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Consider the following definitions:

> 1). Somewhere Locally Bounded:

$\exists p \in X, \exists \epsilon >0, \exists \delta >0, \forall q \in X: d(p,q)< \delta \Rightarrow d(f(p),f(q)) < \epsilon$.

> 2). Locally Bounded:

$\forall p \in X, \exists \epsilon_p < \infty, \exists \delta_p >0, \forall q \in X : d(p,q)<\delta \Rightarrow d(f(p),f(q)) < \epsilon$.

> 3). Continuous:

$\forall p \in X, \exists \epsilon >0 \exists \delta_{p,\epsilon} >0, \forall q \in X : d(p,q)<\delta \Rightarrow d(f(p),f(q)) < \epsilon$.

> 4). Uniformly Continuous:

$\forall \epsilon >0, \exists \delta_\epsilon >0, \forall p \in X, \forall q \in X: d(p,q)< \delta \Rightarrow d(f(p),f(q)) < \epsilon$.

> 5). Lipschitz Continuous:

$\exists C>0, \forall \epsilon >0 \forall p \in X, \forall q \in X: d(p,q)< \frac{\epsilon}{C} \Rightarrow d(f(p),f(q))< \epsilon$.

Alternatively, we can also write

$\forall \delta>0, \forall p,q \in X, d(p,q)<\delta \Rightarrow d(f(p),f(q))< C \cdot \delta$.

What I am asked to do is to find 4 examples of $(X,Y,f)$ showing $(5) \nLeftarrow (4) \nLeftarrow (3) \nLeftarrow (2) \nLeftarrow (1)$.

Here is my attempt at the complete solution:

$(5) \nLeftarrow (4)$.

Not all uniformly continuous functions are Lipschitz. Let us consider a continuous function on a compact set which has a vertical asymptote. Let $f(x)=\sqrt[3]{x}$ on $[-1,1]$. Since $[-1,1]$ is compact, we have uniform continuity. But for $x_n= \frac{-1}{n^3}$ and $y_n = \frac{1}{n^3}$, we have

$\displaystyle {|\frac{f(x_n)-f(y_n)}{x_n-y_n}| = |\frac{\frac{-2}{n}}{\frac{-2}{n^3}}| = n^2}$. However there is no $C$ such that $n^2 \leq C$ for all $n$, so the function $f$ cannot be Lipschitz continuous.

$(4) \nLeftarrow (3)$.

Continuity does not imply Uniform Continuity. Let $f(x)=x^2$. This function is continuous, but not uniformly over $I=(0,\infty)$.

$\forall x_0 \in I, \forall \epsilon >0, \exists \delta >0, \forall x \in I$, such that $|x-x_0|< \delta \Rightarrow |x^2-x_0^2|< \epsilon$.

Let us fix $x_0$ and let $a=x_0+1$ and $\delta = \min(1,\frac{\epsilon}{2a})$. Choose $x\in I$. Assuming $|x-x_0|< \delta$, we have $|x-x_0|<1, \quad x<x_0+1=a.$ So we get

$|x^2-x_0^2|= |(\frac{1}{\delta}+\frac{\delta}{2})^2-(\frac{1}{\delta})^2| = |\frac{1}{\delta^2} +1+ \frac{\delta^2}{4} - \frac{1}{\delta^2}| = 1+ \frac{\delta^2}{4} > 1 = \epsilon$ as required.

$(3) \nLeftarrow (2)$

Locally Bounded does not imply Continuity.

Let $f(x)= \frac{x^2-4}{x-2}$. Then we have that $f(x)$ is locally bounded, say by (1,3), but $f(x)$ is not continuous at 2.

$(2) \nLeftarrow (1)$

For the last implication, I do not know how to start.

This is what I have so far, please feel free to correct any technical errors. I apologize for the very long post, but I have been struggling with this problem for a while now and would like to finally solve it.

As always, any assistance is greatly appreciated. Thanks in advance.

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For your $3)\not\Leftarrow 2)$ example, try the Dirichlet Function: $$ D(x)=\left\{\begin{array}{lc} 1 & :x\in\mathbb{Q} \\ 0 & :x\not\in\mathbb{Q} \end{array}\right. $$ It tends to be great for continuity counterexamples, as it is discontinuous everywhere (and can be easily modified to be continuous, and even differentiable, at a finite or countable number of points). It should be obvious that is is bounded, as $|D(a)-D(b)|\leq 1$ for any $a,b\in\mathbb{R}$.

As for showing that $2)\not\Leftarrow 1)$, you only need to find a function that is not locally bounded at a single point. The easiest example is probably $$f(x)=\left\{\begin{array}{lc} 0 & : x=0 \\ \frac{1}{x} & :x\neq0 \end{array}\right.$$ at $x=0$.

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    $\begingroup$ The issue here is that, technically, $f$ is not defined at those points, so you cannot choose your points $p$ as $2$ or $0$ respectively. $\endgroup$ – Some Math Student Dec 1 '14 at 18:05
  • $\begingroup$ You're right, I've changed my examples. $\endgroup$ – KSmarts Dec 1 '14 at 18:24
  • $\begingroup$ I like your examples, actually, I was thinking of the Dirichlet function after I read the answer from @SomeMathStudent. For the second one, if I was interested in finding an $\epsilon$ and $\delta$ that would make the locally bounded condition fail, would it suffice to choose $x=0$ so that the distance is less than 1? $\endgroup$ – Jamil_V Dec 2 '14 at 15:53
  • $\begingroup$ Since the conditions for locally bounded only require existence, you have to show that it fails for whatever $\epsilon, \delta$ you choose. So you need to show that for any $\epsilon$, you can choose an arbitrarily small $q$ such that $f(q)>\epsilon$. $\endgroup$ – KSmarts Dec 2 '14 at 16:08
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    $\begingroup$ More specifically, let $0<q<\min(\delta,\frac{1}{\epsilon})$. $\endgroup$ – KSmarts Dec 2 '14 at 16:10
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You could try to translate those definitions into a more 'geometrical' form.

$1)$ tells you that there is some fixed point $p$, where the for all $q$ close enough to $p$ the values $f(p),f(q)$ are not too far apart.

$2)$ tells you that for every point $p$, for some $q$ close enough to $p$ the values $f(p),f(q)$ are not too far apart.

So, a way to go about this is to notice that the first one is a purely local property - given a function which is constant on a certain interval and completely arbitrary on the rest, can still fulfill the first condition, whereas the second property is more global - at least, you have the local property everywhere on the domain.

I'd suggest looking at any not locally bounded function (you might want to think about rationals here, anything continuous will not work), and then switch it up with a locally bounded one somewhere - thus, you will get your contradiction.

Also, please note that your counterexample to $3)\Leftarrow 2)$ does not hold - the function $\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2$ is perfectly continuous. Again, I'd recommend separating rational numbers and irrational numbers, and try to find a non-continuous, bounded function.

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  • $\begingroup$ Ok, I think I understand what you mean by local and global properties. However, what do you mean by thinking about rationals? Like in your comment on my counterexample to $(3) \nLeftarrow (2)$, do you mean to define a function as, say 1 if it is rational and 0 if it is irrational? $\endgroup$ – Jamil_V Dec 1 '14 at 17:49
  • $\begingroup$ For the second part, exactly that, yes. You are still bounded, but not continuous anymore. For the first part, you will need to modify this function a bit. $\endgroup$ – Some Math Student Dec 1 '14 at 17:50
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You can fix the counterexample involving $f(x)=\frac{x^2-4}{x-2}$ by specifying that $f(2) = \frac{5}{2}$.

Here is a function that is somewhere locally bounded but not locally bounded: $$ f(x) = \left\{ \begin{array}{cc} x^q & x \mbox{ rational }, x= p/q \mbox{ with gcd}(p,q) = 1 \\ 0 & x \mbox{ irrational } \end{array} \right. $$ $f(x)$ is locally bounded at $x=0$ (or indeed at any point $x=p$ with $0 \leq p < 1$ thus it is somewhere locally bounded. But $f(x)$ is not locally bounded: Consider point $p$ at $x=2$, where for points arbitrarily close to $p$, $f(x)$ is close to $2^n$ for arbitrarily large $n$.

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