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QED
QED
6:03 PM
@Clash figured it out yet?
 
robjohn
robjohn
@JM I've just noticed that that answer was downvoted. Why was that, I wonder?
 
QED
QED
Lenna is a beauty.
 
J. M.
J. M.
@robjohn Yeah, I don't understand why it was downvoted...
 
Clash
Clash
@QED not really, no... I've been staring at it and looking at the answer below, which says you should get 4 equations, which I do not get... I've been tryting to solve it like this, we have for example a(1, ...) + b(0, ...) -x(0, ...) - y(1, ...), and then I write something like a -y= 0 => a=y. Is this one of the four equations I should be trying to get?
 
J. M.
J. M.
@QED ...and now the image processing community adores her. :)
 
QED
QED
6:05 PM
@Clash, try to write it as a matrix multiplied by a vector [a,b,x,y]
 
yoda
yoda
@JM They tracked her down in Sweden and flew her to an image processing conference (I think to commemorate the 25th anniversary or something)
 
QED
QED
hah
I'd not heard that one Yoda
 
Clash
Clash
@QED done, I guess I do gauß now? let me see what I get
 
J. M.
J. M.
@yoda Oh yeah, I read that; she became a fundraiser or something. She had no idea how famous she became...
 
QED
QED
Well don't just eliminate without know what to do once you've done it
but yes, that's the idea
 
yoda
yoda
6:07 PM
 
t.b.
t.b.
I think this has just become one of my favorite titles of a math paper: A separable somewhat reflexive Banach space with nonseparable dual
 
J. M.
J. M.
@yoda BTW: good thing you decided to write an addendum. Congrats on your first "Nice Answer" here!
@tb "somewhat"? :)
 
yoda
yoda
@JM hah, thanks :) I should spend more time here... Math.se was the first site I joined (much before I went to SO).
Plenty to learn :)
 
Clash
Clash
weird, I got a=-y, b=0, x=y and y could be any value. I must have done something wrong. Here is the matrix that I have
1 0 0 -1
1 1 1 -2
0 3 2 -2
-1 1 -1 2
 
QED
QED
@Clash, when you do row operations you've got to also change the variables around
i.e. "a" might be replaced with "a-b"
"augmented matrix"
 
Clash
Clash
6:14 PM
Can I transpose the matrix and do row operations then?
 
robjohn
robjohn
$\left( \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right)$
@Clash Transpose of what you got :-)
 
Clash
Clash
thanks @robjohn, im trying it right now
i pray for my life
 
robjohn
robjohn
If you have Mathematica, I put that matrix through NullSpace[Transpose[a]]
 
Clash
Clash
oh well, that did not work out either... i hate doing stuff without understand why it should work
 
robjohn
robjohn
@Clash what are you trying to do?
You want to do column ops on your matrix or row ops on mine
 
Clash
Clash
6:21 PM
I have just done row operations on your matrix and got y can be anything, x=-2y, b=5y and a=4y
and I should get x=1, y=1, a=1, b=0
 
Matt
Matt
Anyone interested in discussing some problems regarding extrema of three multivariable functions and lagrange multipliers? :)
 
robjohn
robjohn
$
\left[
\begin{array}{rrrr}
-1&0&1&1
\end{array}
\right]
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
=
\left[
\begin{array}{rrrr}
0&0&0&0
\end{array}
\right]
$
The null space of the matrix has 1 dimension
and $\{\left[
\begin{array}{rrrr}
-1&0&1&1
\end{array}
\right]\}
$ is its basis
 
Clash
Clash
@robjohn. thanks! but why did row operations on that matrix fail me to get the desired results?
 
robjohn
robjohn
what row ops did you do? what were you trying to do with the row ops?
 
Clash
Clash
I got the same result as wolframalpha
http://www.wolframalpha.com/input/?i=NullSpace%5B%7B%7B1%2C1%2C0%2C-1%7D%2C%7B0%2C1%2C3%2C1%7D%2C%7B0%2C-1%2C-2%2C1%7D%2C%7B1%2C2%2C2%2C-2%7D%7D%5D
 
t.b.
t.b.
6:33 PM
 
robjohn
robjohn
@Clash You are getting the right side null space there...
@Clash Try the Transpose of the matrix
 
Clash
Clash
transpose the matrix again? won't i have again
1 0 0 -1
1 1 1 -2
0 3 2 -2
-1 1 -1 2
I do get your results with the matrix above. (-a, 0, a, a)
however he has a different answer here: math.stackexchange.com/questions/25371/…: 1, 0, 1, 1
 
Srivatsan
Srivatsan
Can someone help me? I want to plot some graph and I don't have access to Mathematica. [Wolfram|Alpha times out.]
 
QED
QED
what graph do you want to plot?
 
Srivatsan
Srivatsan
I want to plot the region $6x^3y^2 - 4y^3 + 3x^3y - 4x^6 - x^3y^3 \geqslant 0$, where $0 \leqslant x, y \leqslant 1$.
 
J. M.
J. M.
6:39 PM
Eep, inequalities...
 
Srivatsan
Srivatsan
@JM This one is a beauty. [If I have done everything correctly...]
 
QED
QED
 
Srivatsan
Srivatsan
@QED Gnuplot can do such complicated things?
 
QED
QED
we'll see..
 
Srivatsan
Srivatsan
Hm, interesting.
 
J. M.
J. M.
6:41 PM
The "eep" wasn't for the inequality itself; I was thinking that W|A is the only web-based thing I know that does inequalities...
 
Srivatsan
Srivatsan
@JM Aw, that's right. I am overly dependent on WA.
 
QED
QED
"non-integer passed to boolean operator"
lets fix that
 
Srivatsan
Srivatsan
@QED - what does that mean? Which is the boolean operator?
 
robjohn
robjohn
@Clash are you still here?
 
Clash
Clash
yes @robjohn
 
robjohn
robjohn
6:44 PM
To record our row ops, we start with my matrix and an identity matrix:
$
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
\left|
\left[
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right]
\right.
$
Then we subtract the top row from the bottom row on both
 
QED
QED
wel..
it displays it but it closes immediately afterwards
 
robjohn
robjohn
$
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
0 & 1 & 2 & -1
\end{array}
\right]
\left|
\left[
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-1 & 0 & 0 & 1
\end{array}
\right]
\right.
$
 
J. M.
J. M.
@Srivatsan Tsk, I tried out RegionPlot[6*x^3*y^2 - 4*y^3 + 3*x^3*y - 4*x^6 - x^3*y^3 > 0, {x, 0, 1}, {y, 0, 1}] on W|A and it chokes. Unfortunately I'm not on my Mathematica box...
 
Srivatsan
Srivatsan
@QED Um, =).
@JM Yes, it did choke =)
 
robjohn
robjohn
@Clash can you see my matrices?
 
t.b.
t.b.
6:46 PM
@Srivatsan that's the best I can manage at the moment:
 
Clash
Clash
@robjohn yes i can
 
QED
QED
 
t.b.
t.b.
 
QED
QED
is it supposed to look like that?
 
Srivatsan
Srivatsan
@tb Wow, that's nice.
At least for now. :-)
 
J. M.
J. M.
6:46 PM
...and that's what's called independent confirmation, people. :)
 
Srivatsan
Srivatsan
Thanks, both.
QED used gnuplot. What about you, @tb?
 
t.b.
t.b.
I used grapher.app That's the only tool I can handle a bit.
 
robjohn
robjohn
okay, so then we add the second row to the third, and subtract it from the fourth:
$
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -2
\end{array}
\right]
\left|
\left[
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
-1 & -1 & 0 & 1
\end{array}
\right]
\right.
$
 
Srivatsan
Srivatsan
@tb Oh, ok. Thanks.
@QED As to whether it's supposed to look like that, perhaps. I do not have much of an idea =)
I am basically experimenting.
 
robjohn
robjohn
Then we add the third row to the fourth
$
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0
\end{array}
\right]
\left|
\left[
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
-1 & 0 & 1 & 1
\end{array}
\right]
\right.
$
 
Srivatsan
Srivatsan
6:51 PM
@robjohn In case you are interested, the shaded region in that plot is the feasible region for the (AM, GM, HM) triple for $n=3$, where I have normalised the AM to be $1$. The x-axis represents the GM, and the y-axis is the HM. @tb @JM
 
robjohn
robjohn
@Clash: The matrix on the right is what was multiplied on the left of the original matrix to get the matrix on the left
@Srivatsan Now to $n=4$ :-)
 
Srivatsan
Srivatsan
@robjohn I don't know how to do it. My approach does not extend.
 
J. M.
J. M.
@Srivatsan Looks like a banana. Or a slug. Or a bit of both.
 
Srivatsan
Srivatsan
Btw, thanks to @tb and @QED. =)
 
robjohn
robjohn
@Srivatsan Do you know what the bounding curves are?
 
Clash
Clash
6:54 PM
@robjohn thanks! alright, I also have this result, what does this mean? Why did you the last line of the right matrix as solution?
 
Srivatsan
Srivatsan
@robjohn No clue. It'll just be that huge expression set to zero, right?
 
Matt
Matt
Use the gradient to determine a unit vector normal to the graph of 4ycosx-y^2=3 at the point (pi,1)
 
robjohn
robjohn
@Clash because that row vector times the original matrix gives the $0$ vector
 
Matt
Matt
the gradient here would be <-4ysinx,4cosx-2y>, no?
 
Srivatsan
Srivatsan
robjohn, JM: I can tell you what I did. In case you want to play with it. [Also independent confirmation will be nice.]
 
robjohn
robjohn
6:55 PM
@Clash That is given by the row on the bottom of the left matrix
@Srivatsan sure :-)
 
Clash
Clash
@robjohn why? why don't the other lines matter?
 
yoda
yoda
@JM I personally thought it was a hammock at an angle, but I didn't want to follow your comment because I've learned to not use the words "banana" and "hammock" in close proximity =)
 
robjohn
robjohn
@Clash The other lines show that the rank of the original matrix is 3, but it is the last row that is in the null space of the original matrix
 
J. M.
J. M.
@yoda I can see why. :)
 
Clash
Clash
@robjohn ok so we have -1,0,1,1, why is this still different from the original answer? or are both valid? (ps: many thanks)
 
Matt
Matt
6:58 PM
 
Srivatsan
Srivatsan
For $n=3$, the idea is simple. So we have three numbers: $(A, G, H)$ and three unknowns $(a_1, a_2, a_3)$ such that $A$ is the AM of these numbers, $G$ is the GM, and $H$ is the HM. So we can write a cubic equation such that $a_1, a_2, a_3$ are the roots. This cubic turns out to be
$$
x^3 - 3A x^2 + \frac{3G^3}{H} x - G^3 = 0.
$$
Now write out the condition that this cubic has 3 real, nonnegative roots...
 
robjohn
robjohn
@yoda: ooh, gardening.SE. My wife would love that...
@Clash You still need to process that to get the intersection...
 
Clash
Clash
@robjohn ohhhhhh
 
Srivatsan
Srivatsan
@Matt Cool. It timed out for me. Thanks =)
 
yoda
yoda
@robjohn Hey, you should get her to join our site! We certainly could use more folks :)
 
J. M.
J. M.
6:59 PM
Well, I've capped today, and I'm sleepy. See you all later!
 
robjohn
robjohn
@Clash What we have is a linear combinations of the basis vectors of the two subspaces that equal the same thing (i.e. their difference is 0)
 
Srivatsan
Srivatsan
@JM Bye, JM.
 
robjohn
robjohn
@JM Darn you! I couldn't cap yesterday, and that was the closest I've come recently!
@JM and have a good night :-)
 
t.b.
t.b.
Bye J.M.
 
Srivatsan
Srivatsan
@robjohn Boy, yesterday, I should've got like at least 350 reps from upvotes alone. That plus, 60 from accepts. =)
Well, I just thought you might be interested in this fact... =)
Today I am back to normal though
 
robjohn
robjohn
7:03 PM
@Clash: So from that last line, we see that $
\left[
\begin{array}{rrrr}
-1 & 0 & 1 & 1
\end{array}
\right]
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
=
\left[
\begin{array}{rrrr}
0 & 0 & 0 & 0
\end{array}
\right]
$
@Srivatsan :-p :-p :-p
 
Clash
Clash
ok, what do we have to do to get the intersection?
 
robjohn
robjohn
@Clash: the top two rows and the bottom two rows are the bases for the subspaces. Can you see how to get a vector that belongs to both from that matrix equation?
 
Srivatsan
Srivatsan
By the way, @robjohn, the most interesting feature of the plot for me was that the HM cannot be too large or too small compared to the GM. Of course, there is the $HM \leqslant GM$ bound, but there's some reverse relationship between them as well... That is, $HM \geqslant$ some function of $GM$.
 
robjohn
robjohn
@yoda I will definitely suggest it to her.
 
Clash
Clash
@robjohn no i cant :(
 
Asaf Karagila
Asaf Karagila
7:08 PM
I need to draw a minimal diagram. Some V shaped thing.
Any way to do that without packages?
 
Clash
Clash
@robjohn maybe the upper two bases * [-1 0 1 1] (transposed)?
 
robjohn
robjohn
@Clash: do you see that the negative of the top row plus the sum of the bottom two rows = 0?
 
Jonas Teuwen
Jonas Teuwen
@robjohn Yes it is the OU operator.
 
Clash
Clash
@robjohn yes
 
robjohn
robjohn
@JonasTeuwen Am I correct in thinking that there is some boundary where it ceases to be elliptical?
@Clash: So if I take the sum of the bottom 2 rows (which is in one of the subspaces) and compare it to the top row (which is in the other subspace) what do I see?
 
Matt
Matt
7:12 PM
@Srivatsan Funny. For me it's quite fast.
 
Jonas Teuwen
Jonas Teuwen
@robjohn I didn't check that out.
 
Srivatsan
Srivatsan
@Matt My internet connection isn't too great, but I don't see why that would affect the computation time.
 
yoda
yoda
@robjohn yay! \o/
 
Asaf Karagila
Asaf Karagila
Help with the diagram?
Is it possible without external files, and or packages?
 
Clash
Clash
@robjohn it is the first vector inverted and thus also in the first subspace?
 
robjohn
robjohn
7:15 PM
The first row = the sum of the bottom two rows, so that vector is in the intersection of the subspaces.
 
Clash
Clash
oh I see thats why the first coefficient is 1, the 2nd 0, the 3rd 1 and the 4th 1
oh my god, finally it clicked
jesus, sorry for my slowness robjohn, many thanks for your help!
 
robjohn
robjohn
@Clash: $
\left[
\begin{array}{rrrr}
x & y & 0 & 0
\end{array}
\right]
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
$
is a general vector in the first subspace
@Clash: and $
\left[
\begin{array}{rrrr}
0 & 0 & u & v
\end{array}
\right]
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
$
is a general vector in the other subspace
 
Clash
Clash
x=1, y=0, u=1, v=1
 
robjohn
robjohn
@Clash: so if we can find $
\left[
\begin{array}{rrrr}
x & y & u & v
\end{array}
\right]
\left[
\begin{array}{rrrr}
1 & 1 & 0 & -1 \\
0 & 1 & 3 & 1 \\
0 & -1 & -2 & 1 \\
1 & 2 & 2 & -2
\end{array}
\right]
=
\left[
\begin{array}{rrrr}
0 & 0 & 0 & 0
\end{array}
\right]
$
@Clash: we can break that into a sum of a vector in one subspace plus a vector in the other subspace
@Clash: negate one and they are the same.
@Clash That is the choice to get thet two vectors that are equal :-)
 
Clash
Clash
alright, many thanks!
 
robjohn
robjohn
7:21 PM
Does that clear it up?
 
Matt
Matt
In a latex document, is there any way to prevent it from indenting the first line of a new paragraph?
 
robjohn
robjohn
@Matt \noindent at the beginning of the line
 
Matt
Matt
@robjohn Thanks robjohn, and if I want to turn it off document-wide?
 
robjohn
robjohn
@Matt I think you can change the \parindent variable
 
Matt
Matt
@robjohn I'll try that. Ta!
 
t.b.
t.b.
7:25 PM
@Matt \setlength{\parindent}{0cm}
 
robjohn
robjohn
@Matt search for parindent on this page
 
Matt
Matt
@tb Nice, that did the trick, thanks!
 
Jonas Teuwen
Jonas Teuwen
@tb Hmm, I seem to recall that that messes up other stuff as well.
Or was it the line separation that is often used in conjunction with that?
 
t.b.
t.b.
@JonasTeuwen many people set simultaneously a greater value for \parskip and this messes up lists and enumerations and the like.
(after all, one shouldn't fiddle around with these things anyway)
 
Jonas Teuwen
Jonas Teuwen
Yes.
That was it.
 
robjohn
robjohn
7:29 PM
@tb That is why I just use \noindent :-)
Off to feed the dog. bbl
 
Srivatsan
Srivatsan
See you.
 
t.b.
t.b.
later, robjohn
 
Matt
Matt
Bye robjohn.
 
Asaf Karagila
Asaf Karagila
@robjohn You can feed the dog with Matt's dog.
 
Srivatsan
Srivatsan
@tb, cab we discuss the probability question now?
 
t.b.
t.b.
7:43 PM
Sure.
 
Srivatsan
Srivatsan
What is the question by the way? The way you understand it...
 
t.b.
t.b.
I'm not sure but I believe that the question might be asking why some people impose that the probability function is defined on a $\sigma$-algebra and $\sigma$-additive (Kolmogorov's axioms) and some don't require the $\sigma$.
 
Srivatsan
Srivatsan
@tb And I proved that if $P$ is countably subadditive, then $\sigma$-additivity comes for free. Yes?
 
t.b.
t.b.
Provided that you have finite additivity, yes.
 
Srivatsan
Srivatsan
[Just making sure I am not making some implicit assumption that I don't see.]
 
robjohn
robjohn
7:47 PM
Dog fed...
 
Srivatsan
Srivatsan
So there are probability functions that are not countably subadditive?
 
Asaf Karagila
Asaf Karagila
Wooo! Time to go out for burgers! See you guys later.
 
t.b.
t.b.
Enjoy!
 
Srivatsan
Srivatsan
Bye, Asaf
 
robjohn
robjohn
@AsafKaragila I'll be doing the same soon!
 
t.b.
t.b.
7:48 PM
@Srivatsan Yes. A standard example would be a Banach limit which can be interpreted as a finitely additive measure on $\mathbb{N}$.
 
Srivatsan
Srivatsan
@tb I see.
 
t.b.
t.b.
In fact, the dual space of $\ell^\infty$ can be interpreted as the space of (signed) finitely additive measures on $\mathbb{N}$. Under this identification the $\sigma$-additive measures are precisely the elements of the dual space that come from $\ell^1$ and the probabilities are the non-negative sequences summing to $1$.
 
Matt
Matt
Bye Asaf.
 
Srivatsan
Srivatsan
Wait, I am still hung up on Banach limits.
This is clearly a generalisation of limits, yes. But is this some specific example of some more general concept?
Like, is there a Banach limit for general topological spaces?
 
t.b.
t.b.
Well, yes, there is: Tao can explain this better than me. Also, Martin has some nice related answers.
 
Srivatsan
Srivatsan
8:00 PM
@tb Ok. =) I have seen this post, but I didn't understand much of it.
Anyway...
[Now I am reading your comment.]
@tb How exactly do I interpret the Banach limit as a finitely additive measure on $\mathbb N$?
 
t.b.
t.b.
If you have disjoint sets $A, B \subset \mathbb{N}$ and $L$ is our Banach limit, then by linearity $L([A \cup B]) = L([A]+[B]) = L([A]) + L([B])$ and by normalization $L([\mathbb{N}]) = 1$.
..and furthermore $L \geq 0$, of course.
 
Srivatsan
Srivatsan
Oh, my problem was a bit more basic, but I figured out based on your comment. See if I make sense: Given a set $A \subseteq \mathbb N$, the measure of $A$ is $L(\text{characteristic sequence of } A)$.
 
t.b.
t.b.
exactly.
Now if $A$ is a one point set $L([A]) = 0$, and this shows that $L$ isn't $\sigma$-additive.
 
Srivatsan
Srivatsan
Right. Ok, makes sense.
One tangential question: if $A$ has a natural density, then $L(A)$ will be that natural density, right?
I think I can verify this easily for one simple case. When $A = k \mathbb N$, then I can verify that $L(A) = \frac{1}{k}$.
By natural density, I mean $\lim_{n \to \infty} \frac{|A \cap \{1, \ldots, n\}|}{n}$. I guess this is standard terminology, but not sure how standard.
 
t.b.
t.b.
I'm not quite sure if it is true in general, but you can certainly arrange that it holds.
(natural density certainly is very standard terminology)
 
Srivatsan
Srivatsan
8:14 PM
@tb Ok, that is certainly good.
Ok, back to your original comment.
@tb This makes a lot of sense now.
 
t.b.
t.b.
What was my original comment? :)
 
Srivatsan
Srivatsan
@tb The one my next sentence (the line below) points to... =)
The one about the dual space of $\ell^{\infty}$ and $\ell^1$...
 
t.b.
t.b.
I see.
I don't know what exactly the reasons are, but there are people saying that it is more natural to use finitely additive probabilities instead of $\sigma$-additive ones. Probably some sort of Ockham's razor vs. utilitarian argument. I don't know, I never really thought about it. Anyway, my point was that I think the question might have been about this.
 
Dylan Moreland
Dylan Moreland
I thought that as well. A friend of mine posted some answer using Exts and stuff I'm not so comfortable with and then took it down. Not sure why
 
Srivatsan
Srivatsan
@tb Yes, I see that now. I just wish the OP was clearer though =).
 
t.b.
t.b.
8:23 PM
@DylanMoreland Oh, he accidentally proved the wrong direction.
 
Dylan Moreland
Dylan Moreland
It looked good on the face of it.
Aha.
That's a problem.
 
Srivatsan
Srivatsan
And, thanks for the patient explanation, @tb.
 
t.b.
t.b.
The argument he gave was perfectly fine. Although I wouldn't have mentioned Ext.
@Srivatsan You're welcome.
(Always glad when someone listens :))
 
Srivatsan
Srivatsan
@tb Not sure what that means. Who doesn't listen? =)
 
Dylan Moreland
Dylan Moreland
That's a long list.
@tb I wish the OP for that question had shown a few more attempts. There's certainly stuff you can try.
 
Srivatsan
Srivatsan
8:34 PM
Aw, I see. Natural density of a set is just the "Cesàro limit" of its characteristic sequence. Interesting.
 
t.b.
t.b.
@Srivatsan Exactly. By the way at the end of my answer here I give an outline for the easiest construction of a Banach limit I know of.
 
Dylan Moreland
Dylan Moreland
It does seem like something homological would be much cleaner.
 
t.b.
t.b.
It must be a well known property in homological algebra. What I'm really curious about is: Which injective modules $I$ over a ring have the property that $\operatorname{Hom}{({-},I)}$ reflects exactness?
 
Dylan Moreland
Dylan Moreland
Hm, it's an exercise in Lang. I don't know what that tells me.
Another reason why I didn't think about the problem earlier is that I figured Matt would come along and kill it, but that hasn't happened yet.
 
t.b.
t.b.
Well, Matt wasn't around for four hours, and the question was posted about an hour ago. Tetrapharmakon claims in a comment that this article settles the problem. I don't understand exactly why, yet.
 
Dylan Moreland
Dylan Moreland
8:47 PM
Ah, right, one can check. I always thought that was a funny feature.
I'm surprised that you can't turn it off.
 
t.b.
t.b.
Yeah, I'd prefer to be able to turn that off.
 
Dylan Moreland
Dylan Moreland
I originally started visiting this site just so I could figure out if he was near his computer so I could bug him about math or a grant letter that was due in an hour.
I didn't think I'd use the site proper for anything. But questions that you know the answers to are sort of addictive in a way.
 
t.b.
t.b.
That's too true...
 
Srivatsan
Srivatsan
Shoot. I was so confused reading the conversation because I didn't realise Matt refers to ME. :/
This is not the first time, that too =)
 
Dylan Moreland
Dylan Moreland
Ah, sorry.
 
t.b.
t.b.
8:53 PM
As a rule of thumb: Dylan is around and Matt's hat is dimmed, so it must be Matt E we're talking about.
 
Srivatsan
Srivatsan
@DylanMoreland Of course, no problem. It's just so amusing and obvious once I realise.
@tb - what's his MO handle? I cannot find him there...
 
t.b.
t.b.
 
Srivatsan
Srivatsan
Aw, that's a catch. I don't think I could've guessed that anyway...
If it matters: He is really close to posting 400 answers in MO.
 
t.b.
t.b.
@DylanMoreland I don't get it.
 
Dylan Moreland
Dylan Moreland
Yeah, strange.
 
Srivatsan
Srivatsan
9:06 PM
@tb, Re your answer, I have a basic question.
Given $x \in \mathbb \ell^\infty$ and $y \in \ell^1$ (seen as the dual space of $\ell^\infty$), can you show how $y$ acts on $x$?
 
t.b.
t.b.
$x \mapsto \langle x, y \rangle = \sum_{n=1}^{\infty} x_n y_n$.
 
Srivatsan
Srivatsan
@tb Ok, checks out. Thanks.
I'm trying to verify your claim that no $y \in \ell^1$ will satisfy the OP's requirements. [Check.]
 
t.b.
t.b.
Tell me if you want a hint :)
 
Srivatsan
Srivatsan
It's clear, thanks :)
 
Srivatsan
Srivatsan
9:35 PM
I have a clarification about the proof: a positive linear functional maps positive elements to a nonnegative real number, yes?
What are the positive elements of $\ell^{\infty}$?
 
t.b.
t.b.
Yes // sequences with non-negative entries.
 
Srivatsan
Srivatsan
Alright. Thanks.
Neat proof. =)
But it seems kind of coincidental that the all-ones sequences is at a positive distance from $U$ [where $U$ is the space of sequences of the form $Tx-x$ with $x \in \ell^\infty$].
 
Dylan Moreland
Dylan Moreland
I assume QiL is Qing Liu
Guy is an algebraic geometry machine
his book is great too
 
t.b.
t.b.
@Srivatsan I like it, too. I think it's pretty close to Banach's original argument, but I'd need to check. Two more related threads: here and here
@DylanMoreland Zhen has the same suspicion. See also here and here for some confirmation.
@Srivatsan How would you get close to that sequence by sequences of the form $Tx-x$?
 
Srivatsan
Srivatsan
@tb Well, I could "proof" it, but still it feels a bit weird. [I don't have any issue with the other steps.] I guess digesting this thing will take some time, that's all...
 
t.b.
t.b.
9:49 PM
@Srivatsan My intuition is very silly: let's assume $x_1 = 1$ then in order for $Tx - x$ to be close to $(1,1,1,\ldots)$ we must have $x_2 \approx -2$ thus $x_3 \approx 3$ and so on, so the sequence must be unbounded.
 
Srivatsan
Srivatsan
@tb Yes, is it the case that $U$ is precisely the space of sequences with bounded partial sums?
 
robjohn
robjohn
Ick, how did this get burped back up? I know that there was some tag editing and a title change a couple of days ago, but why is it still in the current main page?
 
t.b.
t.b.
@robjohn Mike Jones
 
Srivatsan
Srivatsan
@tb Oh boy.
 
robjohn
robjohn
Ah, I missed his edit 18 min ago.
 
t.b.
t.b.
9:52 PM
@Srivatsan I think so, yes.
(but I'm not sure)
 
Srivatsan
Srivatsan
conundrum, paradox, enigma, thought experiement, fallaciously
too many big words. =)
Well, $Tx-x = (x_2 - x_1, x_3 - x_2, \ldots, x_{n+1} - x_{n}, \ldots)$. So given a sequence $y$, we can "solve" for the right $x$ such that $Tx -x = y$. The solution is $x = x_1(1, 1, 1, \ldots) + (0, y_1, y_1+y_2, y_1+y_2+y_3, \ldots)$. We want this $x$ to be in $\ell^{\infty}$.
 
robjohn
robjohn
@Srivatsan Which is the dual to sequences of bounded variation (I think)
 
Srivatsan
Srivatsan
@robjohn In what sense? Are you using "dual" in a technical sense?
 
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