if $f$ is real valued continuous function on $[0,1]$ and If $∫_0^1$ $f(x)x^n dx=0$ then $f(x)=0$
Here is an argument…
We can prove that if $f$ is a continuous real-valued function on $[0,1]$ and if $∫_0^1$ $f(x)x^n dx = 0$ for all non-negative integers $n$, then $f(x) = 0$ for all $x$ in $[0,1]$.
To prove this, we will use the Weierstrass approximation theorem, which states that any continuous function on a closed interval $[a,b]$ can be uniformly approximated by polynomials. Since $f$ is continuous on $[0,1]$, we can approximate it uniformly by a sequence of polynomials $p_n(x)$.
Let $ε > 0$ be given. Since $p_n$ converges uniformly to $f$ on $[0,1]$, we can choose an integer $N$ such that $|p_n(x) – f(x)| < ε/2$ for all $x$ in $[0,1]$ and for all $n ≥ N$.
Now, we have:
$∫_0^1$ $f(x)x^n dx$ = $∫_0^1$ $[f(x) – p_n(x)]x^n dx$ + $∫_0^1$ $p_n(x)x^n dx$
Am i in right way? Please give solution or hints .. I'm stuck here…
Best Answer
You now have $p_n$ and $f$ and you know that $|p_n(x)-f(x)|<\frac\epsilon2$ for all $x\in[0,1]$. Let's call $e$ the difference function, i.e. $e(x)=f(x)-p_n(x)$.
Also, note that for any polynomial $p$, you have $$\int_0^1 f(x)\cdot p(x) dx = 0$$ because the above integral is just a linear combination of intervals of the type $\int_0^1 f(x)x^mdx$ which are all $0$.
Now, you have
$$\begin{align} \int_0^1 f^2(x)dx &= \int_0^1 f(x)\cdot (p_n(x) + e(x))dx\\ & = \int_0^1 f(x)p_n(x)dx + \int_0^1 f(x)e(x)dx\\ & = 0 + \int_0^1 f(x)e(x) dx \end{align}$$
can you finish your proof from here?