Standard notation for "open integral"

Click For Summary
The discussion centers on finding a standard notation for denoting an "open integral" involving a function defined over an integral. The user seeks clarity on how to express the relationship between functions without implying multiplication of the integral result by another function. They highlight the similarity to convolution but emphasize that their question is more general. The conversation touches on the distinction between different forms of notation and their implications in mathematical expressions. Ultimately, the user is looking for established conventions rather than creating new notation.
qu_bio
Messages
4
Reaction score
0
If I have a function defined over an integral, e.g.

## F(t') = \int_{-\infty}^{\infty} dt \tilde{F}(t,t') ##

is there a standard way to denote this integral as being "Open", that is to say if I write

## H = F(t') G(t) ##

I want this to mean

## H = \int_{-\infty}^{\infty} dt \tilde{F}(t,t') G(t) ##

Rather than

## H = [\int_{-\infty}^{\infty} dt \tilde{F}(t,t') ] * G(t) ##

I could make up some notation, but I'd rather not if one exists!

Many thanks
 
Physics news on Phys.org
I don't know too much about this, but I believe a convolution is a closely related concept.
 
certainly said:
I don't know too much about this, but I believe a convolution is a closely related concept.
Yes I suppose it looks similar if F(t,t') -> F(t-t'), but my question is a little more general.
 
qu_bio said:
Yes I suppose it looks similar if F(t,t') -> F(t-t'), but my question is a little more general.
One more difference:-
##\int_{-\infty}^{+\infty}\bar{F}(t'-t)G(t) dt=[\bar{F}\ast G](t')## not ##F(t')##.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K