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.

  • $\begingroup$ I believe $f(x)=\sqrt x$ works for part (2). $\endgroup$ – Mario Carneiro Dec 10 '12 at 7:03
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    $\begingroup$ So far I haven't gotten any work worth showing. You mean you do not know how to solve (1)? If you spent two minutes thinking about it with a pen and a piece of paper in front of you and if you have the slightest idea of the definitions involved, I find this simply difficult to believe. $\endgroup$ – Did Dec 10 '12 at 7:03
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    $\begingroup$ Once again: you pretend you spent 5 minutes thinking about question (1) with the definition of uniform continuity and the hypothesis of the exercise both written on a slip of paper placed in front of you, and that nothing popped out? $\endgroup$ – Did Dec 10 '12 at 7:09
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    $\begingroup$ All right, let's try to be polite. did is right though, 1) is completely trivial from the definition of uniform continuity. Write it out in full detail with epsilons and all, and it should be simple. 3) also follows directly from the definition of differentiability. (Perhaps more than 5 minutes on a problem is appropriate before it ends up on stackexchange?) $\endgroup$ – Gyu Eun Lee Dec 10 '12 at 7:27
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    $\begingroup$ 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). $\endgroup$ – Jesse Madnick Dec 10 '12 at 8:32

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

  • $\begingroup$ @JonasMeyer Thanks. I've corrected the error and taken the suggestion into account. $\endgroup$ – Gyu Eun Lee Dec 15 '12 at 20:26
  • $\begingroup$ For 1). What if $K=0$ ? Then $\delta =\frac{\epsilon}{0}$ ? Doesn't seem rigth to me. $\endgroup$ – Kasper Dec 17 '12 at 16:28
  • $\begingroup$ If $K=0$ this is trivial because this implies $|f(x)-f(y)|\leq 0<\epsilon$ for any $\epsilon > 0$. $\endgroup$ – Gyu Eun Lee Dec 17 '12 at 22:42
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    $\begingroup$ In other words, $K=0$ implies that $f$ is a constant function, and constant functions are obviously continuous. $\endgroup$ – Gyu Eun Lee Dec 19 '12 at 21:40

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 $ε>0$.


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.


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