Lebesgue integral ?

That picture makes me want to play Marvel vs Capcom 2

>>9153014
ok. whats with my question though ?

>>9153001
Can you express it in math terms?
You have a function f(x) and create a new function
[eqn] g(x)=\frac{1}{\Gamma(x)}\int_0^xf(t)dt[/eqn]
I don't think simply being integrable is enough for the function to belong to a class of functions that admits a basis.
And how would you use Perseval's Identity to "define" scalar multiplication(inner product?)? You mean by defining the inner product as the sum of the products of the coefficients?

>>9153001
>Is that Bullshit or does it make sense ?
It's bullshit.

>>9153040
inner product: yeah thats what I mean.

But write 1/gamma(t) * integral from zero to infinity f(x,t) dx = g(t)

Its not a new function, its kinda just an integral.

>>9153055
Short answer: no.

Long answer: you need some assumption on your function f. Specifically, if F is your integral, you want F over gamma to be in L^2 (or some other Hilbert space).

>>9153040
Do you have an Idea how i can show that it lives in such a complete space ?


I thought like this: the integral is like a summation and f(t) consists of cauchy functions, then its as if you say that the limes in mean or something like that every cauchy function converges to a point in a space L^p, and this space is complete.

because its like a condition for convergence in a complete space, where the this mean converges, its in lives in this space....


sorry, im bad at explaining what i mean

>>9153065
what if f is a cauchy sequence ?

And thanks for the answers so far, its really nice of you.

>>9153066
I'm sorry but you seem really confused about integrals and L^p spaces. If f is an arbitrary function in two variables then integrating over one of them just leaves you with a function in the remaining variable t (what I called F). Your whole question then boils down to whether F/Γ is in L^2 for an arbitrary function F, and the answer is no.

Could you tell me why you're asking this question? What is so special about the gamma function?

>>9153071
Not sure how you would integrate a sequence.

>>9153085
yes i am really confused. why could it not be in L^p with p>=1 ?
>>9153086
e^x is a cauchy sequence, or not ?

>>9153014
GONNA TAKE YOU FOR A RIDE

>>9153095
Pick any function that is not in L^p and call it g. Define F by F=gΓ, then clearly F/Γ is not in L^p (since it's equal to g). Your problem is that F is arbitrary, you need some assumption on it.

e^x is a function, not a sequence.

>>9153105
but e^x =limn-->infty (1+x/n)^n

multiply every term with each other and get a cauchy sequence .. ?

>>9153095
It could, but not without some restrictions on f. If you assume f is L^2 then you could write it in some basis {h_n}, maybe even one that would be nice to integrate but the issue is that when you've integrated and divided by Gamma your inner product might be rough to define.

File: Parallelogram_law[1].png (58KB, 444x592px) Image search: [Google]

58KB, 444x592px

>>9153095
The space L^p with p not equal to do 2 does not have an inner product which is compatible with its distance function. One way to think of it is that there exists a compatible inner product for a metric exactly if the metric satisfies the parallelogram law:
[eqn]2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2[/eqn]
For p not equal to 2 it doesn't.
That an inner product exists (equivalently, that angles make sense) is a very special thing for infinite dimensional vector spaces.

>>9153125
Ok assume the integral divided by the gamma can be written as a cauchy sequence and converges to a function.

Is this function then L^p ?

>>9153120
The sequence (1+x/n)^n is a sequence in n. The limit of this sequence no longer depends on n, it is a function in x.

>>9153131
No, since the limit function could be anything. If it converges to our previous g then it is not in L^p

>>9153141
L^p spaces are complete with respect to the L^p metric, but he was probably talking about convergence pointwise.

>>9153143
yes thats what i mean.

>>9153143
I assumed he was talking about that yes.

>>9153141
and if the limit function is a cauchy sequence ? I mean im not sure if i interpret this correctly....

A convergent sequence is always necessarily a Cauchy sequence. However the converse is not necessarily true. A metric space with the property that any Cauchy sequence has a limit is called complete, see also Cauchy criteria.


property that any Cauchy sequence has a limit

can you maybe give me an example of such a limit function that youre talking about which is in a complete space?

>>9153156
The point I have been trying to get across is that you are starting with arbitrary functions. If you start with a sequence of functions in L^p and if this sequence is Cauchy with respect to the L^p norm (note that the sequence is Cauchy, not the functions themselves), then the limit is in L^p. This is not what you have been asking so far.

>can you maybe give me an example of such a limit function that youre talking about
No problem. You've already figured out yourself that e^x is a limit of a Cauchy sequence, and e^x is not in L^p(R).

>>9153167
Ok thanks. Ill have to try to understand it better but you helped me.