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Is $\delta\left(a+bi\right)=\delta\left(a-bi\right)$?

I wonder whether Dirac Delta (as defined via Fourier transform) is symmetric against the real axis.

We can write Delta function as

$$\delta(z) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itz}\, dt=\delta\left(a+bi\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\cos ax\, dx+\frac{i}{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\sin ax\, dx.$$

The second integral is always zero (using Abel regularization), the first integral does not depend on the sign of $b$. So, $\delta\left(a+bi\right)$ should be equal to $\delta\left(a-bi\right)$.

But this contradicts the fact that $$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$

which depends on the sign of $b$.

I have asked this on Math.Stackexchange, but received no answers.

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    $\begingroup$ What do you even mean by "$\delta(a+bi)$" since it's not a function? What kind of mathematical object would $\delta(a+bi)$ represent? Not a real or complex number, that's for sure. $\endgroup$
    – Nate Eldredge
    Commented Mar 7, 2021 at 20:46
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    $\begingroup$ That doesn't answer my question. The definition you wrote down is an integral that doesn't exist in the usual sense. You may certainly define $\delta$ as a distribution, which is usually what "the Fourier transform definition" does, but then it doesn't make sense to plug a real or complex number into it. $\endgroup$
    – Nate Eldredge
    Commented Mar 7, 2021 at 20:49
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    $\begingroup$ In that case $\delta(i+x)$ is just convenient notation for thinking of the distribution $C^\infty_c \ni \phi \mapsto \phi(-i)$. It doesn't mean you can plug numbers in. Your question reads as if you don't have a clear understanding of what distributions are, and if that's the case, I don't think it's going to be productive to discuss it at an MO level until you do. $\endgroup$
    – Nate Eldredge
    Commented Mar 7, 2021 at 21:27
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    $\begingroup$ @Anixx No they are not. The answer clearly states that this is NOT a distribution, but it can be interpreted in terms of "analytic functionals". And the answer is NOT using the FT, it is saying that for a very particular class of functions you can actually make sense of the RHS in a very specialized context. $\endgroup$
    – Nick S
    Commented Mar 7, 2021 at 22:59
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    $\begingroup$ I’m voting to close this question because it was attracting feedback on MSE, it is just that the OP seems to have misunderstood things (and the question is founded on a misreading of an answer to another of the user's MSE questions) $\endgroup$
    – Yemon Choi
    Commented Mar 8, 2021 at 23:01

1 Answer 1

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There are two aspects of this question.

  1. Is the delta distribution (on the plane) symmetric with respect to complex conjugation? The answer is, of course, a resounding yes. It is even symmetric with respect to reflection in any line through the origin, or any rotation around the origin, indeed under the action of any diffeomorphism of the plane which leaves the origin and satisfies the obvious scaling condition on its derivative there. This is kindergarten level in the theory of distributions.

  2. The second (implicit) question is whether this symmetry can be expressed pointwise, i.e. in terms of values at points. Some comments here seem to subscribe to the very common fallacy that the fact that a distribution need not have a value at each point implies that one can never compute its value at any point. The concept of the value of a distribution at a point was examined in detail by pioneers in the 50's and in fact those distributions which occur in practice tend to have values at most points. Of course, it is even pre-kindergarten level that the delta distribution (defined anywhere sensible--real line, complex plane, euclidean space, differentiable manifold, fractals: take your pick) vanishes everywhere except at the origin and so your formula holds there (with plus or minus sign as you like). If you want to consider the value at the origin (which I take to be the main point of your question), then be aware that you are leaving the mainstream. That isn't necessarily a bad thing but then you have to very precise in specifying in what sense your concepts and formulae are to be understood, something I consistently miss in your prolific posts.

Finally, the frequent occurrence of the Fourier transform in this thread seems to me to be the grandaddy of all red herrings--but that is probably just me.

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  • $\begingroup$ How it is symmetric with respect to rotation if $\delta(i) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-t}\, dt$, which is divergent to infinity, while $\delta(1)=0$? $\endgroup$
    – Anixx
    Commented Mar 9, 2021 at 8:26
  • $\begingroup$ Also, according to the linked result, $\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)\ne f(bi)=\int_{-\infty}^\infty \delta(t-bi)f(t)dt$ $\endgroup$
    – Anixx
    Commented Mar 9, 2021 at 8:32
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    $\begingroup$ If you are seriouly claiming that the delta distribution is NOT symmetric under rotations, then I'm outta here! $\endgroup$
    – bathalf15320
    Commented Mar 9, 2021 at 8:48
  • $\begingroup$ I am judging from the Fourier transform-based definition. Yes, it seems not symmetric under rotations. $\endgroup$
    – Anixx
    Commented Mar 9, 2021 at 8:53

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