# $\epsilon, \delta$ exercise

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?

• 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? – AnyAD Sep 30 '18 at 22:48
• 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 – yoshi Sep 30 '18 at 22:56
• I guess the second delta should actually be an epsilon in your statement. – Alonso Delfín Sep 30 '18 at 23:14
• YES! Sorry, someone just edited – yoshi Sep 30 '18 at 23:23
• @yoshi I think there was a misunderstanding (on my part) of which statement you meant for 'second choice' – AnyAD Sep 30 '18 at 23:37

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.

• I don't know if you read the comments, but you agree also that the second choice implies the first, right? – Ovi Sep 30 '18 at 23:34
• @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 $f.$ – mfl Sep 30 '18 at 23:37
• 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? – yoshi Sep 30 '18 at 23:38
• 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 $\lim_{x\to \pm \infty} f(x)=\pm \infty.$ – mfl Sep 30 '18 at 23:41

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.