I want to prove the theorem:
If $f \in L^1(\mathbb{T})$ and $\hat{f}(n) = 0$ for all $n \in \mathbb{Z}$,then f(x) = 0 for almost everywhere.
This is my current proof,I can only prove that if $f \in L^1(\mathbb{T})$ and bounded case,and I would like everyone to help me check if there are any issues with my proof.
My proof:
Since,
$$\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx} dx = 0 ,n \in \mathbb{Z}$$
So for any trigonometric polynomial p(x),there is :
$$\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)p(x) dx = 0$$
and since the trigonometric polynomials are dense in $L^1(\mathbb{T})$,then for any given $\varepsilon >0$,there is a trigonometric polynomial $p_{\varepsilon}(x)$ satisfies $||f-p_{\varepsilon}||_{1} < \varepsilon$,since the function is bounded then :
$$\begin{aligned} \left|\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) (f(x) - p_{\varepsilon}(x))\right| dx &\leq \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(x) (f(x) - p_{\varepsilon}(x))| dx \\&< M \varepsilon \end{aligned}$$
Therefore,$\int_{-\pi}^{\pi}f^2(x) dx =0 \implies f(x) =0 $ a.e.
- Is my proof correct under the stronger condition?
- How should the original theorem be proven?
Pick some $X \in (-\pi,\pi)$, and let $s_n$ be the continuous function whose graph connects the points $(-\pi,0), (x-{1 \over n}, 0), (x,1), (\pi-{1 \over n}, 1), (\pi, 0)$. Let $p_n$ be a trigonometric polynomial such that $\|s_n-p_n\|_\infty < {1 \over n}$, and let $s=1_{[x,\pi)}$. An application of the dominated convergence theorem gives $\int f s = \int_{-\pi}^x f = 0$. Since this holds for all $x \in (-\pi,\pi)$, the differentiation theorem shows that $f=0$ ae.