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Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be $L^1$ and $x \in \mathbb{R}$. Suppose that $\int_{-1}^0 |(f(x+t)-a)/t| dt < \infty$ and $\int_0^1 |(f(x+t)-b)/t| dt < \infty$ for some $a,b \in \mathbb{C}$. Denote $\mathcal{F}(f,p)=\int_{\mathbb{R}} f(x) e^{-ipx} dx$.

Is the following true (the analog of this) : $$ \lim_{h \to +\infty} \int_{-h}^h F(f,p) e^{ipx} \frac{dx}{2\pi} = \frac{a+b}{2}.$$

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up vote 0 down vote accepted

I think I got it.

Denote $D_h(x) = \pi^{-1} h.sinc(hx) = \frac{\sin(hx)}{\pi x}$. Then : $$\mathcal{G}(f,x) := \int_{-h}^h \mathcal{F}(f,p) e^{ipx} \frac{dp}{2\pi} = \int_{-\infty}^{+\infty} dy f(y) \int_{-h}^h e^{ip(x-y)} \frac{dp}{2\pi} = (f*D_h)(x).$$ Note that $\lim_{h \to +\infty} \int_0^1 D_h(x) dx = 1/2$. Now calculate : $$\mathcal{G}_h(f,x) - \left( b \int_{-1}^0 D_h(x) dx + a \int_0^1 D_h(x) dx \right) = \int_{-\infty}^{-1} f(x-y) D_h(y) dy + \int_{-1}^0 (f(x-y)-b) D_h(y) dy+ \int_0^1 (f(x-y)-a) D_h(y)dy + \int_1^{+\infty} f(x-y) D_h(y) dy.$$ i.e : $$\mathcal{G}_h(f,x) - \left( b \int_{-1}^0 D_h(x) dx + a \int_0^1 D_h(x) dx \right) = \int_{-\infty}^{-1} \frac{f(x-y)}{y} \sin(hy) dy + \int_{-1}^0 \frac{f(x-y)-b}{y} \sin(hy) dy+ \int_0^1 \frac{f(x-y)-a}{y} \sin(hy) dy + \int_1^{+\infty} \frac{f(x-y)}{y} \sin(hy) dy.$$

But $\frac{f(x-y)}{y}$ is in $L^1((-\infty,-1])$, so by Lebesgue's lemma, one has $$\lim_{h \to +\infty} \int_{-\infty}^{-1} \frac{f(x-y)}{y} \sin(hy) dy =0.$$ And same for the others integrals. So RHS goes to $0$.

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