Uniformly continuous and differentiable. Let $D\subseteq \mathbb{R}$ and let $f:D\rightarrow \mathbb{R}$.  We say that the function $f$ is an $\mathcal{L}$-function if there exists some constant $K\geq 0$ for which $\left|f(x)-f(y)\right|\leq K\left|x-y\right|$.
1.) Prove that every $\mathcal{L}$-function function is uniformly continuous on its domain.
2.) Give an example showing that there exist uniformly continuous functions which are not $\mathcal{L}$-functions.
3.) Prove that if $f:(a,b)\rightarrow \mathbb{R}$ is an $\mathcal{L}$-function and is differentiable, then $f'$ is bounded.  
4.) Prove or disprove that a function is an $\mathcal{L}$-function on $(a,b)$ if and only if it is differentiable on $(a,b)$.
Response: So far I haven't gotten any work worth showing.
 A: Edit: Thanks to Jonas for pointing out a mistake, and for suggesting an improvement.
(1) Let $\epsilon > 0$. Choose $\delta = \epsilon/K$. Then if $|x-y|<\delta$, then $|f(x)-f(y)|<K\delta=K\epsilon/K=\epsilon$. Thus $f$ is uniformly continuous.
(2) $f(x)=\sqrt{x}$ on $[0,1]$. To see that $f$ is not an $\mathcal{L}$-function, note that since $f$ is differentiable with derivative $f'(x)=\frac{1}{2\sqrt{x}}$, the difference quotient $\displaystyle \frac{|f(x)-f(y)|}{|x-y|}$ is unbounded. But $f$ is uniformly continuous on $[0,1]$:
For $x=0$ or $y=0$ the problem is trivial. For $x,y>0$, $|\sqrt{x}-\sqrt{y}|\leq \sqrt{|x-y|}$.

Proof: Without loss of generality, take $x > y$. Then
  $\displaystyle x \leq x+2\sqrt{(x-y)y}=(x-y)+2\sqrt{(x-y)y}+y = (\sqrt{x-y}+\sqrt{y})^2$. Taking the square root of both sides and rearranging, we get the desired inequality.

Now taking $|x-y|<\epsilon^2$, we get $|\sqrt{x}-\sqrt{y}|<\epsilon$.
(3) For any $x,y\in(a,b)$, if $f$ is a $\mathcal{L}$-function then $\displaystyle \frac{|f(x)-f(y)|}{|x-y|}\leq K$, so taking the limit as $y \to x$ we get $f'(x) \leq K$.
(4) False.
If $f(x)=x^{-1}$ on $(0,1)$, then $f'(x)=-x^{-2}$, so $f$ is differentiable on $(0,1)$. But $|f'|$ is obviously unbounded, so by (3) it cannot be a $\mathcal{L}$-function. Thus we have a differentiable function on $(0,1)$ that is not a $\mathcal{L}$-function on $(0,1)$.
If $f(x)=|x|$ on $(-1,1)$, then $f$ is not differentiable at $0$. But $f$ is a $\mathcal{L}$-function with $K=1$, since by the reverse triangle inequality $|~|x|-|y|~| \leq |x-y|$.
A: 1)
Let $ε>0$. Choose $δ=?$. Exercises like these are about choosing a right $δ>0$.
We know that there exist a number $K≥0$, such that $|f(x)-f(y)|≤K|x-y|$. (1)
Note that: $|x-y|<δ$ implies that $(K+1)|x-y|<(K+1)δ$. (as $K+1>0$). (2)
Combining (1) and (2) gives: 
$$|f(x)-f(y)|≤K|x-y|<(K+1)|x-y|<(K+1)δ$$
If we manage to get $(K+1)δ$ equal to $ε$, than we are done.
We can't choose $ε$ tough, but we can choose $δ$.
Therefore we choose $δ=\frac{ε}{K+1}$.
Note that $δ>0$ because $K+1>0$ and $ε>0$.
A: You could also say that f(x) = x^2 is continuous since it's a polynomial, and since [0,1] is compact and f is invertible on [0,1], its inverse is continuous on a compact set, and is thus uniformly continuous.
