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I can not conclude this problem, I tried two ways but I can not conclude ...

Let, $$f:(0,+\infty)\rightarrow \mathbb{R}$$ Lipschitz function. Prove that there exists a finite limit: $$\lim_{x \to 0^+}f(x)$$


Let $x_0$ and $x$ two points of the domain of $f$: $$x_0 , x \in (0, + \infty).$$ We know the function is Lipschitz continuous (and thus also continuous): by definition Lipschitz functionwe have: $$|f(x)-f(x_0)|\le L| x-x_0 | $$ from which: $$ \frac {| f (x)-f (x_0)|}{| x-x_0 |} \le L $$ Then, passing to the limit $x \to x_0$ $$\lim_ {x \to x_0} \frac {| f (x)-f (x_0 )|}{| x-x_0 |} = f '(x) \le L $$ but from here I can not prove $\lim_{x \to 0^+}f(x)$


$$ \forall x, y \in(0, + \infty), \quad L| x-y | \le f(x)-f (y) \le L | x-y |$$ or: $$ L | x-y | + f(y) \le f(x) \le L | x-y | + f (y) $$

$$ f (x) \le L | x-y | + f (y) $$ passing to the limit $ x \to 0 ^+ $

$$\lim_ {x \to x_0} f (x) \le L | x-y | + f (y) $$

but from here can I conclude that $\lim_{x \to 0^+}f(x)$ is a finite limit?

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Beware of the phrase "passing to the limit". You cannot pass to the limit in an inequality until you have already shown that both sides actually converge to limits. – Nate Eldredge Dec 21 '11 at 0:18
Hint: Try showing that for any $\epsilon > 0$, we have $\limsup_{x \to 0^+} f(x) \le \liminf_{x \to 0^+} f(x) + \epsilon$. Then it follows that $\limsup_{x \to 0^+} f(x) = \liminf_{x \to 0^+} f(x)$ and so $\lim_{x \to 0^+} f(x)$ exists. – Nate Eldredge Dec 21 '11 at 0:20
Do you mean to write $-L|x - y| \leq \cdots$ at the start of the second attempt? – Dylan Moreland Dec 21 '11 at 0:53
At the end of your first attempt I think you mean $$\lim_{x\to x_0}\frac{\vert f(x)-f(x_0)\vert}{\vert x-x_0\vert}=f'(x_0).$$ Also, consider the the Dyland's comment. – leo Dec 21 '11 at 2:29
up vote 0 down vote accepted

Another way to look at the liminf/limsup approach is to do it as a proof by contradiction... if the limit did not exist, then $\limsup_{x \rightarrow 0+}f(x) = B$ and $\liminf_{x \rightarrow 0+}f(x) = A$, where $A< B$. (Lipschitz functions on $(0,1)$ are immediately bounded so you don't have to worry about $A$ or $B$ being $\pm \infty$.) Then for some small $\epsilon > 0$ you could find $a_n \rightarrow 0$ and $b_n \rightarrow 0$ such that $f(a_n) < A + \epsilon$ and $f(b_n) > B - \epsilon$. So $|f(b_n) - f(a_n)| > B - A - 2\epsilon$, even though $|b_n - a_n|$ goes to zero, which contradicts Lipschitzness.

And you can do this less rigorously without any liminfs or limsups. If the limit doesn't exist, then there are $a_n,b_n \rightarrow 0$ such that $|f(b_n) - f(a_n)| > \delta$ for some fixed $\delta$, despite $|b_n - a_n|$ going to zero. The liminf and limsup just formalizes this idea.

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I'll try to say something about your attempts in a bit, but here's another idea.

Let $\{x_n\}$ be a sequence in $(0, \infty)$ such that $x_n \to 0$. You can use the Lipschitz condition to show that the sequence $\{f(x_n)\}$ is Cauchy and hence convergent to a limit $p$. If we select another sequence $\{y_n\}$ in $(0, \infty)$ such that $y_n \to 0$ and $f(y_n) \to q$, then we'd like to show that $p = q$.

For this, assume that $\varepsilon = |p - q|$ is non-zero and choose an integer $N$ such that $n \geq N$ implies the following:

  1. $|f(x_n) - p| < \varepsilon/3$ and $|f(y_n) - q| < \varepsilon/3$
  2. $x_n, y_n < \varepsilon/(3K)$, where $K$ is a Lipschitz constant for $f$.
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I'd be very interested in other solutions; in particular, I never have the idea, "Oh, let's take a $\liminf$ here", so I might try to flesh out Nate's idea later for my own betterment. – Dylan Moreland Dec 21 '11 at 0:46
I think you mean $\{y_n\}$ s.t. $y_n\to 0$ and ... – leo Dec 21 '11 at 2:39
@leo I thought that would be understood, but you're right: better to restate the conditions. – Dylan Moreland Dec 21 '11 at 2:39

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