$\int_{0}^{1}f(x)x^{k}=c^{k}$ 
*

*Let $\,\mathrm{f}:\left[0,1\right] \to \mathbb{R}_{+}$ be a continous function for which exists $c\in \mathbb{R}$, such that $$\int_{0}^{1}\mathrm{f}\left(x\right)x^{k}\,\mathrm{d}x = c^{k}\,,\qquad
k\in {0,1\ldots n}$$

*Prove that $c\in \left[0,1\right]$.

*If $\,\mathrm{g}:\left[0,1\right] \to \mathbb{R}$ is continuous, find $\int_{0}^{1}\mathrm{f}\left(x\right)\mathrm{g}\left(x\right)\mathrm{d}x$.


I tried using Mean Value Theorem.
 A: You can simply use that as $f$ is positive, as $k<k' \Rightarrow x^{k'} < x^k$ for all $x \in [0,1]$, the sequence $\big(\displaystyle{\int_0^1} f(x) x^k\big)_k$ is decreasing. Thus $c\le 1$.
A: Apparently $c\geq 0$ because of that $f$ being nonnegative.
Now use Stone-Weierstrass to find a polynomial $P$ such that $\|f-P\|_{L^{\infty}}<1$, then
\begin{align*}
c^{k}&=\int_{0}^{1}f(x)x^{k}dx\\
&\leq\int_{0}^{1}|f(x)-P(x)|dx+\int_{0}^{1}|P(x)|dx\\
&\leq\dfrac{1}{k+1}+\dfrac{|a_{0}|}{k+1}+\cdots+\dfrac{|a_{n}|}{n+k+1}\\
&<1
\end{align*}
for large $k$, where $P(x)=a_{0}+\cdots+a_{n}x^{n}$, so $c<1$.
A: a) First, as $f$ is continuous, $f$ is bounded, ie $|f(x)|\leq M$ for all $x\in [0,1]$. Then $$|\int_0^1 f(t)t^kdt|\leq M\int_0^1t^kdt=\frac{M}{k+1}$$, we get that $c^k\to 0$, hence $|c|<1$. (and $c\in [0,1[$ as $f\geq 0$)
b) For any polynomial $P(x)=a_0+\cdots+a_s x^s$, we get immediately that $\int_0^1f(t)P(t)dt=P(c)$.
c) Let $g$ a continuous function. By the Stone-Weiertrass theorem, there exist a sequence of polynomials $P_n$ converging uniformly to $g$ on $[0,1]$. Then $f(t)P_n(t)\to f(t)g(t)$ uniformly, and we get that $\int_0^1f(t)P_n(t)dt\to \int_0^1f(t)g(t)dt$, as as the first integral is $P_n(c)$, and that $c\in [0,1]$, we have $\int_0^1f(t)g(t)dt=g(c)$.
d) Now take $g(t)=f(t)[t-c[$. Then $g$ is continous, ansd we have $\int_0^1f(t)^2[t-c|dt=0$. This show that $f(t)^2|t-c|=0$, $f(t)=0$ for $t\not =c$, and by continuity, for all $t$. Hence $c^k=0$ for $k\geq 1$, and $c=0$, $f=0$.  
A: Since $f(x)\geq 0$, the function $M(k)=\int_{0}^{1}f(x)x^k\,dx $ is log-convex over $\mathbb{R}^+$ by the Cauchy-Schwarz inequality (CS ensures the midpoint-log-convexity and $M(k)$ is continuous with respect to $k$). If $M(k)M(k+2)=M(k+1)^2$ for some $k\in\mathbb{R}^+$ (it actually holds for any $k\in\mathbb{N}$, given $M(k)=c^k$) then CS is attained as an equality, hence $f(x)=0$ almost everywhere on $(0,1)$. The continuity of $f$ then implies $f(x)=0$ and $c=0$, so $\int_{0}^{1}f(x)\,g(x)\,dx = 0$.
A: Necessity of $\boldsymbol{c\lt1}$
If $c\ge1$, then Dominated Convergence says that
$$
\lim_{k\to\infty}\int_0^1f(x)\,\left(\frac xc\right)^k\mathrm{d}x=0
$$
However, we are given that
$$
\lim_{k\to\infty}\int_0^1f(x)\,\left(\frac xc\right)^k\mathrm{d}x=1
$$
Therefore, $c\lt1$.

Integral of $\boldsymbol{f(x)\,g(x)}$
Approximate $\left|g(x)-\sum\limits_{k=0}^na_kx^k\right|\le\frac{\epsilon/2}{1+\int_0^1f(x)\,\mathrm{d}x}$, then
$$
\begin{align}
\left|\int_0^1f(x)g(x)\,\mathrm{d}x-\sum_{k=0}^na_kc^k\right|
&=\left|\int_0^1f(x)\left(g(x)-\sum_{k=0}^na_kx^k\right)\mathrm{d}x\right|\\
&\le\int_0^1f(x)\left|g(x)-\sum_{k=0}^na_kx^k\right|\mathrm{d}x\\[9pt]
&\le\epsilon/2
\end{align}
$$
Since
$$
\left|g(c)-\sum_{k=0}^na_kc^k\right|\le\epsilon/2
$$
the triangle inequality gives
$$
\left|\int_0^1f(x)g(x)\,\mathrm{d}x-g(c)\right|\le\epsilon
$$
Therefore, since $\epsilon\gt0$ is arbitrary,
$$
\int_0^1f(x)g(x)\,\mathrm{d}x=g(c)
$$

Impossibility of $\boldsymbol{\int_0^1f(x)\,x^k\,\mathrm{d}x=c^k}$ for $\boldsymbol{c\gt0}$
Note that if $c\gt0$, then for any $\epsilon\gt0$,
$$
\begin{align}
1
&\ge\int_{c+\epsilon}^1f(x)\left(\frac xc\right)^n\mathrm{d}x\\
&\ge\left(1+\frac\epsilon{c}\right)^{n-k}\int_{c+\epsilon}^1f(x)\left(\frac xc\right)^k\mathrm{d}x
\end{align}
$$
Therefore, letting $n\to\infty$, we get
$$
\int_{c+\epsilon}^1f(x)\left(\frac xc\right)^k\mathrm{d}x=0
$$
and by Monotone Convergence,
$$
\int_c^1f(x)\left(\frac xc\right)^k\mathrm{d}x=0
$$
Therefore, by Dominated Convergence
$$
\begin{align}
1
&=\lim_{k\to\infty}\int_0^cf(x)\left(\frac xc\right)^k\mathrm{d}x\\
&=0
\end{align}
$$
Therefore, we must have $c=0$.

If $\boldsymbol{c=0}$ then $\boldsymbol{f(x)=0}$
The hypotheses then imply that for any $\epsilon\gt0$
$$
\begin{align}
0
&=\int_0^1f(x)\,x\,\mathrm{d}x\\
&\ge\int_\epsilon^1f(x)\,x\,\mathrm{d}x\\
&\ge\epsilon\int_\epsilon^1f(x)\,\mathrm{d}x\\
\end{align}
$$
Since $f$ is continuous and non-negative, $f(x)=0$ for $x\ge\epsilon$. Since $\epsilon\gt0$ was arbitrary, and $f$ is continuous, we have $f(x)=0$ for all $x\in[0,1]$.

Note About $\boldsymbol{c=0}$
The section showing that $\int_0^1f(x)g(x)\,\mathrm{d}x=g(c)$ assumes that $0^0=1$. If we agree with this, then $f=0$ fails $\int_0^1f(x)\,x^0\,\mathrm{d}x=c^0$.
However, since there is not a universally accepted definition of $0^0$, one might say that $0^0=0$ and $f(x)=0$ satisfies the original conditions with $c=0$ and that $\int_0^1f(x)g(x)\,\mathrm{d}x=0$.
