Continuity vs Uniform Continuity? What's the difference?

[Quackers McDuck]

I know the definitions are different; these are the ones I'm using (based on sequences and straight from my book):

Continuity: A function f:D->R is said to be continuous at the point x₀ in D provided that whenever {x_n} is a sequence in D that converges to x₀, the image sequence {f(x_n)} converges to f(x₀). The function f:D->R is said to be continuous provided that it is continuous at every point in D.

Uniform Continuity: A function f:D->R is said to be uniformly continuous provided that whenever {u_n} and {v_n} are sequences in D such that if

lim_n->infin.[u_n - v_n] = 0,

then

lim_n->infin.[f(u_n) - f(v_n)] = 0.

And I've seen examples from the book of functions that are continuous but not uniformly continuous (eg f(x)=x²), but I can't figure out more intuitively what the difference is. It seems like wikipedia is actually wrong:

In mathematics, a function ƒ is uniformly continuous if, roughly speaking, it is possible to guarantee that ƒ(x) and ƒ(y) be as close to each other as we please by requiring only that x and y are sufficiently close to each other.

Wouldn't f(x)=x² satisfy that? So it has to be more than that, right?

[gmalivuk]

It seems like wikipedia is actually wrong:

In mathematics, a function ƒ is uniformly continuous if, roughly speaking, it is possible to guarantee that ƒ(x) and ƒ(y) be as close to each other as we please by requiring only that x and y are sufficiently close to each other.

Wouldn't f(x)=x² satisfy that? So it has to be more than that, right?

No, Wikipedia isn't really wrong, you're just misunderstanding what that statement means. Which is fair, because as with all natural language explanations of formal mathematical statements, there's a bit of ambiguity that crops in.

"As close to each other as we please" means we first set some D>0, and we want f(x) and f(y) to be closer to each other than D.
"x and y are sufficiently close to each other", in this case means that, from D and f alone, we can compute some H>0 such that, if x and y are closer to each other than H, then no matter how big or small x and y are, f(x) and f(y) will be within D of each other.

For simple continuity, the analogue of our H will also tend to depend on x, such as for example with f(t)=t². In this case, our H will need to be smaller as |x| gets bigger, because f(x) and f(x+H) differ by |2H x + H²| = H |2x + H|. So regardless of how small a fixed H is, we can make x big enough to make this value bigger than whatever D we started with. The fact that, once we're told x, we can still find an H that works means it's still continuous, but the fact that we can't use this same H for *all* possible values of x means it's not uniformly continuous.

Edit: I've used a slightly different definition here than yours with sequences, because mine is more like what Wikipedia seems to be talking about.

With sequences, simple continuity requires that the sequences <u_n> and <v_n> actually converge to some value x₀. Uniform continuity is a stronger statement, because it only requires that <u_n> and <v_n> get close to each other, even if they don't actually converge to some particular value. So while both still require |u_n-v_n| to approach zero, simple continuity additionally requires |u_n-x₀| and |v_n-x₀| to approach zero, but uniform continuity doesn't.

For example, u_n = n+1/n and v_n = n-1/n would be applicable to check for uniform continuity, but not simple continuity. A function can be continuous even if we don't say anything at all about this particular pair of sequences. But for uniform continuity, we have to check them, along with all other sequences that approach each other without approaching any fixed value. So, again, uniform continuity is a stronger condition, because it has to work for a wider range of sequences than simple continuity has to work for.

[Quackers McDuck]

Thanks, that helps a lot.

Would it be generalizing too much too say that any function with a derivative going off into positive or negative infinity would not be uniformly continuous, then?

And, if the above is true, are there non-uniform-continuous functions that are also continuous without a slope going to infinity?

[gmalivuk]

Would it be generalizing too much too say that any function with a derivative going off into positive or negative infinity would not be uniformly continuous, then?

Correct. Functions with unbounded derivatives are not uniformly continuous. The proof of this can be shown fairly easily from my x+h account of uniform continuity and the x+h definition of derivatives.

And, if the above is true, are there non-uniform-continuous functions that are also continuous without a slope going to infinity?

Yes, because there are functions which are continuous but nowhere differentiable. I don't remember offhand whether the Weierstrass function is uniformly continuous, but even if it is, we can easily use it to make a function that isn't, by adding it to x². The slope doesn't go to infinity, because the slope isn't even defined anywhere, and yet this is still not uniformly continuous.

[Harg]

I have a comment here. It is not true that functions with unbounded derivatives are not uniformly continuous. Consider f(x)=\sqrt{x} on the interval [0,1]. The derivative diverges towards infinity as one approaches zero, yet the function is uniformly continuous, since it is defined on a compact set.
However, should a function be everywhere differentiable and have a bounded derivative, then it is uniformly continuous (actually Lipschitz, but never mind).

edit: Oooh, wikipedia tells me that the compact, uniform thingy is called the Heine-Cantor theorem

[t0rajir0u]

Compactness is important to keep in mind. The distinction between continuity and uniform continuity only exists over non-compact sets (intuitively because there's a lot of "wiggle room" for bad behavior even out of a continuous function - consider f(x) = 1/x on (0, 1)). As such, uniform continuity is a requirement we impose so as to get lots of nice behavior (like preserving Cauchy sequences) that merely continuous functions don't have.

[gmalivuk]

It is not true that functions with unbounded derivatives are not uniformly continuous. Consider f(x)=\sqrt{x} on the interval [0,1].

Ah indeed. I wasn't thinking about functions on bounded intervals.

[MartianInvader]

Actually, having bounded intervals isn't that important. Take y = \sqrt{x} on all the positives, or y = x^{1/3} on all reals. Since the derivative is bounded away from zero, it's uniformly continuous even though it's a function with unbounded derivative whose domain and range is all reals.

If you want uniform continuity and unbounded derivatives as you go out to infinity, you can get examples by playing cut-and-paste with copies of y = x^{1/3} and y = -x^{1/3} on the interval [-1, 1] (paste them together infinitely many times going out to infinity).

[gmalivuk]

Yeah, you're right. My excuse would be that it's Sunday, and I don't think well on Sundays. Except that falls kind of flat considering the mathing I've been doing in the planet destroyer thread over in Science. So my excuse will instead have to be that I'm just stupid.

The reason x^2 isn't uniformly continuous is that the actual vertical difference between two points a given horizontal distance apart (in the usual way of graphing it) can be made arbitrarily large. But if there's only a small region where the derivative goes above some absolute value, it can still be uniformly continuous there, because the function itself doesn't go to infinity and so the actual difference between f(x) and f(x+h) never gets too large for a given h, even if the derivative is infinite.

[Yakk]

So you can talk about continuity using the concept of having a "modulus of continuity".

This function maps your epsilons to your deltas, and traditionally is called \mu.

A function f:A->B is said to be continuous at a point p if there is a function \mu_p:R+ -> R+, called the modulus of continuity at the point p, such that for all epsilon > 0 and x from A, | x - p | < \mu_p(epsilon) implies | f(x) - f(p) | < epsilon.

A function is said to be uniformly continuous if there exists a modulus of continuity that works at every point in A -- ie, there is a choice of modulus of continuity that is uniform.

Note that most epsilon-delta proofs end up implicitly creating a modulus of continuity -- this definition just makes it explicit.