If $\int_{0}^{\pi}t^{n}f(t)dt=0$ for all $n\in\{0\}\cup \mathbb N$ then prove that $f \equiv0$

Question

Let, $f:[0,\pi ]\to \mathbb R$ be a continuous function such that $f(0)=0$.

If $\int_{0}^{\pi}t^{n}f(t)dt=0$ for all $n\in\{0\}\cup \mathbb N$ then prove that $f \equiv0$.

Since, $f$ is continuous & $t^{n}$ is also continuous & $t^{n}$ keeps the positive sign throughout the interval $[0,\pi]$, so using $1^{st}$ Mean-Value-Theorem of Integral calculus $\exists$ $\xi\in [0,1]$ such that, $$\int_{0}^{\pi}t^{n}f(t)dt=f(\xi)\int_{0}^{\pi}t^{n}dt=f(\xi).\dfrac{\pi^{n+1}}{n+1}.$$

Using given condition we get $\exists$ $\xi \in[0,1]$ such that, $f(\xi)=0.$ From here how we can conclude that $f\equiv0$ ?

Or any other way to prove this?

duplicate question and a solution here avoiding W-approx thm. — sciona, Jan 15 '15 at 17:42

score 3 · Accepted Answer · answered Jan 15 '15 at 17:32

3

Hint: Weierstrass approximation (by polynomials) of continuous functions in compact interval and $\displaystyle \int_0^{\pi} f^2(t)\,dt = 0 \implies f = 0$, tell me if you need more details.

answered Jan 15 '15 at 17:32

sciona

3,710

I want details & where $f(0)=0$ need? – Empty Jan 15 '15 at 17:35
@Panja.S. If $f(0) = 0$ we may approximate $f$ with a polynomial $P(x) = \sum\limits_{k=1}^{n} a_k x^k$ (i.e., without constant terms) uniformly in the interval $[0,\pi]$. I'll add more steps in the solution. – sciona Jan 15 '15 at 17:38
@Panja.S. sorry (should have noticed this before) duplicate from the related column. – sciona Jan 15 '15 at 17:42
If $t^{n}$ is replaced by any other continuous function (like as $\sin x$, $\cos x$ ect.. then are the results valid? – Empty Jan 15 '15 at 17:44
@Panja.S. The basis needs to satisfy a few properties see Stone-Weierstrass Theorem. – sciona Jan 15 '15 at 17:46
How can you do that? Can you please add more steps? – Oct 04 '17 at 19:45

If $\int_{0}^{\pi}t^{n}f(t)dt=0$ for all $n\in\{0\}\cup \mathbb N$ then prove that $f \equiv0$

1 Answers1