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I'm going through an example in my textbook, and they mention that the Fourier transform $F(z)=\int_{-\infty}^\infty f(t)e^{-2\pi i z t}dt$ can be bounded by some $c/(1+x^2)$ for $f$ being bounded and $C^\infty$. More concisely, $|F(x+iy)|\leq c/(1+x^2)$ for $y$ being bounded, and it's supposed to follow by integration by parts. But I'm doing this out and running into a few problems; namely, by using integration by parts twice and letting $u=f$ and then $u=f'$, I have something like this: $$|\hat{f}(z)|=\left|-\frac{f(t)}{2\pi i z}e^{-2\pi izt}|_{t=-\infty}^\infty-\frac{f'(t)}{(2\pi i z)^2}e^{-2\pi izt}|_{t=-\infty}^\infty+\int_{-\infty}^\infty \frac{f''(t)}{(2\pi i z)^2}e^{-2\pi izt}dt \right|.$$ Letting $z=x+iy$, I don't see how this is $O(c/(1+x^2))$. I also know that I can change the contour of integration (that is, change $t$ to $t+is$), but even with these added terms I'm not seeing it. Would anybody have any thoughts? Thanks in advance.

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1 Answer 1

You need to assume some decay/integrability of $f$ and its derivatives in addition to $C^\infty$. Then your first terms after integration by parts are bounded. In the last term you can put the $1/z^2$ out of the integral and you are basically done.

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Thanks, got it. –  Peter C. Mar 22 '12 at 17:10
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