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Integral question on a polynomial

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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.
 
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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.
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.

Regarding your problem, I'm not sure that such a function exists. If we ignore the f(x) part, we get ##\int_0^1 x dx = \frac 1 2, \int_0^1 x^2 dx = \frac 1 3, \dots, \int_0^1 x^n dx = \frac 1 {n + 1}##, and so on.
 

LCKurtz

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Have you thought about using the Cauchy-Schwartz inequality to disprove it?
 
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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.
A function ##f## is in ##L_2[0,1]## if ##\int_0^1 |f|^2\, dx## exists.

Have you thought about using the Cauchy-Schwartz inequality to disprove it?
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 $$1 \leq \frac{1}{2n+1}\int_0^1|f|^2\, dx.$$
But I'm unsure how this is not true.
 

LCKurtz

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What happens if n is large?
 
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What happens if n is large?
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

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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##?
No. You are using an indirect argument, supposing there such an ##f##. It doesn't get to change and it has a finite ##\|f\|##.
 
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Gotcha! Thanks. I liked your above comment so if anyone searches their attention is drawn to that one!
 

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