If the fourier transformed of f is odd, is f odd? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:54:40Z http://mathoverflow.net/feeds/question/19997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd If the fourier transformed of f is odd, is f odd? Nicolò 2010-03-31T20:00:36Z 2010-06-22T11:25:10Z <p>Let $f\in L^1(R)$ such that $F(f)$ is odd, where $F$ is the Fourier transform. Can I then say that $f$ is odd?</p> <p>If $F(f)$ is odd, then </p> <p>$\int \cos(x\xi) f(x) dx = 0 \:\:\forall \xi\in R$</p> <p>Can I deduce from it that $f$ is odd?</p> http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd/20000#20000 Answer by Johannes Hahn for If the fourier transformed of f is odd, is f odd? Johannes Hahn 2010-03-31T20:33:48Z 2010-03-31T20:33:48Z <p>Look at $L^2$ first: In $L^2$ the FT is diagonalizable. The space of odd functions $\in L^2$ is the direct sum `$Eig(\mathcal{F},+i)\oplus Eig(\mathcal{F},-i)$ of the eigenspaces of $\mathcal{F}$ with respect to the eigenvalues $+i$ and $-i$. Because eigenspace are mapped into themselves, $\mathcal{F}$ maps odd functions to odd functions.</p> <p>In particular this is true for all $f\in L^1\cap L^2$ and by continuity it is true for all $f\in L^1$. (In fact an analogue statement is true for all tempered distributions.)</p> <p>EDIT: Oh, I just saw that you asked for the other direction. Using the same argument you can show that $\mathcal{F}f$ odd $\implies f=\mathcal{F}^3(\mathcal{F}f)$ odd is true for all $f\in L^2$. This time I'm not quite sure if it is possible to extend this from $L^1\cap L^2$ to $L^1$, but maybe the result for $L^1\cap L^2$ is useful for you too.</p> http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd/20042#20042 Answer by Robin Chapman for If the fourier transformed of f is odd, is f odd? Robin Chapman 2010-04-01T06:26:24Z 2010-04-01T06:26:24Z <p>Write $f^-(x)=f(-x)$ etc. Then $F(f^-)=F(f)^-$. If $F(f)$ is odd then $$0=F(f)+F(f)^-=F(f+f^-).$$ The only $L^1$ function with zero Fourier transform is $0$ so that $f+f^-=0$, that is, $f$ is odd.</p> http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd/29077#29077 Answer by Zen Harper for If the fourier transformed of f is odd, is f odd? Zen Harper 2010-06-22T11:25:10Z 2010-06-22T11:25:10Z <p>Yes. Using <em>tempered distributions</em> it's immediately obvious, since $f = C \mathcal{F}^3 (\mathcal{F} f)$ for a constant $C$, and $\mathcal{F}$ maps odd <em>distributions</em> into odd <em>distributions</em>.</p> <p>The missing details of the proof are just simple exercises in distributions.</p> <p>Distribution theory is very useful for Fourier transform questions like this, since Fourier inversion works perfectly.</p>