# Definition of uniform convergence written as math symbols

1. Apr 7, 2009

### Gerenuk

Hi,

how would I write out the definition of "uniform convergence" of a function f(x,y) with as few a possible words and using symbols like $\forall$?

2. Apr 10, 2009

### rochfor1

Uniform convergence?...for that you need a sequence of functions. You can do uniform continuity with just one though. Let me know which one you mean, and I can help.

3. Apr 10, 2009

### Gerenuk

A sequence has index $n$ as in $f_n(x)$. My index is $y$ so I wonder if
$$\lim_{y\to a}f(x,y)=g(x)$$
converges uniformly. I'd like to express the definition of uniform convergence with the correct mathematical symbols without using many word. So quiet abstract.

4. Apr 10, 2009

### rochfor1

Ok. That's a bit odd...usually uniform convergence is discussed in the context of sequences (or nets) but here it goes...

$$\forall \varepsilon > 0 \exists \delta > 0$$ so that if $$| y - a | < \delta, | f ( x, y ) - g( x ) | < \varepsilon \forall x$$.

The uniform part is that $$\delta$$ doesn't depend on x.

Last edited: Apr 10, 2009
5. Apr 11, 2009

### Gerenuk

Oh thanks. I probably would have messed up the order.
Is there a way to use logic arrows instead of "so that if"?
Does the $\forall x$ have to be at the end?

6. Apr 11, 2009

### HallsofIvy

I'm not sure why you would want to use "logic arrows" but, yes, the $\forall x$ must be at the end.

Ordinary convergence:
$\forall x$, given $$\epsilon> 0\exists \delta> 0[/itex] $|x-a|< \delta\rightarrow |f(x)- L|< \epsilon$ Uniform convergence: Given $\epsilon> 0\exists \delta> 0$ $|x-a|< \delta\rightarrow |f(x)- L|< \epsilon, \forall x$ In the first, how we choose $\delta$, given $\epsilon$, may depend upon x- that's why it is mentioned first. In the second, the same $\delta$ will work no matter what x is. 7. Apr 11, 2009 ### maze Usually the idea of convergence xn->x is that, if you give me a tolerance epsilon, I can give you an N so all later points xi (i>N) are within the tolerance of x. On the other hand, for functions fn(x) there might arise a situation where I could find the N for every x, but the minimum N's can be made larger and larger by choosing different values of x. In fact, for any N you choose it might happen that fn(x) is not within the tolerance for the majority of the x's. In this sense, the question: "find me a N so that the fi is close to f" cannot be satisfactorily answered. Thus the notion of uniform convergence, and convergence in a norm. 8. Apr 11, 2009 ### rochfor1 I know I'm late back to the party but HallsofIvy is absolutely right, you lose uniformity if you move the [tex]\forall x$$ further up.

9. Apr 11, 2009

### isabelle

One reason for a purely symbolic definition is that many basic results in real analysis can be demonstrated using predicate calculus i.e. symbolic logic. This is not usually taught in math courses, but since I studied logic before real analysis I got to apply it all the time. HallsofIvy definitely gave the kind of formulation that would be appropriate for the blackboard in a math classroom: it reads somewhat like english, and the semantics are correct. But for applying the replacement rules of symbolic calculus (imagine a systematic computer program) it saves a lot of work to input things in a standard form.

Ordinary Convergence:

$$(\forall x)(\forall \epsilon)(\exists \delta)[\epsilon > 0 \rightarrow [(\delta > 0) \wedge (|x-a|< \delta) \rightarrow |f(x)- L|< \epsilon]]$$

Uniform Convergence:

$$(\forall \epsilon)(\exists \delta)(\forall x)[\epsilon > 0 \rightarrow [(\delta > 0) \wedge (|x-a|< \delta) \rightarrow |f(x)- L|< \epsilon]]$$

I know this may look strange, but I promise it is totally standard in any textbook on predicate calculus. One of the key features is that 'for all x' need not appear at the end of the formula (which it never does in predicate logic form) for uniform convergence, it just needs to appear after the 'for all epsilon, there exists delta.' For me it is easier to view the difference in this way, after I got used to the admittedly less natural-language formal use of symbolic notation.

Last edited: Apr 11, 2009
10. Apr 12, 2009

### Gerenuk

Looks good!
Does the statement change if I remove the arrow?
$$(\forall \epsilon)(\exists \delta)(\forall x)[(\epsilon > 0) \wedge (\delta > 0) \wedge (|x-a|< \delta) \rightarrow |f(x)- L|< \epsilon]$$

Is it bad practice to include the inequalities in the front?
$$(\forall \epsilon>0)(\exists \delta>0)(\forall x)[|x-a|< \delta \rightarrow |f(x)- L|< \epsilon]$$

11. Apr 13, 2009

### CRGreathouse

I don't agree with that. You just need to have $\forall x$ to the right of $\exists\delta.$ In fact, in the formal logical notation I learned, a trailing $\forall v$ isn't even meaningful (for any variable v).

12. Apr 13, 2009

### CRGreathouse

That works.

No, that's fine. Formally, you can think of
$$\forall v>a\;\phi$$
as shorthand for
$$\forall v\;(v>a\rightarrow\phi),$$
just like
$$\exists v\;\phi$$
might be considered shorthand for
$$\neg\forall v\;\neg\phi.$$

Of course this sort of rewriting messes with the prenex normal form isabelle was so kind to provide.

13. Apr 13, 2009

### rochfor1

CRGreathouse:

Of course you're right. As long as the $$\forall x$$ is after the $$\exists$$ it's good. As for the logic bit, I've studied a lot of analysis but no logic, so you're probably right.

14. Apr 13, 2009

### CRGreathouse

No worries, I was just pointing out the difference between informal and (one version of) formal.

15. Apr 13, 2009

### poutsos.A

You got a very good point there.In the examples that follow in continuity and uniform continuity,a complete formalized definition is shown .The definition are done for one variable and you can try for several variables.

Let f:A----->B AND:

1) f is continuous at aεA IFF(<====>)

$$\forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x( x\in A\wedge |x-a|<\delta\Longrightarrow|f(x)-f(a)|<\epsilon))]$$

2) f is continuous over A IFF(<====>

$$\forall\ a\forall\epsilon[a\in A\wedge\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x( x\in A\wedge |x-a|<\delta\Longrightarrow|f(x)-f(a)|<\epsilon))]$$.

3) f is uniform continuous over A IFF(<=====>

$$\forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x\forall y( x\in A\wedge y\in A\wedge |x-y|<\delta\Longrightarrow |f(x)-f(y)|<\epsilon))]$$

Here the logical symbols are explicitly mentioned ,and when we want to negate the definitions,that is, when we want to prove that f is not continuous or uniform continuous ,this can be done in a safe way by using the laws of logic.

For example what must we do to prove that f is not continuous over A?

16. Apr 13, 2009

### Gerenuk

I justed wanted to copy the equation to my notes and suddenly it all doesn't make sense to me. A function of two variables f(x,y) is supposed to converge uniformly to a function g(x). The above statement doesn't incluce two variables and seems wrong if I do include them?!

I suggest
$$\lim_{y\to a} f(x,y)&=g(x)\qquad \Leftrightarrow\qquad (\forall \epsilon>0)(\exists \delta>0)(\forall x)[|y-a|<\delta \rightarrow |f(x,y)-g(x)|< \epsilon]$$
which is different from above (different variables $\forall x$ and $|y-a|<\delta$)??!!

Does the concept of uniform convergence make sense for a function of one variable only?

17. Apr 13, 2009

### CRGreathouse

No... it's just that there are a half-dozen or so different versions of uniform convergence. The usual one is for a sequence of functions.

18. Apr 14, 2009

### Gerenuk

Sure. In any case it must be a function of two variables (the index in the case of a sequence of functions).

In physics I usually don't care about th correctness of interchanged $\lim$, but uniform convergence is in principle required so I was interested.