Xrfgtjtk

Suppose $f$ is continuous on $[a,b]$ and $int_a^bf(x)p(x)dx=0$ for every polynomial $p$ with $p(a)=0=p(b)$.

Show that $f$ is identically is zero?

This question is different than If $f$ is continuous on $[a , b]$ and $int_a^b f(x) p(x)dx = 0$ then $f = 0$

Because we have assumption on the end points of $p$.

A more simple-minded approach might be to say: Consider
$$ frac{1}{B(p,q)} int_0^1 x^{p-1} (1-x)^{q-1} f(x) , dx $$
with $B$ denoting the Beta function. Let $p,q to infty$ with $frac{p}{p+q}=mu$ fixed, and the integral is $f(mu)$ in the limit, which will need to be zero. As this works for any $mu$, we have $f=0$.

The set of all polynomials $p colon mathbb{R} rightarrow mathbb{R}$ with $p(a) = p(b)=0$ form a sub-algebra $mathcal{P}_0$ of the space $C_0((a,b))$, the space of real-valued continuous functions on $(a,b)$ which vanish at $a$ and $b$.

1. The polynomial $p(t) := (t-a)(t-b)$ has the property that $p(x) ne 0$ for all $x in (0,1)$, i.e. $mathcal{P}_0$ vanishes nowhere.


2. 
   $mathcal{P}_0$ seperates points, because for $x neq y$ in $(0,1)$ we can define $p(t) := (t-a)(t-b)(t-y)$. Then $p(x) ne 0 = p(y)$.

By the Stone-Weierstraß theorem in the locally compact version the space $mathcal{P}_0$ is dense in $C_0((a,b))$. Since $f$ is bounded, we can conclude that $int_a^b f(x)p(x) , d x =0$ for all $p in mathcal{P}_0$ already implies that
$$int_a^b f(x) g(x) , dx =0 quad text{ for all } g in C_0((a,b)).$$
Now any indicator function of an interval $[c,d]$ with $a<c<d<b$ can be approximated by a sequence of functions from $C_0((a,b))$. Hence we get
$$tag{1}int_c^d f(x) , dx =0.$$
Now the set $mathcal{A}:={A in mathcal{B}((a,b)) colon int_B f(x), dx =0 }$ is a Dynkin-system, which contains a generator of the Borel-$sigma$-Algebra. Therefore we have already $mathcal{A} = mathcal{B}((a,b))$. As known $int_B f(x) , dx =0$ for all $B in mathcal{B}((a,b))$ implies that $f(x) =0$ almost everywhere. Since $f$ is also continuous and nullsets don't contain any interval, we get already $f(x) =0$ for all $x in [a,b]$.

Lemma: Assume $f$ is continuous on $[a,b]$ and $f(a)=0=f(b).$ Then there exists a sequence of polynomials $p_n$ with $p_n(a)=0=p_n(b)$ such that $p_nto f$ uniformly on $[a,b].$

Proof: By Weierstrass there are polynomials $q_n$ converging uniformly to $f$ on $[a,b].$ Let $l_n$ be the linear function joining $(a,q_n(a)),(b,q_n(b).$ Then $p_n=q_n-l_n$ does the job.

Proof for the given problem: Given $a<a'<b'<b,$ define $g$ this way: on $[a,a']$ $g$ is the line joining $(a,0)$ and $(a',f(a'));$ on $[a',b'],$ $g=f;$ on $[b',b]$ $g$ is the line joining $(b',f(b'))$ and $(b,0).$ Find polynomials $p_n$ converging to $g$ uniformly as in the lemma. Then

$$0= int_a^bfp_n = int_a^{a'}fp_n +int_{a'}^{b'}fp_n + int_{b'}^b fp_n $$ $$to int_a^{a'}fg+int_{a'}^{b'}f^2+ int_{b'}^b fg$$

Now let $a'to a^+, b'to b^-$ to see $0=int_{a}^{b}f^2.$ This proves $fequiv 0.$