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Let $k \to f(k)$ be a function and define its Fourier transform as $$ \hat{f} (u) = \int_{-\infty}^{\infty} e^{iux} f(x) dx $$ if $\hat{f} (u)$ is integrable we can get back $f$ by doing the inversion $$ f(k)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iux} \hat{f}(u) du $$ now a source I am reading indicates (if I understand it correctly) that this can be done if and only if $$ \lim_{k\to -\infty} f(k) =0 $$ is this true? If so how come?

htd
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    The condition you need is that $f$ and $\hat{f}$ are integrable. If $\hat{f}$ is integrable, then the condition $\lim_{|k| \to \infty} f(k) = 0$ follows from the Riemann-Lebesgue Lemma, but this condition alone is not sufficient (it doesn't even guarantee $f \in L^1$, for instance). –  Sep 21 '14 at 11:20
  • Either you misunderstood, or misquoted, or the source is incorrect. Why not give a direct quote or reference? –  Sep 21 '14 at 15:28
  • Thanks! I did indeed misunderstand it. If you like you may make some kind of answer and I will accept since you made me figure it out. – htd Sep 22 '14 at 08:16

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