Let $f : [0;1] \to \mathbb{R}$ be a continuous function such that $f(0) = 0$. Which of the following statements are true?

Question

Let $f : [0;1] \to \mathbb{R}$ be a continuous function such that $f(0) = 0$. Which of the following statements are true?
a. If $\int_ 0^{\pi} f(t) \cos nt\, dt = 0,$ for all $n \in {0} \cup \mathbb{N}$, then $f= 0$.
b. If $\int_ 0^{\pi} f(t) \sin nt\, dt = 0,$ for all $n \in \mathbb{N}$, then $f= 0$.
c. If $\int_ 0^{\pi} f(t)\,t^n\, dt = 0,$ for all $n \in \{0\} \cup \mathbb{N}$, then $f= 0$.

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My thoughts:-
(a) it is not true as we can take $f(t)=\sin{nt}$.
(b)no idea. (c)it is true.

Does my answers correct?
and what about (b)

Answer 1

All the three are correct.

c. is true due to Stone-Weierstrass Theorem.

For a. extend $f$ to $[-\pi,\pi]$ as an even function. Then, if a. holds for $\cos nx$ in $[-\pi,\pi]$, by symmetry, and for $\sin nx$ since $f$ is even.

For b. extend $f$ to $[-\pi,\pi]$ as an odd function.

For a. and b. we use the fact that: If $f:[-\pi,\pi]\to\mathbb R$ is continuous and $$ \int_{\pi}^\pi f(x)\,\cos nx\,dx=\int_{\pi}^\pi f(x)\,\sin nx\,dx=0, \quad\text{for every}\,\, n\in\mathbb N, $$ then $f$ is constant.

Answer 2

Parseval's identity shows that a) and b) are true: For a) extend $f$ as an even function; For b), extend $f$ as an odd function.

c) is correct

we don not need $f(0)=0$

there exist $M\gt0$ such that $|f(x)|\lt M$

By Stone-Weierstrass Theorem, $\forall \epsilon \gt0$, there exist a polynomial $P(x)$ such that

$$\int_0^\pi|f(x)-P(x)| dx\lt \epsilon.$$

Since $\int_ 0^{\pi} f(t)\,t^n\, dt = 0 $, we have $\int_0^\pi f(x)P(x) dx=0$, so

$$\int_0^\pi f^2(x) dx = \int_0^\pi f^2(x) dx - \int_0^\pi f(x)P(x) dx = \\ \int_0^\pi f(x) (f(x) - P(x) )dx\leq \int_0^\pi |f(x) (f(x) - P(x) )|dx \\ \leq M \int_0^\pi |f(x) - P(x) |dx \lt M\epsilon$$

Hence, $$\int_0^\pi f^2(x) dx =0$$

so $f\equiv 0$