Let $f:[-1,1] \to \mathbb{R}$ be a continous function such that
$\int_{-1}^1 f(x) x^{2n}dx=0$
for all $n\ge 0$ . Which of the following statements is necessarily false ?
$(1) \int_{-1}^1 f(x)^2 dx=\int_{-1}^1 f(-x)^2 dx$
$(2) (\sup _{x\in [-1,1]}f(x))+(\inf _{x\in [-1,1]} f(x))=0 $
$(3) f(0) \neq 0$
$(4) f(1/2)f(-1/2) \le 0$
My thinking :-
I aim to show that $f$ is necessarily an odd function.
We have, by substituting $u=-x$
$\int_{-1}^1 f(x)x^{2n}dx=\int_{-1}^1 f(-u)u^{2n} du=0$
$\Rightarrow \int_{-1}^1 (f(x)+f(-x)) x^{2n}dx=0$
$\Rightarrow \int_{0}^1 (f(x)+f(-x)) x^{2n}dx=0, \forall n\ge 0 \quad (*)$
Since $g(x)=f(x)+f(-x), x\in [-1,1]$ is an even function.
The restriction $g\big|_{[0,1]}$ is continous and fortunately by this result (Uniform approximation by even polynomial) can be uniformly approximated by a sequence of even polynomials.
But by $(*)$, it is provable that the restriction $g\big|_{[0,1]}=0 $ (zero function) .
Due to the even nature of $g$ , we have $g(x)=0, \forall x \in [-1,1]$ i.e $f(x)=-f(x)$
So $(3)$ is necessarily false, $(1)$ and $(4)$ are necessarily true .
$(2)$ may be true by taking $f(x)=\sin x$ and may be false by taking $ f$ any non-zero constant function.
Are my works correct? Any alternative ideas / solutions will be appreciated.
Thanks a lot for your time !
The essential step is to show that $f$ is an odd function, and you got that right.
You also correctly concluded that (3) is false and (4) is true for any such function.
(1) is true for any (continuous or integrable) functions, as can be seen by substituting $x=-u$.
(2) is true for any odd function, because $$ \sup _{x\in [-1,1]}f(x) = \sup _{x\in [-1,1]}f(-x) = \sup _{x\in [-1,1]}(-f(x)) = -\inf _{x\in [-1,1]}f(x) $$ Your example of a “non-zero constant function” does not work because such a function cannot be odd.