The discussion revolves around the existence of a function \( f \in L_2[0,1] \) that satisfies the integral condition \( \int_0^1 x^n f(x) \, dx = 1 \) for all natural \( n \). Participants are exploring concepts related to functional analysis and properties of integrable functions.
The discussion is ongoing, with various interpretations and approaches being explored. Some participants express uncertainty about the existence of the function and the implications of the Cauchy-Schwartz inequality. There is no explicit consensus, but several productive lines of reasoning are being examined.
Participants are grappling with the definitions and properties of functions in \( L_2 \) spaces, as well as the implications of the integral condition across natural numbers. The discussion includes considerations of limits and the behavior of functions as parameters change.
Please remind me (it's been a really long time since I've dealt with this stuff) what ##f \in L_2[0,1]## means. One of my topology books used ##\mathcal l^2##, but this had to do with Hilbert spaces and sequences that were square-integrable.joshmccraney said:Problem Statement: Does there exist a function ##f \in L_2[0,1]## such that for all natural ##n##, we have $$\int_0^1 x^n f(x) \, dx = 1$$
Relevant Equations: Nothing comes to mind.
At first I was thinking about using the dirac delta function ##\delta(x-1)##, but then I recalled ##\delta \notin L_2[0,1]##. Any ideas? I'm thinking no such function exists.
A function ##f## is in ##L_2[0,1]## if ##\int_0^1 |f|^2\, dx## exists.Mark44 said:Please remind me (it's been a really long time since I've dealt with this stuff) what ##f \in L_2[0,1]## means.
The Cauchy-Schwartz inequality for this case would be $$\int_0^1| f x^n| \, dx \leq \int_0^1|f|^2\, dx\int_0^1|x^n|^2\, dx = \frac{1}{2n+1}\int_0^1|f|^2\, dx$$ and if we assume such a function ##f## exists, then we have $$1 \leq \frac{1}{2n+1}\int_0^1|f|^2\, dx.$$LCKurtz said:Have you thought about using the Cauchy-Schwartz inequality to disprove it?
But isn't it possible ##n\to\infty \implies f \to \infty##, and so balances? Like if ##f = n##, or would this imply ##f \notin L_2##?LCKurtz said:What happens if n is large?
No. You are using an indirect argument, supposing there such an ##f##. It doesn't get to change and it has a finite ##\|f\|##.joshmccraney said:But isn't it possible ##n\to\infty \implies f \to \infty##, and so balances? Like if ##f = n##, or would this imply ##f \notin L_2##?