So I have two pertinent properties of Fourier Transformations:

(1) F($\dot{f}$) = ik$\hat{f}$

(2) F(c) = c$\sqrt{2\pi} \delta(k)$

So suppose we have two functions f and g that differ by a constant: f = g + c. Then by (1) we have that

F($\dot{f}$)=F($\dot{g}$) $\rightarrow \hat{f} = \hat{g}$.

On the other hand by (2) $\hat{g}$ = $\hat{f}$ + c$\sqrt{2\pi} \delta(k)$ But from the previous conclusion, we would then have that 0 = c$\sqrt{2\pi} \delta(k)$, which I suppose is true a.e.

But I'm a bit bothered by this. Is there something wrong with my assumed properties? Or is it just the case that the Fourier transform of two functions that differ by a constant are a.e. equal?

EDIT So upon discussing this with people around my department this was pointed out to me that to go from ik$\hat{f}$ = ik$\hat{g}$ to $\hat{f}$ = $\hat{g}$ we must divide by k, but k may be zero. Moreover, if f = g + c then

$ik\hat{f}$ = ik$\hat{g}$ and $\hat{f}$ = $\hat{g} + \sqrt{2\pi}\delta(k) \rightarrow ik\hat{f} = ik\hat{f} + ik\sqrt{2\pi}\delta(k) \rightarrow 0 = k\delta(k)$

Now, I don't know how to evaluate this last expression at 0 but this seems like it could be possible?


The Fourier transform can be extended to distributions - like the Dirac function - and in this sense the transform of a constant exists - it is a Dirac impulse. Likewise, the transformation of a Dirac impulse yields a constant. Both formulas are correct but the second one is valid for the more generalized form of the Fourier transform which also applies to distributions. If the Fourier transforms of the derivatives of two functions are equal, then we cannot conclude that the Fourier transforms of the original functions are equal because of a possible Dirac impulses at 0 (which corresponds to a constant in the original function).

  • $\begingroup$ Okay, is it valid to keep the derivative property above (property (1)) if we just now assume equality in the sense of being true a.e.? Or should one simply disregarding property 1 completely when considering more general F.T.? $\endgroup$ – Fractal20 Apr 17 '13 at 15:45
  • $\begingroup$ The rule about the derivative is correct. What is not correct is the conclusion that two functions are identical if the Fourier transforms of their derivatives are the same. $\endgroup$ – Matt L. Apr 17 '13 at 16:57
  • $\begingroup$ so do you mean that analogously to my edit section? Thanks, I didn't see that before. Would say it is possible to say that if two functions have the same derivative, then their F.T.'s are equal except possibly at 0? $\endgroup$ – Fractal20 Apr 17 '13 at 20:08
  • $\begingroup$ Yes, I think you're right about that. It is indeed true that $k\delta(k)=0$. $\endgroup$ – Matt L. Apr 17 '13 at 20:24
  • $\begingroup$ would you happen to know how to show $k\delta(k) = 0$ for $k=0$? While it seems reasonable, especially for tying up the loose ends above, I wouldn't know how to show that. $\endgroup$ – Fractal20 Apr 17 '13 at 21:20

Fourier transform can be seen as a linear operator, in the sense that $$\mathcal{F}[a+b](\xi)=\mathcal{F}[a](\xi)+\mathcal{F}[b]{\xi}$$

The Fourier transform of $g(x)=f(x)+c$ exists only if $f$ is integrable and $c=0$, for the functions needs to be integrable and a constant function is not if its different f0rom $0$

  • $\begingroup$ Sorry, I wasn't explicit about assuming integrability. I guess I was considering a more general case since I am considering the fourier transformation of a constant to exist in a distributional sense. But with this in mind, I was using the linearity to get the above results. Thanks for the reply. $\endgroup$ – Fractal20 Apr 17 '13 at 15:44

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