# Abstraction

royalchicken
Posers and Puzzles 22 Oct '05 22:03
1. royalchicken
CHAOS GHOST!!!
22 Oct '05 22:031 edit
Following on something I just read:

Here's one that's a total triviality if you know what an inner product space is and quite difficult if not:

If f:[0,pi] --> R is continuous then the square of the integral of f(x)sin(x) over the domain of f is at most the integral of f^2(x)sin(x) over the same interval, times some constant. Find the smallest constant for which this inequality holds for all such f.

There are all sorts of similar inequalities, but I chose this one because it is hard but not impossible to do straight out, without the abstraction which makes it very easy.
2. PBE6
Bananarama
01 Nov '05 20:16
Originally posted by royalchicken
Following on something I just read:

Here's one that's a total triviality if you know what an inner product space is and quite difficult if not:

If f:[0,pi] --> R is continuous then the square of the integral of f(x)sin(x) over the domain of f is at most the integral of f^2(x)sin(x) over the same interval, times some constant. Find the smallest ...[text shortened]... is hard but not impossible to do straight out, without the abstraction which makes it very easy.
Hey royalchicken, I tried looking up "inner product space" on mathworld.wolfram.com, and it said the following:

"A vector space together with an inner product on it is called an inner product space."

It also gave an example which seemed pertinent:

"3. The vector space of real functions whose domain is an closed interval [a,b] with inner product:

= int[f*g*dx] (a|b)"

It seems to me that this is saying if you have two functions f and g, and their inner product (calculated the same way as the dot product = x^2 + f*g?) is equal to the above integral when evaluated from a to b, then you have an inner product space. But not any f(x) will produce an inner product space in the question as written, will it?

I need a hint. ðŸ˜•
3. royalchicken
CHAOS GHOST!!!
01 Nov '05 23:56
The integral you give takes the place of the dot product in the 'space' of functions defined on [a,b]. In other words, the fact that that collection of functions is what's known as a vector space over the real numbers, and this vector space has the same structural properties as more familiar 'space', we can treat the functions as though they are vectors in space.

I'm not going to give a hint for the inner product space approach (beyond saying 'Cauchy-Schwarz'ðŸ˜‰ unless you really want it; I posted this because I wanted to see if anyone can do it straight-up calculus style, because I can't.
4. 15 Nov '05 04:27
Originally posted by royalchicken
The integral you give takes the place of the dot product in the 'space' of functions defined on [a,b]. In other words, the fact that that collection of functions is what's known as a vector space over the real numbers, and this vector space has the same structural properties as more familiar 'space', we can treat the functions as though the ...[text shortened]... d this because I wanted to see if anyone can do it straight-up calculus style, because I can't.
hello
no clue, but long time no see.
i just droped by the fourms to remind myself how to format for italics
say you.
isn't the Cauchy horizon the bondry in a space-time continum between where one can traval to in a time machien and where one can not? if so, i've heard they can be quite odd/leathal.
5. royalchicken
CHAOS GHOST!!!
15 Nov '05 14:22