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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.

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    In general, when trying to prove a statement like "If A, then B," you should write down the definitions involved in A, and also the definitions involved in B. Then see if you can play around with these definitions to go from A to B. This helps for both (1) and (3).2012-12-10

3 Answers 3

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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)|. 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 $|~|x|-|y|~| \leq |x-y|$.

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    In other words, $K=0$ implies that $f$ is a constant function, and constant functions are obviously continuous.2012-12-19
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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 $ε>0$.

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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.