1. ## continuity

Let f:R->R, suppose there is $\L\geqslant 0\$ such that for all
$\
x,y \in \mathbb{R}
\
$
$\
\left| {f(x) - f(y)} \right| \leqslant L\left| {x - y} \right|
\
$

show that f is continuos.

2. Originally Posted by mms
Let f:R->R, suppose there is $\L\geqslant 0\$ such that for all
$\
x,y \in \mathbb{R}
\
$
$\
\left| {f(x) - f(y)} \right| \leqslant L\left| {x - y} \right|
\
$

show that f is continuos.
Have you made any attempt? Have you seen this condition before? It's known as the Lipschitz condition. It comes up a lot in analysis and topology.
It is powerful as it specifies a condition on a function which is stronger than being continuous but not as strong as being differentiable (comes in handy for Picard's Theorem and the study of weak solutions of PDEs)
I recommend googling it and then attempting the question yourself.

If you get stuck here is a pointer in the right direction:

Spoiler:
Choose $\delta = \frac{\epsilon}{L + 1}$

Hope this helps.

pomp.

3. Spoiler:

If we have $0 then it makes sense to pick what you suggested, since it's $L>0$ ('cause $L=0$ makes no sense), and then under this we pick $\delta=\frac\epsilon{L}.$

4. Spoiler:
Suppose, doesn't really matter though. I chose to have L+1 so that the inequality is strict.