Help Understanding Why Function is Continuous I have read that because a function $f$ satisfies
$$
|f(x) - f(y)| \leq |f(y)|\cdot|x - y|
$$
then it is continuous. I don't really see why this is so. I know that if a function is "lipschitz" there is some constant $k$ such that
$$
|f(x) - f(y)| \leq k|x - y|.
$$
But the first inequality doesn't really prove this because the |f(y)| depends on one of the arguments so isn't necessarily lipschitz. So, why does this first inequality imply $f$ is continuous?
 A: Fix $y=x_0$. Then $|f(x) - f(x_0)| < |f(x_0)|\cdot|x - x_0|$ for all $x$. In particular, $f$ is Lipschitz near $x_0$ and this suffices for $f$ to be continuous at $x_0$.
Note that the condition $|f(x) - f(y)| < |f(y)|\cdot|x - y|$, which presumably holds only for $x\ne y$, implies that $f$ is never zero and so you can divide the given $\varepsilon$ by $f(x_0)$.
Edit: For the updated question, you need to argue as follows. If $f$ has a zero, then $f=0$ everywhere. Otherwise, $f$ is never zero and the locally Lipschitz argument given above works. But now my answer reduces to the one given by Martin...
A: For your condition to make sense, you need $\leq$ rather than $<$. Otherwise, as Johannes noted, the zero function fails it.
So, below, I assume $\leq$ in your condition. Note that if $f(y)=0$ for any $y$, then $f$ is identically zero. So we restrict to nowhere zero functions.
A Lipschitz function is uniformly continuous. Your condition is weaker than that, but it still implies continuity.
Formally, if you fix $y$ and choose any $\varepsilon>0$. Let $\delta=\varepsilon/|f(y)|$. So, if $|x-y|<\delta$, we have
$$
|f(x)-f(y)|\leq |f(y)|\;|x-y|\leq|f(y)|\;\delta=\varepsilon.
$$
So $f$ is continuous at $y$.
A: You're right that the dependence on $y$ means this inequality isn't like the Lipschitz condition.  But the same proof will show continuity in both cases.  (In the Lipschitz case you get uniform continuity for free.)  Here's how:
Let $y\in\operatorname{dom} f$; we want to show $f$ is continuous at $y$.  So let $\epsilon > 0$; we want to find $\delta$ such that, if $|x-y| < \delta$, then $|f(x) - f(y)| < \delta$.  Let's choose $\delta$ later, when we figure out what it ought to be, and just write the proof for now: if $x$ is such that $|x-y| < \delta$ then
$$
  |f(x) - f(y)|
  < |f(y)|\cdot|x-y|
  < |f(y)|\delta
  = \epsilon
$$
The first step is the hypothesis you've given; the second step is the assumption on $|x-y|$; the last step is just wishful thinking, because we want to end up with $\epsilon$ at the end of this chain of inequalities.  But this bit of wishful thinking tells us what $\delta$ has to be to make the argument work: $\delta = |f(y)|^{-1}\epsilon$.
(If $f$ were Lipschitz, the same thing would work with $|f(y)|$ replaced with $k$, and it would yield uniform continuity because the choice of $\delta$ wouldn't depend on $y$.)
(Oh, and a technical matter: the condition you've stated only makes sense for $x\ne y$; otherwise the LHS is at least $0$ but the RHS is $0$, so the strict inequality cannot hold.  But this doesn't affect the argument for continuity; you just assume at the right moment that $x\ne y$.)
