For a Lebesgue integrable functionon $f$ defined on $\mathbb{T}=[0,2\pi]$
$\forall k\in\mathbb{Z}\quad\int_\mathbb{T}f(x)e^{-ikx}\frac{dx}{2\pi}=0$
Is this sufficient to conclude that $f(x)\equiv0 $ almost everywhere?
I just begin to learn this part. I have no idea about this.
Thanks in advance.
The Fourier series of $\chi_{[a,b]}$ converges to $\chi_{[a,b]}$ in such a way that $$ \int_{\mathbb{T}}\chi_{[a,b]}(x)f(x)=\lim_{N\rightarrow\infty} \int_{\mathbb{T}}\left(\sum_{n=-N}^{N}\int_{\mathbb{T}}\chi_{[a,b]}(t)e^{-int}dt e^{inx}\right)f(x)dx \\ = \lim_{N\rightarrow\infty} \sum_{n=-N}^{N}\int_{\mathbb{T}}\chi_{[a,b]}(t)e^{-int}dt \int_{\mathbb{T}}f(x)e^{inx}dx $$ And the right side is $0$ under the assumption that $\int_{\mathbb{T}}f(x)e^{inx}dx=0$ for all $n\in\mathbb{Z}$. So, for such an $f$, $$ \int_{T}\chi_{[a,b]}(x)f(x)dx = 0, \;\;\; \forall a,b. $$ You can then use the Lebesgue differentiation theorem to conclude that $f=0$ a.e. by differentiating with respect to $a$ or with respect to $b$.