Functional Analysis Study Group

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Galois in the Field

09:58

@Huy How integral do you think the measure theory appendix is? Maybe I will just refer to it as I need it. If we are all going to study together, should we pick parts of the text to work on?

Huy

@GaloisintheField: As said, I'll go through the chapters one by one, until chapter 4, and then I'll decide which I'll take on next. I think it's quite integral, it pretty short though, so why skip it?

Galois in the Field

@Huy Oh I have been reading it and I get a feeling I'll forget it all, but that is true, it is short

Galois in the Field

10:25

@Huy Chapters one by one means doing all the exercises btw?

Huy

@GaloisintheField: I think the most important results are monotone and dominated convergence, the rest is mostly notation.

Galois in the Field

Oh god I haven't done analysis in awhile and I am having trouble with what I assume is the easiest thing in world, exercise 1.1

I am going to say it is unique, but I don't know how to show it

Huy

@GaloisintheField: do you know how standard "it is unique" proofs work?

Galois in the Field

I can't recall

Huy

also, theoretically, you could skip over the first chapter, it's just a bunch of (cool) examples where functional analysis can be applied. the actual function analysis starts in chapter 2, but you can start in chapter 1 of course

say there's a function $f$ that has certain properties and you want to show it's unique. usually you assume $g$ is another such function and then you use the properties that $f$ and $g$ both satisfy to show $f = g$

or you assume they're different at some point, and then you get a contradiction

Galois in the Field

10:33

Okay I'll give it a shot

I let $e_1(x) = \frac{f(x)+f(-x)}{2}$ and assumed this was not equal to $e_2(x)$. Then:

$$f(x) = e_2(x) + \frac{f(x)-f(-x)}{2}$$
$$\implies e_2(x) = \frac{f(x)+f(-x)}{2} = e_1(x)$$

so they are equal

Hence unique

Huy

@GaloisintheField: For this proof, you already assume $o(x) = \frac{f(x)-f(-x)}{2}$, no?

Galois in the Field

Oh damn, I misunderstood the question as show that the even part was unique puh

Huy

no, you need to show the decomposition into odd and even part is unique :)

Galois in the Field

Okay haha, I promise I won't be asking trivial stuff soon ahaha

Huy

don't worry about it.

where do you study, btw?

Galois in the Field

10:43

Part time at the university of Melbourne

Huy

Galois in the Field

It's a high rated university in Australia

Huy

btw, at the end of the book, there are hints for selected exercises

if you ever try to solve one that you really have no idea how to tackle, maybe you find something useful there

Galois in the Field

Oh that is really good, thanks

Huy

if you don't see how to do it after 15 minutes, just tell me

Galois in the Field

10:49

Okay thanks, hopefully I'll get it :)

\begin{align}
f(x) =& e_1(x)+o_1(x) = e_2(x) + o_2(x)\\
\implies& e_1(x)-e_2(x) = o_2(x)-o_1(x)
\end{align}

Fix $x>0$

Then:

\begin{align}
e_1(x)-e_2(x) =& o_2(x)-o_1(x)\\
\implies& e_1(-x)-e_2(-x) = o_2(-x)-o_1(-x)\\
\implies& e_1-e_2 = o_1(x)-o_2(x)\\
\implies& o_1(x)-o_2(x)=o_2(x)-o_1(x)\\
\implies& o_1(x)=o_2(x)\\
\implies& e_1(x)-e_2(x)=0\\
\implies& e_1(x)=e_2(x)
\end{align}

Done

And hence uniqueness and since the given $e$ and $o$ satisfy the requirements it follows they are equal(by uniqueness) to the ones here

Huy

yeah that should work

Galois in the Field

Now I feel a little better haha

Huy

2 hours later…

Galois in the Field

13:11

What time are you usually on btw @Huy?

Huy