Let us say we know that the Fourier transform of the function $f(x)$ exists, which means that: $$\int^{\infty}_{-\infty}|f(x)|dx<\infty$$ However, this does not mean that $f(x)\rightarrow 0$ as $x \rightarrow \pm\infty$ (according to this question anyway Does an absolutely integrable function tend to $0$ as its argument tends to infinity?). So I cannot see why the value of $$[fe^{-ikx}]^{\infty}_{-\infty}=0$$ Please can someone explain?
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1Where are you getting $\left.f(x)e^{ikx}\right|_{-∞}^∞ =0$? is this supposed to be Riemann-Lebesgue? – Calvin Khor Sep 25 '15 at 08:12
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@CalvinKhor It could be if we can guarantee the inverse Fourier transform exists. – Quantum spaghettification Sep 25 '15 at 08:18
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Thats not in your assumptions. I am asking for context, because I don't know where you would get this expression from, much less ask it to be 0. – Calvin Khor Sep 25 '15 at 08:20
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@CalvinKhor it comes from when deriving the Fourier Transform for $f'(x)$ and the relationship that this is given by $ikF(k)$ where $F(k)$ is the Fourier transform of $f(x)$. – Quantum spaghettification Sep 25 '15 at 08:25
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Integration by parts? This only holds for nice enough functions, or if you switch to a distribution. – Calvin Khor Sep 25 '15 at 08:30
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By the way it is easy to see that an absolutely integrable function need not converge to 0; take $f = ∑0^∞ \Bbb 1{\left(n,n+\frac{1}{n^2}\right)}$. – Calvin Khor Sep 25 '15 at 08:32
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@CalvinKhor Thanks. By 'nice enough functions' do you basically mean a function that tends to 0 as $x\rightarrow \pm \infty$ or is there more to it then that? – Quantum spaghettification Sep 25 '15 at 08:36
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Yeah, thats it. – Calvin Khor Sep 25 '15 at 10:42
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To justify the partial integration, it suffices to have sequences $(x_n)_n, (y_n)_n $ with $x_n\to\infty $ and $y_n \to -\infty $ with $f\cdot e^{ikx}\bigg |_{y_n}^{x_n} \to 0$ (why?).
Existence of such sequences can be shown as follows: We have $$ \infty > \int |f (x)|\,dx = \sum_{n \in \Bbb {Z} } \int_{(n,n+1)} |f(x)|\,dx =\sum_n \int_0^1 |f (x+n)|\,dx =\int_0^1 \sum_n |f (x+n)|\,dx. $$ The steps above can be justified using monotone convergence.
Now, a function with finite integral is finite almost everywhere, so that we get $$ \sum_{n \in\Bbb {Z}} |f (x+n)|<\infty $$ for almost all $x \in (0,1) $. Since the terms of a convergent series are a null sequence, this easily implies existence of the desired sequences.
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