ – yoshi Sep 30 '18 at 23:38
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    $\begingroup$ Consider $f(x)=\tan x$ in the domain of $\tan$ and $0$ otherwise. It is unbounded but $\lim_{|x|\rightarrow \infty}|f(x)| $ doesn't exist. You need $\lim_{x\to \pm \infty} f(x)=\pm \infty.$ $\endgroup$ – mfl Sep 30 '18 at 23:41
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The second choice implies the first, but the first choice does not imply the second. Consider $f(x)=1/x$. Then $f$ is unbounded but $\lim_{|x|\to \infty} f(x)=0$.

The problem statement can be interpreted as: ``For any $\epsilon>0$, there is a distance $\delta$ so that all $x$ further than $\delta$ from $1$ have value at least distance $\epsilon$ from $f(1)$.

Notice that $f(x)=1/x$ is an unbounded function, but does not satisfy the given property which you can see by taking $\epsilon=2$.

On the other hand, notice that for any $\epsilon>0$ there is $\delta$ so that

$$\epsilon<|f(x)-f(1)| \leq |f(x)|+|f(1)|$$

for all $|x-1|\geq \delta$. That is to say, $|f(x)|\geq \epsilon- |f(1)|$ whenever $|x-1|\geq \delta$. Thus, $\lim_{|x|\to \infty} |f(x)|=\infty$. On the other hand, if $\lim_{|x|\to \infty} |f(x)|=\infty$ then given any $\epsilon$ there is $N$ sufficiently large so that if $|x|\geq N+1$ then $|f(x)|\geq \epsilon+|f(1)|$. Using the reverse triangle inequality you can then rewrite the latter inequality as $$\epsilon\leq |f(x)|-|f(1)|\leq |f(x)-f(1)|.$$ (I am being a little sloppy here, but there is only a little work to make everything super tight and formal).

To see that the third condition is not equivalent, let $A$ be any non-measurable set of $[0,\infty)$ and define $f(x) = x\chi_A -x\chi_{\mathbb{R}\setminus A}$, where $\chi$ is the indicator function. Then $f$ has the desired property but is not integrable.

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