Application of weierstrass approximation to prove $f(x)=0$

Question

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...

Answer 1

Since $f$ is continuous on $[a,b]$, there exists a sequence of polynomials $(p_n)_n$ such that $\rho(p_n,f)\to 0 $ as $n \to \infty$. Since $p_n = \sum_{k=0}^{n}b_kx^{k}$ for some scalars $b_1,\dots,b_n$, we have that $$\int_a^{b} f(x)p_n(x)dx = \sum_{k=0}^{n}b_k \int_{a}^{b}f(x)x^{k}dx = 0$$ Let $g_n = f-p_n$. Since $p_n$ converges uniformly to $f$, $g_n$ converges uniformly to $0$ and $fp_n$ converges uniformly to $f^2$. Since $f$ is continuous on a compact interval, its bounded and hence the sequence $h_n=fg_n$ converges uniformly to $f\cdot 0\equiv 0$. In fact, $$\rho(h_n,f)=\sup_{x \in [a,b]}{|f(x)g_n(x)|}\leq ||f||_{\infty}\rho(g_n,0) \to 0$$ and $$\rho(f^2,fp_n)=\sup_{x \in [a,b]}|f^2(x)-f(x)p_n(x)|\leq ||f||_{\infty}\rho(f,p_n)\to 0$$ Finally, note that $h_n$ is a sequence of continuous bounded functions converging uniformly to the $0$ function, hence we have $$I =\int_{a}^{b}f^2(x)dx = \int_{a}^{b}\lim_{n \to \infty}f(x)(f(x)-p_n(x)+p_n(x))dx=\lim_{n \to \infty}\int_{a}^{b}h_n(x)+f(x)p_n(x)dx$$ or equivalently, $$\int_{a}^{b}f^2(x)dx = \lim_{n \to \infty}\int_{a}^{b}h_n(x)dx+\lim_{n \to \infty}\int_{a} ^{b}f(x)p_n(x)dx=0$$ Since $f^2$ is continuous and no-negative, we conclude that $f^2\equiv 0 \iff f \equiv 0$.

Answer 2

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?