Exercise 4.4.9:
A function $f: A \rightarrow \mathbb{R}$ is called Lipschitz if there exists a bound $M > 0$ such that
$$\left|\frac{f(x) - f(y)}{x-y}\right| \leq M$$
for all $x,y \in A$.
a) Show that if $f: A \rightarrow \mathbb{R}$ is Lipschitz, then it is uniformly continuous on $A$.
b) Is the converse true? Prove or provide a counter example.
Just throwing up the notes that were given in class about the homework.
$$\left| f(x) -f(y) \right| \leq C \left|x-y\right|$$ if $0 < C < 1$.
Banach's Fixed Point Theorem:
- prove $f$ is uniformly continuous
- define $(X_n)$ as follows. Let $X_1 \in \mathbb{R}$ be arbitrary. $X_n = f(X_{n-1}) \rightarrow$ then $(X_n)$ is cauchy.
- Then $(X_n)$ converges to a point $X_n$ which is a fixed point of $f$.
What about something like:
$\begin{equation*}
\begin{split}
\left|\frac{f(x) - f(y)}{x - y}\right| & \leq M \\
& \Rightarrow \left|f(x) - f(y)\right| \leq M\left|x - y\right| \\
\end{split}
\end{equation*}$
Then let $\delta = \frac{\epsilon}{M}$ ...
I was thinking about your answer, and I feel like we need to be sure without a doubt that $|f(x) - f(y) < M|x-y|$
rather than $|f(x) - f(y)| \leq M |x-y|$ . With the $\delta = \frac{\epsilon}{M}$, we gotta make sure it ends up with $|f(x) - f(y)| < \epsilon$ ecause otherwise I don't think the proof works.
$$\delta = \frac{\epsilon}{M}$$ together with $$|f(x) - f(y)| < \delta$$ gives you exactly the result you're talking about.
(remember that in general $a \leq bc$ and $c < d$ implies $a \leq bc < bd$) when $a,b,c,d \geq 0$
@jmacdona is correct. Since we begin by assuming $\left| x-y \right| < \delta$ we can produce $M\left| x-y \right| < M\delta$.
I tend to agree with @jmacdona and @jmincey. We are given that $\left| \frac{f(x) - f(y)}{x-y} \right| \leq M$, thus $$\left|f(x) - f(y)\right| \leq M\left| x-y \right|. $$ We don't need to have that this inequality is strictly less than, we chose our $\delta$ so that we can use the definition: that is that whenever $\left| x-y \right| < \delta$, then $\left|f(x) - f(y)\right| < \epsilon.$
You guys have swayed me, I agree with y'all.
I'm with you on part A! Now how about part B... is it still an epsilon/delta proof? Just reversed...?
@ediazloa
Think about functions involving even radicals. (ex $f(x) = \sqrt[4]{x}$). We can certainly find an interval on which $f$ is uniformly continuous. But, can we satisfy the definition of Lipschitz?
I agree with @jmacdona, i think finding a counter example is the easier option
@jmacdona That is the way I went with it too. Glad to know I'm not the only one.
I went this route as well!