Let $f\in L^{2} (\mathbb R)$ with $\lim_{t\to \pm \infty} f(t)=0.$ Put, $F_{n} (x)= \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt, \ (n=1,2,...).$ Fix $\alpha \in (0, \infty)$ and we define $H_{n}(x)$ as follows:
$$\frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt = (e^{-i\alpha x}- e^{i\alpha x})F_{n}(x) + H_{n}(x).$$
My Question: How to show $H_{n}(x) \to 0$ as $n\to \infty$ in $L^{2}(\mathbb R)$ ? (I guess some where we need to use Parseval' identity( Plancherel); but I am bit confused, how to use it.)
Edit: @LutzL, pointed out below, and in MO, for the same, I have tried little in that direction:
$$H_{n}(x)= \frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt -(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt$$ we may write, $H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt-(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt;$ Or, we may write,
$H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt- \frac{1}{2\pi}\int_{-n-\alpha}^{n+\alpha} e^{itx} f(t+\alpha) dt + \int_{-n+\alpha}^{n+\alpha} e^{itx} f(t-\alpha) dt= I_{1}-I_{2}-I_{3}+I_{4};$
But then,(using one of the above form), we need to show: $\lim_{n\to \infty}\int_{\mathbb R} |H_{n}(x)|^{2} dx=0.$ Now If I interchange the limit and integral (this may not be possible, here, I am doing without rigours), and then I can take limit inside the square, and then after taking limit, each integrals, $I_{i}, (i=1,2,3,4)$, will be becomes over $\mathbb R$; then I think, it will cancels each other;
Is this o.k or am I missing something, still, I don't see, where we need to use asymptotic behaviours of $f$ (and what does it mean ?, here is $f$ is arbitrary member of $L^{2}$) ? I am wondering, is it possible to get solution without, Plancherel or parsevel somewhere ?
Copying the MO answer, You can use that
$$\int_{-n}^ne^{itx} f(t+α)\,dt=\int_{-n+α}^{n+α}e^{i(t-α)x}f(t)\,dt =e^{-iαx}\int_{-n+α}^{n+α}e^{itx}f(t)\,dt$$
so that
$$\int_{-n}^ne^{itx} f(t+α)\,dt-e^{-iαx}\int_{-n}^{n}e^{itx}f(t)\,dt =e^{-iαx}\left[\int_{n}^{n+α}-\int_{-n}^{-n+α}\right]e^{itx}f(t)\,dt$$
and analogously
$$\int_{-n}^ne^{itx} f(t-α)\,dt-e^{iαx}\int_{-n}^{n}e^{itx}f(t)\,dt =e^{iαx}\left[\int_{n-α}^{n}-\int_{-n-α}^{-n}\right]e^{itx}f(t)\,dt$$
so that in total
$$|H_n(x)|\le \left[\int_{-n-α}^{-n+α}+\int_{n-α}^{n+α}\right]|f(t)|\,dt\le \sqrt{4α}\sqrt{\left[\int_{-n-α}^{-n+α}+\int_{n-α}^{n+α}\right]|f(t)|^2\,dt}$$
and because of $f\in L^2(\mathbb R)$ we have that
$$\sum_{n\in\mathbb Z}\int_{n-α}^{n+α}|f(t)|^2\,dt<\infty\implies \lim_{|n|\to\infty}\int_{n-α}^{n+α}|f(t)|^2\,dt=0.$$
But one can get more than pointwise convergence. Shifting the integration variables the other direction one can define
$$h_n(t)=\begin{cases} f(t-α)&t\in[n,n+α)\\ f(t+α)&t\in[n-α,n)\\ -f(t-α)&t\in[-n,-n+α)\\ -f(t+α)&t\in[-n-α,-n)\\ 0&\text{ everywhere else} \end{cases}$$
so that $H_n(x)=\int_{-\infty}^\infty h_n(t)e^{ixt}\,dt$ and by Plancherel identity
$$\|H_n\|_2^2=2\pi\|h_n\|_2^2=2\pi\int_{-α}^α(|f(n+t)|^2+|f(-n+t)|^2)\,dt$$
and from dominated convergence or other asymptotic arguments (see comments) $\lim_{n\to\infty}\|H_n\|_2=0$ follows.