2
$\begingroup$

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?

  • 0
    Now it's a different question.2012-03-12

3 Answers 3

2

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...

2

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 $y$.

  • 2
    @Martin, but the condition is not required of *all* continuous functions. Rather, *if* a function satisfies the condition, *then* it is continuous, but not necessarily conversely. So, yes, the zero function is continuous but does not satisfy the condition.2012-03-12
1

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 $x\ne y$.)

  • 0
    Oh, the question has been edited so that this doesn't make sense anymore. And I see you've given your own answer anyway.2012-03-12