/sci/ - Science & Math

subtract them to get

<span class="math">\int_0^1 (f(y)-g(y))K(x,y)\,dy = g(x)-f(x)[/spoiler]

multiply by K(x,z), then integrate to get

<span class="math">\int_0^1 \int_0^1 (f(y)-g(y))K(x,y)K(x,z)\,dy\,dz = f(x)-g(x)[/spoiler]

Since K is positive this implies <span class="math">f(y)-g(y)[/spoiler] is zero.

Nope. Please insert a card. Please try again.

Homework threads are against the rules. Please stop posting them.

(USER WAS WARNED FOR THIS POST)

>>4338169
Go to hell.

Sincerely, mathfag.

it's pinned, ya know..
.>>4338169

>>4338169
You must be new here...

>>4338371
First week here. So what?

reported for homework enjoy your ban

>>4338507
someone does not know what the thumbtack means huh?

Sage.
Didn't mean to bump this thread

>>4338533
What are you babbling on about?

>>4338546
countersage

>>4338547
You sad cunt

>>4338546
How does no one remember you?

>>4338554
What do you mean? I think my friends, family members, lover, and (some) professors do.

>>4338558
>implying any professors remember their students.

>>4338562
As a pre-med you have to do a lot of sucking up/off to get those letters of recommendations so, yes they remember me.

>>4338566
>implying your blowjobs are THAT memorable

>>4338575
>implying mine aren't

>>4338580
You don't usually hear of engineering as a premed field.

>>4338586
;) the times are changing

>>4338605
Huh. Well, my doctor might hit on me, but if I'm a cyborg, I might just be okay with that tradeoff.

>>4338609
>2012
>not being a gynoid

Step your shit up

>>4338616
Gynoids aren't yet advanced enough to be me.

>that feel when we traded EK threads for blackman threads

>>4338656
You're welcome

I love this problem. The solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint, positive-definite operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (positive-definiteness follows easily from the fact that <span class="math">K[/spoiler] is strictly positive and continuous on its compact domain, and therefore bounded away from 0).

A well-known result from functional analysis:
If <span class="math">\mathcal{A}[/spoiler] is a compact, self-adjoint, positive definite operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction (this is a consequence of <span class="math">\mathcal{A}[/spoiler] being self-adjoint, and therefore having orthogonal eigenfunctions: two positive functions cannot possibly be orthogonal).


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

I love this problem. The solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint, positive-definite operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (positive-definiteness follows easily from the fact that <span class="math">K[/spoiler] is strictly positive and continuous on its compact domain, and therefore bounded away from 0).

A well-known result from functional analysis:
If <span class="math">\mathcal{A}[/spoiler] is a compact, self-adjoint, positive definite operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_\mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction (this is a consequence of <span class="math">\mathcal{A}[/spoiler] being self-adjoint, and therefore having orthogonal eigenfunctions: two positive functions cannot possibly be orthogonal).


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

[oops. I posted a solution and then deleted it after I realized it was incomplete.]

This solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (self-adjointness requires that <span class="math">K[/spoiler] be symmetric). <span class="math">\mathcal{K}[/spoiler] is also a positive operator in the sense that <span class="math">u>0\mathrm{ a.e. }\Rightarrow \mathcal{K}u>0[/spoiler] a.e.

A well-known result from functional analysis (Krein-Rutman):
If <span class="math">\mathcal{A}[/spoiler] is a compact, positive operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_\mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction.

[Here's a link to a proof: http://www.worldscibooks.com/etextbook/5999/5999_chap1.pdf ]

>>4338788
[cont'd]


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact, self-adjoint and positive-definite with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

>>4338788
[cont'd]

Note that <span class="math">\mathcal{K}^2[/spoiler] is compact, self-adjoint and positive with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

>>4338797
Why is is K^2 compact?

>>4338805
The composition of a bounded operator and a compact operator is compact.

>>4338806
Then why is K compact?

>>4338797
>Note that K^2 is compact, self-adjoint and positive
oops. Disregard the "self-adjoint". It isn't needed, and it would only obviously be true if <span class="math">K(x,y)=K(y,x)[/spoiler] for all <span class="math">x,y\in[0,1][/spoiler].

>>4338813
<span class="math">\mathcal{K}[/spoiler] maps <span class="math">L^2[/spoiler] bounded sets to precompact sets:

Suppose <span class="math">u_n[/spoiler] is a bounded sequence in <span class="math">L^2(\Omega)[/spoiler], <span class="math">||u_n||_{L^2}\leq C[/spoiler].

Using the Arzelà–Ascoli theorem, you can show that <span class="math">Ku_n[/spoiler] must have an <span class="math">L^2[/spoiler]-convergent subsequence.

>>4338837
Sorry, I'm getting myself confused with something else (it's late — that's my excuse). You can show that <span class="math">\mathcal{K}[/spoiler] is the limit of finite rank operators. I'll omit the details, but any book that discusses compact operators will include this as its first or second example. All that's required is that <span class="math">||K||_{L^2([0,1]\times[0,1])}<\infty[/spoiler].

>>4338877
Hey, looks like you got yourself a tripcode :)
Also good job with the proof, which is clearly beyond my limited abilities in analysis.

>>4339176

Quality post. Five stars. Would read again.

>>4339757
And your post is good because you saged a sticky, or because your off-topic is bitter while his is friendly? I'm really interested.

>>4339176

>>4339757

>/sci/ today

Fuck this, go spew your "umad? huehuehue" shit, I don't give a fuck, because I am mad. I can't fucking enjoy a good Putnam QotD without you autistic fucks ruining every thread possible.

Fuck you /sci/.

I think there is a more basic-calculus way to do this:

>>4340470

Holy shit, at least do it in word.

>>4340163
>>4339757

samefag detected

>>4340480

Bah, it is plenty legible.

>>4340470
I don't see any reason you can assume <span class="math">K(x,y)=A(x)B(y)[/spoiler]. Not all functions of two variables separate in this way (<span class="math">\cos(x+y)[/spoiler], for instance).


>>4339176
>Hey, looks like you got yourself a tripcode :)
Yes, and you can probably guess where I got the idea to enclose a short, relevant word between two /'s.

>>4340517
We'll never guess, a8se

>>4338788
>>4338797
Screenshotted and put into my fap folder.