0

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 $p$.

Saeed
  • 145
  • What have you tried? If you're stuck, try something easier. Like what if you knew that $f$ was a polynomial? – Arthur Dec 09 '18 at 18:23
  • Moreover, you might want to take a look at the following (almost duplicate) questions linked here – saz Dec 09 '18 at 18:28
  • 2
    @norfair It's not exactly a duplicate because the OP is imposing that $p(a)=p(b)=0$. – saz Dec 09 '18 at 18:28

3 Answers3

2

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 $f=0$.

Richard Martin
  • 1,691
  • 7
  • 8
  • I like the probability-inspired solution! I wouldn't really say that's simple-minded though... – zoidberg Dec 09 '18 at 18:56
  • Well it's like saying that I can make a polynomial look like a delta function, which I dare say would be frowned upon in an Analysis class! – Richard Martin Dec 09 '18 at 19:31
  • Yep, which interpreted probabilistically, is some sort of law of large numbers statement. "Simple-minded" to me is a somewhat self-deprecating statement. Maybe you meant it's simple and elegant, which I definitely agree with. The type of argument p4sch gives is certainly the more standard "analysis" one. – zoidberg Dec 09 '18 at 19:36
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 $x \in [a,b]$.

p4sch
  • 7,863
  • Isn't there a way not to use sub-algebra? I am taking real analysis and want to solve it with that knowledge. – Saeed Dec 16 '18 at 00:03
0

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 $p_n\to f$ 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 $f\equiv 0.$

zhw.
  • 107,943
  • We have no assumption on $f$ in the problem statement, except its continuity. How can you assume $f(a)=0=f(b)$? – Saeed Dec 15 '18 at 23:47
  • Also, could you tell me what you mean by the line joining $(a,q_n(a))$ and $(b,q_n(b))$? Is it $l_n=t(a,q_n(a))-(1-t)(b,q_n(b))$ for all $t \geq 0$? – Saeed Dec 16 '18 at 00:00
  • I did not assume $f(a)= f(b)=0.$ However, g(a)= g(b)=0,$ so the lemma applies to $g.$ Your second question: What is the equation of the line joining two given points? You know the answer to that. – zhw. Dec 16 '18 at 04:55