# The limit of an interval

1. Oct 6, 2005

### quasar987

I've tried all the possible ways I could think of to tackle this problem, but found no satisfactory answer. I would like to show that

$$\lim_{\epsilon \rightarrow 0} (t-\epsilon, t+\epsilon) = \{t\}$$

Is it possible?!

Last edited: Oct 6, 2005
2. Oct 6, 2005

### HallsofIvy

One difficulty is defining what you mean by
$$\lim_{\epsilon \rightarrow 0} (t-\epsilon, t+\epsilon)$$!
I don't believe there is any standard definition unless you mean
then intersection of all intervals of the form (t-&epsilon;, t+&epsilon;),
where the intersection is over all possible values of epsilon,
In that case, it should be easy to show that the limit is {t}.

If x is not t, then find an ε less than |x-t|.

3. Oct 6, 2005

### quasar987

Precisely, the problem is the following: knowing that $\delta_n \rightarrow 0$, I want to show that for f integrable and continuous at t, and D_n(x) Dirichlet's kernel (a sequence of function with certain properties),

$$\lim_{n\rightarrow \infty} \int_{t-\delta_n}^{t+\delta_n}f(x)D_n(x)dx = f(t)$$

We have that

$$\lim_{n\rightarrow \infty} \int_{t-\delta_n}^{t+\delta_n}f(x)D_n(x)dx = \lim_{n\rightarrow \infty} \mu \int_{t-\delta_n}^{t+\delta_n}D_n(x)dx$$

where I made use the the mean value theorem for integral (because the properties of D_n(x) allow me to), which means that

$$\mu \in \left[\inf_{x\in [t-\delta_n,t+\delta_n]} f(x), \sup_{x\in [t-\delta_n,t+\delta_n]} f(x) \right]$$

Furthermore, it has been established that the integral of D_n (for any n) about any interval around t is 1. So,

$$\lim_{n\rightarrow \infty} \int_{t-\delta_n}^{t+\delta_n}f(x)D_n(x)dx = \lim_{n\rightarrow \infty} \mu$$

I must show that mu --> f(t), so I must show that in the limit, the interval $[t-\delta_n, t+\delta_n]$ shrinks to {t}.

Would the definition of the limit of an interval you suggested be accepted in view of a proof of this statement?

Last edited: Oct 6, 2005
4. Oct 6, 2005

### Hurkyl

Staff Emeritus
Before I say anything else... you do realize that &mu; is a function of n, right?

Last edited: Oct 6, 2005
5. Oct 6, 2005

### quasar987

Yeah, otherwise the limit of mu would be just mu. If it was the sentence

that made you believe I didn't know mu was a function of n, it was a mistake. Of course, I meant "I must show that mu --> f(t)". I have corrected that in the original post now.

Last edited: Oct 6, 2005
6. Oct 7, 2005

### Hurkyl

Staff Emeritus
Well, that sentence coupled with the approach you're taking to the problem. Why do you think proving that the "limit" of the interval is a singleton will solve the problem?

7. Oct 7, 2005

### quasar987

Because

$$\mu \in \left[\inf_{x\in [t-\delta_n,t+\delta_n]} f(x), \sup_{x\in [t-\delta_n,t+\delta_n]} f(x) \right]$$

If

$$\lim_{n\rightarrow \infty} [t-\delta_n,t+\delta_n] = \{t\}$$

then

$$\lim_{n\rightarrow \infty} \mu \in \left[\inf_{x\in \{t\}} f(x), \sup_{x\in \{t\}} f(x) \right] = [f(t),f(t)]=\{f(t)\}$$

This is the argument Lea uses in her "math for physicists" book. I'm trying to make it rigourous.

Last edited: Oct 7, 2005
8. Oct 9, 2005

### Hurkyl

Staff Emeritus
Ick, that's a horrible way of looking at it, I think.

What you need to look at are the upper and lower bounds for &mu;n.

9. Oct 9, 2005

### benorin

You might try

$$\lim_{n\rightarrow \infty} \int_{t - \delta_{n}}^{t + \delta_{n}} f(x)dx = \lim_{n\rightarrow \infty} \int_{- \infty}^{+ \infty} f(x) \chi_{[t- \delta_{n} , t + \delta_{n}]}(x)dx$$

10. Oct 9, 2005

### benorin

You might try

$$\lim_{n\rightarrow \infty} \int_{t- \delta_n}^{t+ \delta_n} f(x)dx = \lim_{n\rightarrow \infty} \int_{-\infty}^{+\infty} f(x) \chi_{[t- \delta_n,t + \delta_n]} (x)dx = \lim_{n\rightarrow \infty} \int_{-\infty}^{+\infty} f(x) (\chi_{[t- \delta_n,t)} (x) + \chi_{[t,t + \delta_n]} (x))dx$$

where $$\chi_{[a,b]} (x)=\left\{\begin{array}{cc}1,&\mbox{ if } x\in [a,b]\\0, & \mbox{ otherwise } \end{array}\right$$ is the charistic interval of [a,b]. Then use the Lebesgue Dominated Convergence Theorem to show that the limit goes to f(t).

Last edited: Oct 9, 2005
11. Oct 9, 2005

### quasar987

We are not allowed to use anything that has "Lebesgue" in its name.

But why is it a horrible way to look at it? And what does that mean for a start? Is my way of doing it definately wrong, or you simply don't like it?

What do you have in mind when you say "the upper and lower bounds of mu"? Aren't those just 'sup f(x)' and 'inf f(x)' respectively?

12. Oct 9, 2005

### Hurkyl

Staff Emeritus
Because, IMHO, it doesn't resemble what's going on in the proof at all.

One big red flag should be that Lea's argument doesn't involve the continuity assumption on f at all, but that's a very necessary assumption. (Try letting f be a step function with a discontinuity at t)

It is important to observe that the interval in the domain descends to the singleton [t, t], but that last line is far too optimistic. You need to show directly that the lower and upper bounds on &mu; descend to the singleton [f(t), f(t)]. (And then apply the squeeze theorem)

(Yes, when I said the bounds of &mu;, I meant that sup and inf)

13. Oct 9, 2005

### quasar987

Ok, I see what you mean with the "step function counter-exemple". What I need to show is that

$$\lim_{n\rightarrow \infty} \inf_{x\in [t-\delta_n,t+\delta_n]} f(x) = \lim_{x\rightarrow t} f(x)$$

Just to make sure I'm not frying my brain cells for nothing:

does proving that necessitates fiddling around with epilons and deltas? Namely, using the fact that f is continuous at t to obtain a relationship btw epsilon and delta and then using those to show my limit on the infinum?

14. Oct 10, 2005

### Hurkyl

Staff Emeritus
I think that's the easiest way to do it.

15. Oct 10, 2005

### quasar987

I came up with this:

From the continuity of f at t, we get that $\forall \epsilon > 0$, there exists a $\delta (\epsilon) > 0$ such that $f(0\leq |x-t|<\delta(\epsilon)) \subset (f(t)-\epsilon, f(t)+\epsilon)$

Then, note that

$$\lim_{n \rightarrow \infty} \delta_n = 0 = \lim_{\epsilon \rightarrow 0} \delta (\epsilon)$$

Where the epsilon and delta are those from the continuity of f at t.

So I want to show that

$$\lim_{\epsilon \rightarrow 0} \inf_{x \in [t-\delta (\epsilon),t+\delta (\epsilon)]} f(x) = f(t)$$

This is the case because

$$\inf_{x\in [t-\delta (\epsilon),t+\delta (\epsilon)]} f(x) \in f(0\leq |x-t|\geq \delta (\epsilon)) = f([t-\delta (\epsilon),t+\delta (\epsilon)])$$

When we take epsilon --> 0, we end up (following the definition of a limit of an interval given by HallsofIvy) with

$$\lim_{\epsilon \rightarrow 0} \inf_{x\in [t-\delta (\epsilon),t+\delta (\epsilon)]} f(x) \in \{f(t)\}$$

Please tell me what you think.

Last edited: Oct 10, 2005
16. Oct 11, 2005

### benorin

Limit of a sequence of sets

Yes, it is possible. The most applicable notion of a limit of an interval is that of a limit of a sequence of sets, see Halmos' book MEASURE THEORY for an excellent treatment of the topic. In brief: Let A_n denote a sequence of sets. Define

(ignore the commas, just to preserve spacing)

,,,,,,,,,,,,,,,,,infty infty
,,,,,,,,,,,,,,,,,,,, _
lim sup A_n := | | |_| A_m
n->infty,,,,,,,,,,n,,,m=n

and

,,,,,,,,,,,,,,,,infty infty
,,,,,,,,,,,,,,,,,,,,,,,, _
lim inf A_n := |_| | | A_m
n->infty,,,,,,,,,n,,,m=n

Then,

lim A_n = lim inf A_n = lim sup A_n
n->infty,,,n->infty,,,,n->infty

Rather, see the following web page (near the bottom): http://en.wikipedia.org/wiki/Lim_inf

so let d_n -> 0 as n->infty and put A_n = [t - d_n, t+ d_n].

Enjoy!

17. Oct 11, 2005

### Hurkyl

Staff Emeritus
It's not necessarily true that $\delta(\epsilon) \rightarrow 0$, and you don't need that assumption anyways.

Remember that you're trying to show that $\lim_{n\rightarrow \infty} \inf_{x\in [t-\delta_n,t+\delta_n]} f(x) = f(t)$. Just write out the epsilon-delta definition of this limit.

18. Oct 11, 2005

### quasar987

I think I've got it...

First make the limit change $n \rightarrow \infty \Rightarrow \delta_n \rightarrow 0$

From the continuity of f at t, we get that $\forall \epsilon > 0$, there exists a $\delta ^* (\epsilon) > 0$ such that $0<|x-t|<\delta ^* \Rightarrow |f(x)-f(t)|<\epsilon$

We want to show that

$\forall \epsilon > 0, \exists \delta >0$ such that $0<|\delta_n |=\delta_n <\delta$ implies

$$\left| \inf_{|x-t| \leq \delta_n}f(x) - f(t) \right| < \epsilon$$

First, we note that

$$0<|x-t| \leq a \Rightarrow \left| \inf_{|x-t| \leq a} f(x) - f(t)\right| \leq |f(x)-f(t)|$$

and that if we set $\delta = \delta ^*$, then since $\delta_n < \delta ^*$, for $|x-t|\leq \delta_n < \delta ^*$,

$$\left| \inf_{|x-t| \leq \delta_n}f(x) - f(t)\right| \leq |f(x)-f(t)|< \epsilon \ \ \ \blacksquare$$

Last edited: Oct 11, 2005
19. Oct 11, 2005

### quasar987

Except that

$$0<|x-t| \leq a \Rightarrow \left| \inf_{|x-t| \leq a} f(x) - f(t)\right| \leq |f(x)-f(t)|$$

is false.

Last edited: Oct 11, 2005