Suppose $f : [-\pi, \pi] \to \mathbb C$ is a continuous function such that for all $m \in \mathbb Z $ the Fourier coefficient $\hat f(m) = 0.$ Let $p(x) = \sum_{k=0}^{\infty}a_ke^{ikx}$ be a trig polynomial.
Which of the following is true?
- $\int_{-\pi}^{\pi}|f(x)e^{ikx}|dx =0$.
- $\sum_{k=0}^{\infty}a_k\int_{-\pi}^{\pi}|f(x)e^{ikx}|dx =0$.
One of the following questions is to show $f=0$ so I guess both are true, but I'm trying to show this is true in order to prove $\int_{-\pi}^{\pi}|f(x)|^2dx=0$ by using the fact that for any Riemann integrable function there exists a sequence of trig polynomials $(p_n(x))_{n=1}^{\infty}$ such that
$$ \int_{-\pi}^{\pi}|f(x) - p_n(x)|dx \xrightarrow {n \to \infty} 0$$
And so since $f$ is bounded (say by $M$) then we have:
$$ \int_{-\pi}^{\pi}|f(x)|^2dx \leq M \int_{-\pi}^{\pi}|f(x)-p_n(x)|dx + \int_{-\pi}^{\pi}|f(x)p_n(x)|dx $$
There exists $N \in \mathbb N$ such that for all $N \leq n$ we have:
$$ \int_{-\pi}^{\pi}|f(x)|^2dx < \epsilon + \int_{-\pi}^{\pi}|f(x)p_n(x)|dx $$
For any trigonometric polynomial $P$, we have \begin{align} \int_{-\pi}^{\pi}|f(x)-P(x)|^2\,dx&=\int_{-\pi}^{\pi}|f(x)|^2\,dx-2\text{Re}\left(\int_{-\pi}^{\pi}f(x)\overline{P}(x)\,dx\right)+\int_{-\pi}^{\pi}|P(x)|^2\,dx\\ &= \int_{-\pi}^{\pi}|f(x)|^2\,dx+\int_{-\pi}^{\pi}|P(x)|^2\,dx, \end{align} where the cross term vanishes since $\hat{f}=0$. Now, take a sequence of trigonometric polynomials $P_n$ which converge to $f$ in $L^2$ (actually, for continuous $f$, we can do better and approximate uniformly using Weierstrass’ approximation theorem). This implies \begin{align} 0&=2\int_{-\pi}^{\pi}|f(x)|^2\,dx, \end{align} and hence $\int_{-\pi}^{\pi}|f(x)|^2\,dx=0$.
Note also that although you only asked about continuous $f$, it actually holds for any $f\in L^p$ with $1\leq p<\infty$, because we can first approximate $f\in L^p$ using a continuous $\phi$ (this is where we use $p\neq \infty$). By suitably cutting off the function $\phi$ near the endpoints we can ensure $\phi$ vanishes at (and in fact near) the endpoints while still being a good $L^p$ approximation to $f$ (the reason for cutting off, is so that we get something (which extends to be) periodic). Now, using the Stone-Weierstrass approximation theorem, we can uniformly (and hence in $L^p$) approximate $\phi$ by a sequence of trigonometric polynomials. Thus, every $f\in L^p([-\pi,\pi])$ can be approximated by trigonometric polynomials in the $L^p$ norm.
A-fortiori we can say both (i) and (ii) are true, but you shouldn’t put the absolute values in too early. For example, it might have been conceivable that $\int_{-\pi}^{\pi}|f(x)P_n(x)|\,dx\neq 0$, even though $\int_{-\pi}^{\pi}f(x)P_n(x)\,dx=0$ (of course, the argument above rules out this possibility from the assumption $\hat{f}=0$, but a-priori, we couldn’t have known this).
Edit: Argument using only $L^1$ approximation Anyway, if you only know the $L^1$ approximation fact, then we have to argue slightly differently: for any continuous $f:[-\pi,\pi]\to\Bbb{C}$ and any trigonometric polynomial $P$, we have $\int_{-\pi}^{\pi}f(x)P(x)\,dx=0$, so \begin{align} \int_{-\pi}^{\pi}|f(x)|^2\,dx&=\left|\int_{-\pi}^{\pi}|f(x)|^2\,dx-\int_{-\pi}^{\pi}f(x)P(x)\,dx\right|\\ &\leq\int_{-\pi}^{\pi}\left|f(x)\overline{f}(x)-f(x)P(x)\right|\,dx\\ &\leq M\int_{-\pi}^{\pi}|\overline{f}(x)-P(x)|\,dx, \end{align} where I denote $M=\sup\limits_{x\in [-\pi,\pi]}|f(x)|$. Now, take a sequence of trigonometric polynomials which approximate $\overline{f}$ in $L^1$. Then, the RHS vanishes. Hence, $\int_{-\pi}^{\pi}|f(x)|^2\,dx=0$.
Edit: Generalities
Note that the versions above are not the most general. In the first, I assumed $f\in L^2$, while in the second, I assume $f$ is continuous (though that could have been weakened to $f\in L^{\infty}\cap L^1$). But in fact, the more general theorem is that if $f\in L^1$ is such that $\hat{f}=0$, then $f=0$ a.e.
To argue in this case, note that by hypothesis, $\int_{-\pi}^{\pi}f(x)P(x)\,dx=0$ for all trigonometric polynomials $P$. By the Stone-Weierstrass approximation theorem, for any continuous function $g:[-\pi,\pi]\to\Bbb{C}$ with $g(-\pi)=g(\pi)$, we can find a sequence of trigonometric polynomials $P_n$ which uniformly approximate $g$. So, \begin{align} \left|\int_{-\pi}^{\pi}f(x)g(x)\,dx\right|&=\left|\int_{-\pi}^{\pi}f(x)g(x)\,dx-\int_{-\pi}^{\pi}f(x)P_n(x)\,dx\right|\\ &\leq\int_{-\pi}^{\pi}|f(x)|\,dx\cdot \sup_{x\in [-\pi,\pi]}|g(x)-P_n(x)|, \end{align} and the RHS vanishes as $n\to\infty$. Hence, $\int_{-\pi}^{\pi}f(x)g(x)\,dx=0$. By a suitable variant of the classical “fundamental lemma of calculus of variations”, it follows that $f=0$ a.e.