Does there exist a continuous f ,integration of $\int_{0}^{1} x^n f(x)= 1$ for all $n$

Question

Does there exist a continuous function $f: [0,1]\to[0, \infty )$ such that $$\int_{0}^{1} x^n f(x)= 1$$ for all $n$.

Here is what I tried.

As $f$ is a continuous function on the compact set $[0,1]$ , there exists $m,M \in \mathbb{R}$, $m \leq f(x) \leq M$.

Now applying integration to both side we have $\frac{m}{n+1} \leq 1 \leq \frac{M}{n+1}$ i.e. $m \leq n+1 \leq M$ for all $n\geq 1$.

This contradicts the fact that the set $\mathbb{N}$ is unbounded. It would be great help if you would verify my solution. If the solution is wrong a hint will be greatly appreciated.

Answer 1

This is absolutely correct!

Well done!

Though the contradiction is easier to see than what you said. Just notice that

$$\lim_{n\to \infty} \frac {M}{n+1}=0$$

so you can not have

$$1\leqslant \frac {M}{n+1}$$

for all $n$.