Chris Tynan

Posted on Monday, 08 March, 2004 - 08:50 pm:

I'm just trying to get to grips with the more rigorous method of proving the continuity of functions, I would appreciate it if somebody could help me out.

Prove that $1 / (1 + x^{4})$ is continuous.

Well, considering $| x - x_{1} | < δ$ and $| f (x) - f (x_{1}) | < ε$ , we see that $| 1 / (1 + x^{4}) - 1 / (1 + x_{1}^{4}) | < ε$ given $ε$ . This rearranges to $| x_{1}^{4} - x^{4} | / | 1 + x^{4} | | 1 + x_{1}^{4} | < ε$ . The denominator is clearly $\geq 1$ so it remains to find a value of $δ$ with the numerator $< ε$ .

Therefore we have $| x^{4} - x_{1}^{4} | < ε$ and $| x - x_{1} | < δ$ . But $| x^{4} - x_{1}^{4} | = | x - x_{1} | | x_{1}^{3} + x_{1}^{2} x + x_{1} x^{2} + x^{3} | < δ | x_{1}^{3} + x_{1}^{2} x + x_{1} x^{2} + x^{3} |$ .

I get stuck here, I can't see how we can choose $δ$ such that $| x - x_{1} | < δ \Rightarrow | x^{4} - x_{1}^{4} | < ε$ .

Thanks,

Chris

David Loeffler

Posted on Monday, 08 March, 2004 - 09:07 pm:

Try sticking to proving it to be continuous at some particular chosen point $x$ . Suppose we know that

| x' - x | < | x |

. Then

| x' | < 2 | x |

| x'^{3} + x'^{2} x + x' x^{2} + x^{3} | < (8 + 4 + 2 + 1) | x |^{3} = 15 | x |^{3} < 1 + 15 | x |^{3}

So if we take $δ = min (| x |, ε / (1 + 15 | x |^{3}))$ , then for any $x'$ such that $| x' - x | < δ$ , we have

$| x^{4} - x'^{4} | < (15 | x |^{3}) δ < ε$ .

HOWEVER themoral of this should be that actually proving continuity of anything by hand is a gruesome activity. What is much more sensible is to have a toolkit of known theorems: if $f$ , $g$ are continuous at $x$ , then $f + g$ and $f \times g$ are so, and if $f (x) \neq 0$ then $1 / f$ is also continuous. That will make your life a lot easier.

David

David Loeffler

Posted on Monday, 08 March, 2004 - 09:09 pm:

(NB. Of course the above only works if x =/= 0, but the case x = 0 is easy to deal with on its own.)

Chris Tynan

Posted on Monday, 08 March, 2004 - 09:21 pm:

In my initial post, I thought I assumed (although I didn't explicitly state it) that

x_{1}

was fixed. What happens in the case where

| x' - x | > | x |

(I apologise if this is a triviality and I just don't understand.) Also, when we have

δ = min (| x |, ε / 1 + 15 | x |^{3})

, what is special about the case when

| x | < ε / 1 + 15 | x |^{3}

, and how does this imply

15 | x |^{4} < ε

Thanks for your other advice, the book I am reading (what is mathematics) has not mentioned these extra formulae, although I agree they would seemingly make the problem a lot easier.

Chris

Chris Tynan

Posted on Monday, 08 March, 2004 - 09:23 pm:

Oh I take back the case when

| x | < ε / 1 + 15 | x |^{3}

. This implies that

15 | x |^{4} + 15 | x |^{3} < ε

as such

15 | x |^{4} = 15 | x |^{3} δ < ε

. (I think?)

Chris Tynan

Posted on Monday, 08 March, 2004 - 10:07 pm:

In fact I also think I understand that the other case does not need to be considered, since we have

| x' - x | < δ \leq | x |

| x' - x | < | x |

, for all

x

satisfying

| x' - x | < δ

David Loeffler

Posted on Monday, 08 March, 2004 - 10:45 pm:

Chris: In your initial post, you may have assumed x₁ was fixed, but you didn't actually use this assumption. (This is why I misinterpreted your post and in my reply used x fixed and x' variable).

Had your approach worked, you would actually have succeeded in proving something rather stronger than continuity, known as 'uniform continuity' - the difference between the two is quite subtle, and is essentially that in ordinary continuity your d(e)can depend on x; in uniform continuity it doesn't - there has to be one dfor each e which then works for all x.

If you are considering functions on a closed bounded interval, then ordinary continuity at all points of the interval actually implies uniform continuity on the interval, but this is not true if the interval isn't closed (think x = [0, pi) and f(x) = tan x) or isn't bounded (think f(x) = x² on

Latex image click or follow link to see src

).

Now, your function is in fact uniformly continuous on all of

, and you can always get away with d=e, but this requires a bit more work to prove - it's morally easy by looking at the derivative, but you need enough theorems about derivatives to rigorously justify your use of them.

David

Chris Tynan

Posted on Monday, 08 March, 2004 - 11:04 pm:

I think I understand, despite having just first read about this and studied it seriously recently. So, as I understand it, your proof is a proof of ordinary continuity of f(x)? Is it possible to show that if we assume |x'-x| < e we can likewise show |x'⁴ -x⁴ | < e? I can't see an obvious way of going about this from looking at it, surely it would revolve around showing |x'³ +x'² x + x'x² + x³ | < 1 when |x'-x| < e, or is there a simpler way? (I'm interesting in your use of the derivative, I just can't see it, perhaps a hint?)

How can we show ordinary continuity for all points in a closed bounded interval => uniform continuity? I imagine it is possible to construct a function which covers the interval from which we can derive uniform convergence, i.e. d being a function of e.

I have to admit that I'm extremely poorly-informed on this topic, know very little, and to be honest, it quite baffles me at the minute, so I apologise if my questions seems a little stupid.

Thanks,

Chris

Kerwin Hui

Posted on Monday, 08 March, 2004 - 11:18 pm:

You cannot jump from |x'-x|< e to |x'⁴ -x⁴ |< e without any assumption on what x is. (In fact this is only true for |x|< (1/4)^1/3 -c, some c(e).)

Hint for using the derivative to get uniform continuity - use the mean-value theorem.

The 'usual' proof of continuity on closed bounded interval -> uniform continuity goes as follows: Suppose false, then for some e and any choice of d, we have two points at most d apart with the image at least e apart. Now take a sequence of d's tending to zero, and take the appropriate pair. There is a subsequence converging to some point and this yields a point of discontinuity - contradiction.

Kerwin

Chris Tynan

Posted on Monday, 08 March, 2004 - 11:38 pm:

Kerwin, I think I understand your proof of ordinary continuity on a closed bounded interval => uniform continuity. Is the idea that for all values in the interval, we must be able to find a d such that for a e satisfying |f{x'}-f(x)| < e we can have |x'-x| < d? I understand that if we were not to have uniform continuity, then we can find two points d apart, and, supposing the statement is false, we have that their image is over e apart. Likewise we can choose a smaller d that satisfies this, and so on, so our sequence of ds tends to a particular point, but this point is a discontinuity? I apologise if I've merely regurgitated your proof.

I'm not sure I fully understand what is going on. I am assuming, although I don't know why, that uniform continuity means continuity for every point on the interval, which seems to be implied by proving there is ordinary continuity for each point in that interval.

As for the mean value theorem, that baffles me a bit, I'll read up (get some sleep!) and try the problem again tomorrow.

Thanks for your help once again,

Chris

Chris Tynan

Posted on Tuesday, 09 March, 2004 - 06:47 pm:

To continue this, today I tried to prove that if n is a positive integer, then xⁿ is continuous.

Let's take some point x = a, then we have |xⁿ -aⁿ | < e => |x-a||x^n-1 +ax^n-2 +a² x^n-3 ...+aⁿ | < e.

Setting |x-a| < |a| => |x| < 2|a| => |x-a||x^n-1 +ax^n-2 +a² x^n-3 ...+aⁿ | < d|(2ⁿ -1)a^n-1 | where |x-a| < d. Now if we take d = min(|a| , e/(|(2ⁿ -1)a^n-1 |+1) then we have d|(2ⁿ -1)a^n-1 | < e => |xⁿ -aⁿ | < e as required. (Clearly we have d < a as well, so we may assume that |x-a| < a?)

Can we extend this and then say that all polynomials are continuous, using if f and g are continuous, f + g is continuous?

I've yet to do the uniform continuity problem, apologies, I'll have a look soon.

Chris

roko mijic

Posted on Wednesday, 10 March, 2004 - 01:39 pm:

yes, that's the way to do it! better still, prove first that f, g continuous => fg continuous, prove x is continuous, and then save a lot of bother!

Chris Tynan

Posted on Wednesday, 10 March, 2004 - 07:47 pm:

Is this easy to prove though? (I imagine that it could well be). I approached this by taking a fixed

ε

with

| f (x) - f (a) | < ε

and

| g (x) - g (a) | < δ

giving us intervals of

| x - a | < δ_{1}

and

| x - a | < δ_{2}

. But upon multiplying this gives

f (x) g (x) + f (a) g (a) - \dots

not

f (x) g (x) - f (a) g (a)

. (Surely we are trying to show that

| f (x) g (x) - g (a) f (a) | < μ

where

| x - a | < δ

for any value of

μ

no matter how small?)

If so then I guess we could say $f (x) < f (a) + ε$ and likewise for $g (x) \Rightarrow | f (x) g (x) - f (a) g (a) | < | ε (f (a) + g (a)) + ε^{2} |$ . If we can show that we can make the number in the bracket then surely we are done, with $| x - a | < min (δ_{1}, δ_{2})$ . I doubt this method is the right approach, and doubt further that it gives the answer, but it was the way I approached the problem. I can't see how to do the problem.

Chris

David Loeffler

Posted on Wednesday, 10 March, 2004 - 08:16 pm:

When I did 1A Analysis, our lecturer used to say 'A mathematician is someone who knows that

0 = 1 - 1

.' Try writing

$| f (x) g (x) - f (a) g (a) | = | f (x) g (x) - f (a) g (x) + f (a) g (x) - f (a) g (a) | \leq | g (x) | | f (x) - f (a) | + | f (a) | | g (x) - g (a) |$ .

Now, if $x$ is close enough to $a$ , then $f (x)$ isn't too big, so we can make this small. You know that there's some $δ$ that makes $| f (x) - f (a) | < ε / [2 (2 g (a) + 1)]$ or something silly like that, and you can also make $δ$ sufficiently small that $| g (x) | < 2 | g (a) | + 1$ . That should make the first part $< ε / 2$ ; you can do the same to the second, and if your $δ$ is sufficiently small for all your constraints to be simultaneously satisfied, then your expression will be less than $ε / 2 + ε / 2 = ε$ , and we're safe.

David

Chris Tynan

Posted on Wednesday, 10 March, 2004 - 08:29 pm:

Oh thanks David! I didn't spot the obvious removal of |g(x)| and |f(x)|! Yes, I also understand the bit about making the delta sufficiently small.

Thanks for your help, David, Kerwin and Roko, it's helped to make things seem a bit more clearer to me now.

Chris