we associate to every $f$ $\in$ $L^1(T)$ a function $\hat f$ on $Z$ defined by
$\hat f $ $=$ $\frac {1}{2\pi}$ $\int_{-\pi}^{\pi}$ $f(t)$$e^{-int} dt$ $(n \in Z)$.
It is easy to prove that $\hat f $ $\to$ $0$ as $|n|$ $\to$ $\infty$, for every $L^1$. For we know that $C(T)$ is dense in $L^1(T)$ and that the trigonometric polynomials are dense in $C(T)$. If $\epsilon$ $\gt$ $0$ and $f$ $\in$ $L^1(T)$, this says that there is a $g$ $\in$ $C(T)$ and a trigonometric polynomial $P$ such that $||f-g||_1$ $\lt$ $\epsilon$ and $||g-p||_{\infty}$ $\lt$ $\epsilon$.
Since
$||g-p||_1$ $\leq$ $||g-p||_{\infty}$
it follows that $||f-p||_1$ $\lt$ $2\epsilon$; and if $|n|$ is large enough (depending on P), then
$|\hat f(n) |$ $=$ $|\frac {1}{2\pi} \int_{-\pi}^{\pi}$ { $f(t) - p(t) $ } $e^{-int} dt $ | $\leq$ $||f-p||_1$ $\lt$ $2\epsilon$.
I don't understand why is $|\hat f(n)|$ equal of $|\frac {1}{2\pi} \int_{-\pi}^{\pi}$ { $f(t) - p(t) $ } $e^{-int} dt $ ?
Any help would be appreciated.
The identity is not true for all $n$ but just for sufficiently large ones. Since $p$ is a trigonometric polynomial, it has a finite degree $N$. If $|n|>N$, then $e^{int}$ is orthogonal in $L^2$ to every term in $p$, and thus with $p$ itself. In other words, $$\int_{-\pi}^\pi p(t)e^{-int}dt=0.$$ More explicitly, $p$ takes the form $$p(t)=\sum_{|k|\leq N}a_ke^{ikt}$$ for some complex coefficients $a_k$. Recall the well-known orthogonalities $$\int_{-\pi}^\pi e^{imt}e^{-ikt}dt=2\pi\delta_{m-k}$$ (the Kronecker delta). Thus, using that $|n|>N$, $$\int_{-\pi}^\pi p(t)e^{-int}dt=\sum_{|k|\leq N}a_k\int_{-\pi}^\pi e^{ikt}e^{-int}dt=2\pi\sum_{|k|\leq N}a_k\delta_{k-n}=0.$$ The point is that since $|n|>N$, we can never have $k=n$ in the final sum, so every term is zero.