Real analysis polynomial approximation proof

Question

Let be $f:[a,b] \rightarrow \Bbb{R} $ continous with $\int_a ^b f(x)p(x)dx = 0$, $\forall p(x)$ polynomial.

Using Stone-Weiestrass theorem, show that:

$f(x)=0,\ \forall x\in [a,b]$.

First, I'll asume this refers to

Theorem (Weiestrass Aproximation)

$\forall f \in \mathscr C[a,b], \exists$ $\{p_n\}_\Bbb N$ shuch that

$\{p_n\}_\Bbb N$ converges to $f(x), \forall x\in [a,b]$

but still, I can't find a way to prove it.

Answer 1

Easy facts: 1. If $g_n \to g$ uniformly on $[a,b]$ and $f$ is bounded on $[a,b],$ then $g_nf\to gf$ uniformly on $[a,b].$ 2. If $f_n$ is a sequence of Riemann integrable functions on $[a,b],$ and $f_n \to f$ uniformly on $[a,b],$ then $f$ is Riemann integrable on $[a,b]$ and $\int_a^b f_n \to \int_a^b f.$

In our problem, Weierstrass shows (we don't need Stone here, with all due respect), there is a sequence of polynomials $p_n$ such that $p_n\to f$ uniformly on $[a,b].$ Thus by the above, $p_nf \to f^2$ uniformly on $[a,b].$ Therefore $\int_a^b fp_n\to \int_a^b f^2.$ Since each $\int_a^b fp_n = 0,$ we have $\int_a^bf^2 = 0.$ The continuity of $f$ then implies $f\equiv 0.$

Answer 2

Sketch: suppose $f$ is not identically zero and $p$ is a polynomial which is uniformly close to $f$. It is still a polynomial, so

$$\int_a^b f(x) p(x) dx = 0$$

by assumption. We can "add and subtract" to rewrite this same integral in terms of the small quantity $f-p$:

$$\int_a^b f(x) p(x) dx = \int_a^b p(x)^2 dx + \int_a^b (f(x)-p(x)) p(x).$$

Now that second term on the right side of the last equation goes to zero as $p \to f$, while the first term instead converges to $\int_a^b f(x)^2 dx > 0$. Thus the left side is zero while the right side is positive, which is a contradiction.

Sprinkle epsilons as needed and you have a rigorous proof.

Answer 3

Let $\varepsilon > 0$. Choose a polynomial $p$ such that for all $x \in [a,b]$, $|p(x) - f(x)|<\varepsilon$. Then \begin{align*} \varepsilon^2(b-a) &≥ \int_a^b(f(x)-p(x))^2dx\\ &= \int_a^bf(x)^2dx+\int_a^bp(x)^2dx-2\int_a^bf(x)p(x)dx\\ &= \int_a^bf(x)^2dx+\int_a^bp(x)^2dx\\ &≥ \int_a^bf(x)^2dx ≥ 0\\ \end{align*} This means $\int_a^bf(x)^2dx = 0$, but if $f(x)$ was non-zero somewhere but also continuous you know that $\int_a^bf(x)^2dx > 0$. Hence $f(x) = 0$ for all $x \in [a, b]$.