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?

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 $x\ne y$.)
