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If f is a radial distribution, then why is this equal to zero? watch

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    Let f\in\mathcal{D}'(\mathbb{R}^{2}) be a radial distribution. Recall that f is radial if f\circ R_{\theta}=f, where R_{\theta} is the rotation matrix in the plane through angle \theta. I want to show that

    \displaystyle x\frac{\partial f}{\partial y}-y\frac{\partial f}{\partial x}=0.

    Indeed, for \phi\in\mathcal{D}(\mathbb{R}^{2  }), we have

    \begin{aligned}\displaystyle

\langle x\frac{\partial}{\partial y}f,\phi\rangle-\langle y\frac{\partial}{\partial x}f,\phi\rangle&=\langle yf,\frac{\partial}{\partial x}\phi\rangle-\langle xf,\frac{\partial}{\partial y}\phi\rangle

\\

&= \langle f,y\frac{\partial}{\partial x}\phi\rangle-\langle f,x\frac{\partial}{\partial y}\phi\rangle

\\

&= \langle f \circ R_{\theta},y \frac{\partial}{\partial x} \phi\rangle-\langle f \circ R_{\theta}, x \frac{\partial}{\partial y} \phi\rangle

\\

&= \langle f,(y\frac{\partial}{\partial x}\phi)\circ R_{-\theta} \rangle-\langle f,(x\frac{\partial}{\partial y}\phi)\circ R_{-\theta} \rangle.

 

\end{aligned}

    However, I don't really see where this is going. Does anyone have an idea?
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    (Original post by RamocitoMorales)
    ..
    Not very familiar with distributions, but if I go straight to the integral, we have that

    I = \dfrac{\partial}{\partial \theta} \int f(x, y) \phi(R_\theta(x, y)) dx dy = 0 (where R_\theta(x, y) sends (x, y) to (x, y) rotated by theta.

    We can make the change of variables (x, y) \to R_\theta(x, y), which is area perserving and so we end up with

     I =  \dfrac{\partial}{\partial \theta} \int f(R_{-\theta}(x, y)) \phi(x, y) dx dy

    By taking \dfrac{\partial}{\partial \theta} inside the integral and using that it must = 0 for all possible \phi, we deduce

    \dfrac{\partial}{\partial \theta}f(R_{-\theta}(x, y)) = 0

    and I think the result should follow from here.

    I'm not 100% sure this is right, and even if the idea is generally OK you'll have to rephrase it in terms of bra/ket etc. which I'm not familiar with (had about 3 lectures on it 30 years ago).
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    (Original post by DFranklin)
    Not very familiar with distributions, but if I go straight to the integral, we have that
    Spoiler:
    Show


    I = \dfrac{\partial}{\partial \theta} \int f(x, y) \phi(R_\theta(x, y)) dx dy = 0 (where R_\theta(x, y) sends (x, y) to (x, y) rotated by theta.

    We can make the change of variables (x, y) \to R_\theta(x, y), which is area perserving and so we end up with

     I =  \dfrac{\partial}{\partial \theta} \int f(R_{-\theta}(x, y)) \phi(x, y) dx dy

    By taking \dfrac{\partial}{\partial \theta} inside the integral and using that it must = 0 for all possible \phi, we deduce

    \dfrac{\partial}{\partial \theta}f(R_{-\theta}(x, y)) = 0

    and I think the result should follow from here.



    I'm not 100% sure this is right, and even if the idea is generally OK you'll have to rephrase it in terms of bra/ket etc. which I'm not familiar with (had about 3 lectures on it 30 years ago).
    I appreciate the effort, but this is incorrect for a few reasons. Firstly, f is not locally integrable, so we cannot write it in the integral form (although, you did admit to that yourself).

    Secondly, we would not have \phi(R_{\theta}(x,y)), but rather \phi(R_{-\theta}(x,y)); since R_{\theta} is defined such that \langle f\circ R_{\theta},\phi\rangle=\langle f,\phi\circ R_{-\theta}\rangle.

    Finally, the derivatives are \partial_{x} and \partial_{y} respectively, as opposed to \partial_{\theta} as you have used in your argument.

    So I'm still unsure.
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    (Original post by RamocitoMorales)
    I appreciate the effort, but this is incorrect for a few reasons. Firstly, f is not locally integrable, so we cannot write it in the integral form (although, you did admit to that yourself).
    Perhaps I was unclear. What I was trying to do was say "this is why it's true with integrals", you should be able to do something analogous for distributions. In other words, I assumed you would know enough to fix this yourself.

    Secondly, we would not have \phi(R_{\theta}(x,y)), but rather \phi(R_{-\theta}(x,y)); since R_{\theta} is defined such that \langle f\circ R_{\theta},\phi\rangle=\langle f,\phi\circ R_{-\theta}\rangle.
    My problem is I don't know what you actually mean by \circ here, but in any event it should be obvious that the sign of theta is irrelevant. (i.e, maybe I got the sign wrong, but it doesn't affect the argument).

    Finally, the derivatives are \partial_{x} and \partial_{y} respectively, as opposed to \partial_{\theta} as you have used in your argument.
    Yes, but you can write \dfrac{\partial}{\partial \theta} f(R_\theta(x, y)) in terms of f_x and f_y and if you do so, you should end up with something looking very like the condition you're trying to show.

    To get to the "meat" here: I think it fairly likely that a proof of this is going to involve 2 real steps:

    (a) Justifying that \dfrac{\partial}{\partial \theta} f(R_\theta(x, y)) = 0

    (b) Showing that this implies the desired condition. (i.e. if you can assume normal partial differentiation rules, you can just use the chain rule appropriately).

    Now, I'm not going to pretend to know how to do this properly for distributions, but I would expect both proofs to be analogous to what you do for f being a normal function. Moreover, I can't see what you were doing ever working since it never seems to use the actual given fact that f is radial.
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    OK, having had a quick read of the wiki page on distribution theory, you may need to do part of (a), (b) on the test function phi and then go go back over to the left hand part of the braket to rephrase in terms of partials on f instead of partials on phi. But I'm pretty sure you can make this work.
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    (Original post by DFranklin)
    Perhaps I was unclear. What I was trying to do was say "this is why it's true with integrals", you should be able to do something analogous for distributions. In other words, I assumed you would know enough to fix this yourself.
    I do apologise; but I haven't had a chance to properly try it yet. I will do so quite promptly though!

    (Original post by DFranklin)
    My problem is I don't know what you actually mean by \circ here, but in any event it should be obvious that the sign of theta is irrelevant. (i.e, maybe I got the sign wrong, but it doesn't affect the argument).
    In case you are interested, (\phi\circ R_{\theta})(x,y)=\phi(R_{\theta}  (x,y)), i.e. the composition of functions.

    (Original post by DFranklin)
    Yes, but you can write \dfrac{\partial}{\partial \theta} f(R_\theta(x, y)) in terms of f_x and f_y and if you do so, you should end up with something looking very like the condition you're trying to show.
    I must admit that I am dumbfounded over how we can express \partial_{x} and \partial_{y} as \partial_{\theta}

    (Original post by DFranklin)
    To get to the "meat" here: I think it fairly likely that a proof of this is going to involve 2 real steps:

    (a) Justifying that \dfrac{\partial}{\partial \theta} f(R_\theta(x, y)) = 0

    (b) Showing that this implies the desired condition. (i.e. if you can assume normal partial differentiation rules, you can just use the chain rule appropriately).

    Now, I'm not going to pretend to know how to do this properly for distributions, but I would expect both proofs to be analogous to what you do for f being a normal function. Moreover, I can't see what you were doing ever working since it never seems to use the actual given fact that f is radial.
    I think the first step is getting to that step, to be honest. Again, if you are interested, differentiation for distributions is defined as \langle f',\phi \rangle=-\langle f,\phi'\rangle . Moreover, multiplication by functions is defined as \langle \psi f,\phi \rangle= \langle f, \psi \phi \rangle.

    Also, I thought I did use the fact that f was radial by writing f=f\circ R_{\theta} in the third line or so.

    In any case, we have something like:

    \langle x \cdot \partial_{y}f,\phi \rangle-\langle y \cdot \partial_{x}f,\phi \rangle=\langle\partial_{y}f,x \cdot \phi \rangle-\langle\partial_{x}f,y \cdot \phi \rangle

    = \langle f, y \cdot \partial_{x} \phi \rangle- \langle f, x \cdot \partial_{y} \phi \rangle = \langle f\circ R_{ \theta}, y \cdot \partial_{x} \phi \rangle- \langle f\circ R_{ \theta}, x \cdot \partial_{y} \phi \rangle

    = \langle f, (y\cdot \partial_{x} \phi) \circ R_{-\theta}-(x\cdot \partial_{y} \phi)\circ R_{-\theta} \rangle

    OR

    =\langle f, (y\cdot \partial_{x} \phi- x \cdot \partial_{y} \phi )\circ R_{-\theta} \rangle.

    (Original post by DFranklin)
    OK, having had a quick read of the wiki page on distribution theory, you may need to do part of (a), (b) on the test function phi and then go go back over to the left hand part of the braket to rephrase in terms of partials on f instead of partials on phi. But I'm pretty sure you can make this work.
    Now, indeed, we have everything on the test function side; but it's hard for me to make sense of this if f\notin L^{1}_{\text{loc}}(\mathbb{R}^{2  }). Certainly, if I could make the substitution (x,y)=R_{\theta} (x,y), then it would be a lot more plausible. I have a feeling that the question is potentially not well-posed; it is from an exercise in A Guide to Distribution Theory and Fourier Transforms by Robert Strichartz (which actually has a few "dodgy" exercises), in case you were wondering.
 
 
 
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