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  • 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.

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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 $c=0$, $f=0$.

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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 $c\le 1$.

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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 $c<1$.

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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 $\int_{0}^{1}f(x)\,g(x)\,dx = 0$.

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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 $\int_0^1f(x)g(x)\,\mathrm{d}x=0$.

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