Hello. Im not a mathematician. I have a question, if my interpretation is right.
I have an integral of a function, divided by a gamma function. It is a new function.
I interpret it like this: because I can integrate the function I can write the new function in a complete space without the loss of generality and write it as a linear combination of the basis functions. Then I can use the Parseval Identity to do scalar multiplication in that space, right ?
Is that Bullshit or does it make sense ?
That picture makes me want to play Marvel vs Capcom 2
ok. whats with my question though ?
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?
>Is that Bullshit or does it make sense ?
It's bullshit.
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.
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).
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
what if f is a cauchy sequence ?
And thanks for the answers so far, its really nice of you.
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?
Not sure how you would integrate a sequence.
yes i am really confused. why could it not be in L^p with p>=1 ?
e^x is a cauchy sequence, or not ?
GONNA TAKE YOU FOR A RIDE
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.
but e^x =limn-->infty (1+x/n)^n
multiply every term with each other and get a cauchy sequence .. ?
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.
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.
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 ?
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.
No, since the limit function could be anything. If it converges to our previous g then it is not in L^p
L^p spaces are complete with respect to the L^p metric, but he was probably talking about convergence pointwise.
yes thats what i mean.
I assumed he was talking about that yes.
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?
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).
Ok thanks. Ill have to try to understand it better but you helped me.