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Nonnegative linear operators on L^1 and convergence
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eugene
science forum Guru


Joined: 24 Nov 2005
Posts: 331

PostPosted: Wed Jun 28, 2006 11:04 am    Post subject: Nonnegative linear operators on L^1 and convergence Reply with quote
Let T_n : L_1[0,1] -> L_1[0,1] be a sequence of nonnegtaive(i.e. f >= 0
implies T_n(f) >= 0) operators such that || T_n || <= 1 and T_n(f) -> f
in L_1[0,1] norm for f(x) = x and f(x) = 1. (Here = means idenatically
equal to). Prove that T_n(f) -> f for all f from L_1[0,1].

Do you have any ideas, hints fot this problem.

Thanks
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A N Niel
science forum Guru


Joined: 28 Apr 2005
Posts: 475

PostPosted: Wed Jun 28, 2006 12:56 pm    Post subject: Re: Nonnegative linear operators on L^1 and convergence Reply with quote
In article <1151492687.287739.67230@d56g2000cwd.googlegroups.com>,
eugene <jane1806@rambler.ru> wrote:

Quote:
Let T_n : L_1[0,1] -> L_1[0,1] be a sequence of nonnegtaive(i.e. f >= 0
implies T_n(f) >= 0) operators such that || T_n || <= 1 and T_n(f) -> f
in L_1[0,1] norm for f(x) = x and f(x) = 1. (Here = means idenatically
equal to). Prove that T_n(f) -> f for all f from L_1[0,1].

Do you have any ideas, hints fot this problem.

Thanks


If you know what the theorem is called, you can look it up in a
textbook. If you were told the theorem, but not its name, presumably
they wanted you to think of a proof by yourself.
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eugene
science forum Guru


Joined: 24 Nov 2005
Posts: 331

PostPosted: Wed Jun 28, 2006 1:33 pm    Post subject: Re: Nonnegative linear operators on L^1 and convergence Reply with quote
A N Niel wrote:
Quote:
In article <1151492687.287739.67230@d56g2000cwd.googlegroups.com>,
eugene <jane1806@rambler.ru> wrote:

Let T_n : L_1[0,1] -> L_1[0,1] be a sequence of nonnegtaive(i.e. f >= 0
implies T_n(f) >= 0) operators such that || T_n || <= 1 and T_n(f) -> f
in L_1[0,1] norm for f(x) = x and f(x) = 1. (Here = means idenatically
equal to). Prove that T_n(f) -> f for all f from L_1[0,1].

Do you have any ideas, hints fot this problem.

Thanks


If you know what the theorem is called, you can look it up in a
textbook. If you were told the theorem, but not its name, presumably
they wanted you to think of a proof by yourself.

Thank you very, very helfpul answer.

I haven't heard about exactly this theorem with L_1. I've heard about
korovkin's theorem which deal with C(I) for compact interval I and a
sequence of positive operators on C(I).
An in the Korovkin's theorem, as far as i remember, it is the condition
that T_n(x^i) _> x^i
for i = 0,1,2. But here we have L_1 and i = 0,1.
Is there some kind of a more generalized version of Korovkin's theorem
that i haven't heard about ?

Thanks
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David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Wed Jun 28, 2006 4:04 pm    Post subject: Re: Nonnegative linear operators on L^1 and convergence Reply with quote
On 28 Jun 2006 04:04:47 -0700, "eugene" <jane1806@rambler.ru> wrote:

Quote:
Let T_n : L_1[0,1] -> L_1[0,1] be a sequence of nonnegtaive(i.e. f >= 0
implies T_n(f) >= 0) operators such that || T_n || <= 1 and T_n(f) -> f
in L_1[0,1] norm for f(x) = x and f(x) = 1. (Here = means idenatically
equal to). Prove that T_n(f) -> f for all f from L_1[0,1].

Do you have any ideas, hints fot this problem.

I didn't at first, but I think I have a plan.

First, since the T_n are uniformly bounded it suffices
to show that T_n f -> f for f in some dense subspace of L^1.
By linearity it's enough to show that T_n f -> f when
f is the characteristic function of [0,a]. Fix a, and
let f be that characteristic function.

Let g be the characteristic function of [a,1]. Then
T_nf + T_n g -> 1. Since T_n f >= 0 and T_n g >= 0,
also ||T_n f|| <= a and ||T_n g|| <= 1-a it follows
that int T_n f -> a.

Now choose a small number d. Choose A and B so that
the function

h(x) = Ax + B

satisfies h(a) = 1, h(a+d) = 0. So in particular
f <= h, hence T_n f <= T_n h. But T_n h -> h.
So, defining pos(y) = y for y > 0, pos(y) = 0
for y <= 0, the fact that T_n f >= 0 and
T_n f <= T_n h shows that

int_(a+d)^1 pos(T_n f) -> 0

(this is because h < 0 on (a+d,0), so that
int_(a+d)^1 pos(T_n h) -> 0, and 0 <= T_n f <= T_n h.)

Similarly f <= 1 shows that

int_0^a pos(T_n f - 1) -> 0.

So except for parts with small L^1 norm, T_n f
is supported on [0,a], satisfies 0 <= T_n f <= 1,
and has integral tending to a; hence T_n f -> f.

Quote:
Thanks


************************

David C. Ullrich
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