Somebody help me save this solution

[benorin]

Ok, I have an almost solution, but it fails one requirement at one point for one function !

Problem: ] Find continuous functions $f_{n}: [0,1] \rightarrow [0,\infty),\forall n\in\mathbb{N}$ such that,

i. $f_{n}(x) \rightarrow 0,\forall x\in [0,1]$ as $n \rightarrow \infty,$

and

ii. $\int_0^1 f_{n}(x)dx \rightarrow \infty$, as $n \rightarrow \infty,$

but such that

iii. $g(s):=\mbox{sup} \left\{ f_{n}(s) : n\in\mathbb{N}\right\} = \frac{1}{s}$ on $s\in (0,1]$ so $\lim_{t \rightarrow 0^{+}} \int_t^1 g(s)ds = \infty .$

My (almost) Solution: Let

$$\chi_{A}(x)=\left\{\begin{array}{cc}0,&\mbox{ if } x \in \mbox{NOT}(A)\\1, & \mbox{ if } x\in A\end{array}\right.$$

denote the characteristic function of the set A, where $\mbox{NOT}(A)$ is the complement of A.

Put $$f_{1}(x)=\frac{1}{x} \chi_{(0,1]}(x),$$ and let
$$f_{n}(x)=\frac{1}{nx} \chi_{(\frac{n-1}{n},1]}(x),$$ for $$n\geq 2$$.

Then $$\left\{ f_{n}(x) \right\}$$ satisfies properties i, ii, and iii, except that $f_1$ is not continuous (from the left) at x=0.

Please save it!

 

[matt grime]

why do you even need to define f_1 like that? there is a rathe simple and obvious way to make a function continuous on [0,1] whose sup on that interval is 1, which is after all all you need to do. I've not checked everything else is satisfied, in fact they don't appear to be. since none of the f_n appear to be continuous at x=(n-1)/n

 

[shmoe]

$$f_n$$ has a discontinuity at (n-1)/n for all n. ii) isn't satisfied either, $$f_n$$ is bounded by 1/(n-1) nad is supported on ((n-1)/n,1], an interval of length 1/n so the integrals go to zero as well.

You're trying to capture the sup part in one function, don't.

As a suggestion, ignore the continuity part for now. Break the interval (0,1] into a union of compact intervals where the integral of 1/x over these intervals goes to infinity (as the index of the interval increases). Define f's using this. Then try to satisfy the continuity part by adjusting the f's.

 

[benorin]

Thanks guys! Duh, I'm so tired, I have mathematical tunnel-vision...

P.S. what's tex for not an element of (the negation of $\in$)?

 

[matt grime]

\nin

the latex for negation of \foo is almost always \nfoo

 

[shmoe]

Alternatively you can bypass the "not" by being lazy:

$$\chi_{A}(x)=\left\{\begin{array}{cc}1,&\mbox{ if } x\in A\\ 0, &\mbox{ else }\end{array}\right$$

 

[benorin]

$$\nin$$ = \nin

 

[Gokul43201]

Don't know why that doesn't work, but there's a "not fun" way to do it :

$$\in \hspace{-2ex}\slash$$

Until someone has a nicer solution, that should work.

 

[matt grime]

it could be that \nin is some AMS latex variant and not supported here. it's not as though all latex extensions are, though having xymatrix would be nice, let's see if it works:
$$\xymatrix{ A \ar[r]^{\alpha} & B }$$
evidently not,

 

[shmoe]

What about \notin

$$\notin$$

yep.

 

[AKG]

\not\foo should work in general:

$$\not\in\ \not\exists\ \not =$$

 

[AKG]

I would try to go about this problem constructing the f_n as pictured below. You'd want the pink point to have limit 0 as n goes to infinity (so that f_n(x) -> 0 as n -> infinity, for all x), but have it go to infinity very slowly. On the other hand, you want the blue point to go towards the y-axis very quickly, and the green point to go slowly. This will make sure that for every x, 1/x is the value of some f_n because if you make the blue point go quickly and the green point go slowly, there will always be some overlap between consecutive f_n's so you'll be sure to hit all the 1/x. Moreover, doing it this way, you should be able to have the area under the graphs grow as n increases, which should ensure that the integral of the functions as n goes to infinity is infinity.

 

[benorin]

Doesn't $g(s):=\mbox{sup} \left\{ f_{n}(s) : n\in\mathbb{N}\right\} = \frac{1}{s}\Rightarrow \exists n_{0}\in\mathbb{N}\mbox{ such that } f_{n_{0}}(s)= \frac{1}{s}, \forall s\in (0,1] ?$

Could this particular function be continuous at s=0, or am I delusional?

 

[shmoe]

Doesn't $g(s):=\mbox{sup} \left\{ f_{n}(s) : n\in\mathbb{N}\right\} = \frac{1}{s}\Rightarrow \exists n_{0}\in\mathbb{N}\mbox{ such that } f_{n_{0}}(s)= \frac{1}{s}, \forall s\in (0,1] ?$

No it doesn't. $$f_1(x)=x^{-1}\chi_{(0,1)}(x)$$ but for all n>1 take $$f_n(x)=1$$ for all x. It's no problem producing a continuous example either.

Could this particular function be continuous at s=0, or am I delusional?

If f(x)=1/x on (0,1], then it must be discontinuous at x=0 (assuming it's defined there).

 

[benorin]

Is this more lucid?

So
$g(s):=\mbox{sup} \left\{ f_{n}(s) : n\in\mathbb{N}\right\} = \frac{1}{s}\Rightarrow \forall s\in (0,1], \exists n_{s}\in\mathbb{N}\mbox{ such that } f_{n_{s}}(s)= \frac{1}{s} \geq f_{n}(s) \forall n\in\mathbb{N}?$

Where the subscript of $n_{s}$ emphasizes the notion that n varies with s.

Was that more lucid?

 

[shmoe]

Given that condition i) holds, then yes that's true you must actually hit 1/s with one of the functions and the "sup" could be replaced with "max".

 

[benorin]

Thank you shmoe, matt grime, AKG, et. al. I think I've got a better preception of the problem, and I'll post my next solution for your collective criticism.

 

[benorin]

I would try to go about this problem constructing the f_n as pictured below. You'd want the pink point to have limit 0 as n goes to infinity (so that f_n(x) -> 0 as n -> infinity, for all x), but have it go to infinity very slowly. On the other hand, you want the blue point to go towards the y-axis very quickly, and the green point to go slowly. This will make sure that for every x, 1/x is the value of some f_n because if you make the blue point go quickly and the green point go slowly, there will always be some overlap between consecutive f_n's so you'll be sure to hit all the 1/x. Moreover, doing it this way, you should be able to have the area under the graphs grow as n increases, which should ensure that the integral of the functions as n goes to infinity is infinity.

This suggestion is thus far my only clue, and best hope. It is unfortunately impossible, (assuming that the curves from 0 to the blue dot, and the segment from the green dot to pink dot are linear segments).

Do the curve fitting, make the integral go to zero, and retain the upper bound I require three sequences which satisfy impossible inequalities.

I mean not to complain. Thank you.

I can't figure what this problem has to do with the subject matter contained in the chapter (being Positive Borel measures, the Riesz Representation Theorem, and the like). Any thoughts?

 

[AKG]

Would it help to have f_n(0) not be 0 for all n, rather let it be some number that approaches 0 very slowly? Actually, no, I suppose it wouldn't. It would only contribute an extra triangle of decreasing area.

 

[AKG]

Actually, I think something similar might work, however, it seems so similar to the previous suggestion that either I'm wrong that this one will work or I'm wrong that the previous one doesn't work. Suppose my original suggestion for f_n was a linear map on [0,1/n]. We want to add something on to this, but let everything to the right of 1/n remain the same. Now the line segment from (0,0) to (1/n,n) has slope n². Draw a line parallel to this line segment, intersecting the y-axis at n^-1/4. Also, draw a line parallel to the x-axis, intersecting the y-axis at n. Now the y-axis, the two lines we just drew, and the segment from (0,0) to (1/n,n) form a trapezoid. I believe that the height of this trapezoid is on the order of n^-3/4. The segment (0,0) to (1/n,n) has length on the order n. So the area is on the order of n^1/4. This goes to infinity as n goes to infinity, as desired.

On the other hand, over half of this trapezoid is the triangle I suggested before. But that triangle would have a base of n^-1/4 and a height of 1/n, so it's area would be 0.5n^-5/4 which goes to 0 as n goes to infinity. So what's wrong here?

 

[benorin]

I thinking I need a sequence of functions bounded above by 1/x and converging non-uniformly to the dirac delta at x=0 so that the limit of the integral is zero, yet... crap: I don't know. This problem is... <groan>.

 

[benorin]

I made a typo in the original post: my bad.

ii. should read $\int_0^1 f_{n}(x)dx \rightarrow 0$

instead of $\int_0^1 f_{n}(x)dx \rightarrow \infty$

 

[AKG]

I think some modification of the thing in post #12 should work. If you look at each f_n having three parts: 2 triangles and the 1 area bounded above by 1/x, then make the area in the middle go from x=1/(n+1) to 1/n. This will contain an area of log(ⁿ⁺¹/_n) which of course goes to 0 as n goes to infinity. Instead of having the left-triangle have one vertex at (0,0), give it a vertex at (x_n,0) where x_n is so close to 1/(n+1) that the area of the triangle goes to 0:

0.5(n+1)(1/(n+1) - x_n)
= 0.5(1 - x_n(n+1))

So you want to choose x_n such that 0 < x_n < 1/(n+1) but such that x_n(n+1) -> 1 as n -> infinity. Try x_n = (n + 1 + n^-1)^-1. Do something similar for the right side triangle.

 

[benorin]

I had a similar idea, only I used fixed end points (that time), but I had the problem that the right leg of the right triagle went above 1/x for certian values of n using the [1/(n+1) , 1/n] interval for the portion under 1/x.

I am thinking that this way too much work without using any theorems, then again, it is a grad class and my prof. tweaked the question from the text. I'll check out your lastest post: thank you very much AKG!

 

[AKG]

The slope of the 1/x curve at x=1/n is -n². If you are at (1/n,n) going down at a rate of -n², you will hit the x-axis at x=2/n. Choose a point for the right-most point of your right triangle between 1/n and 2/n. You shouldn't be choosing something bigger than 2/n in the first place. I would recommend something like (n - n^-1)^-1. The area of this triangle would then go to zero as n increases.

 

[benorin]

nice.

What do you mean "my message is too short"?