Integral question on a polynomial

[member 428835]

Homework 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.

 

[Mark44]

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]

Have you thought about using the Cauchy-Schwartz inequality to disprove it?

 

[member 428835]

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]

What happens if n is large?

 

[member 428835]

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]

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 $\|f\|$.

 

[member 428835]

Gotcha! Thanks. I liked your above comment so if anyone searches their attention is drawn to that one!