Existence of functions which satisfy some conditions

Does there exists a continous function $f: [0,1]\rightarrow [0, \infty)$ such that: $$\int_{0}^{1}x^{n}f\left(x\right)dx=1$$ for all $n>0$ ?

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No because by the Cauchy-Schwarz inequality, $\int_0^1 x^n f(x)dx\leq \sqrt{\int_0^1 x^{2n}dx}\sqrt{\int_0^1 f^2(x)dx}$,where the second factor is constant whereas the first factor tends to zero. This shows there is not even an $L^2$ function with such a property.

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For you, too: spikedmath.com/170.html – Phira May 13 '14 at 16:14
Thanks a lot... – Mikhail Katz May 13 '14 at 18:28

Suppose for contradiction that such a function exists.

Without resorting to Cauchy-Schwarz, since $f$ is continuous on a closed/compact interval, it is bounded by some positive $M$.

Now $$1=\left|\int_{0}^1f(t)t^n dt \right| \leq \frac{M} {n+1}$$

The RHS goes to $0$ whereas the LHS is $1$.

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I think you mean compact interval. – Gautam Shenoy May 13 '14 at 9:12
Well, if by interval, you mean a closed and bounded interval then yes. You see, even $\mathbb{R}$ is a closed interval. However you made the necessary changes. – Gautam Shenoy May 13 '14 at 9:16
spikedmath.com/170.html – Phira May 13 '14 at 16:13
@Phira I hope he's not mad anymore. – LeGrandDODOM May 13 '14 at 16:30

No. Since $f$ is continuous, There exist a sequence $P_n$ of polynomials such that converges to $f$ uniformly on $[0,1]$. Therefore $$\int_{0}^{1}x^{n}f\left(x\right)dx=\lim_{n\to \infty}\int_{0}^{1}x^{n}P_{n}\left(x\right)dx$$ Now let $$P_{n}(x)=a_nx^n+a_{n-1}x^{n-1}+....+a_0$$ convert left integral to a sum and show that it does not converge to $1$.

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