# An identity on $\delta(x+y)$?

I am dealing with some derivatives on Dirac delta function, $(x\partial_x+y\partial_y)\delta(x+y)$. Consider the following integral ($f(x,y)$ is bounded) and perform integration by parts, \begin{align} &\int f(x,y)(x\partial_x+y\partial_y)\delta(x+y) dxdy \\ =& -\int \left(2f(x,y)+x\partial_xf(x,y)+y\partial_yf(x,y)\right)\delta(x+y) dxdy \end{align}

Now I can also perform a coordinate transformation into $\{u, v\}$, $$x=\frac{u+v}{\sqrt{2}},\quad y=\frac{u-v}{\sqrt{2}}$$ \begin{align} &\int f(x,y)(x\partial_x+y\partial_y)\delta(x+y) dxdy \\ =& \int \tilde{f}(u,v)(u \partial_u+v\partial_v)\delta(\sqrt{2}u) dudv \\ =& \int \tilde{f}(u,v) \sqrt{2}u \frac{\partial}{\partial \sqrt{2}u} \delta(\sqrt{2}u) dudv \\ =& -\int \tilde{f}(u,v)\delta(\sqrt{2}u) dudv\\ =&-\int f(x,y)\delta(x+y) dxdy. \end{align}

Comparing both it seems that \begin{align} \int\left(2f(x,y)+x\partial_xf(x,y)+y\partial_yf(x,y)\right)\delta(x+y)&=\int f(x,y)\delta(x+y)\\ \int \left(f(x,y)+x\partial_xf(x,y)+y\partial_yf(x,y)\right)\delta(x+y)&=0 \end{align}

Is the above derivation correct? I have tested the above identity using many $f(x,y)$ and it seems it is correct. If it is true, I wonder if there is another proof (direct proof?) of this identity. Thank you very much!

Update: the above identity cannot be true for bounded functions. As Bertram pointed out, it does not work for $1$. However, does it work for functions with compact support?

• How do you define $\delta(x+y)$? It is a distribution, and not a function, right? (If you give me a definition, I think I can help you). Commented Feb 26, 2018 at 14:05
• Yes, it is defined as a distribution. Commented Feb 26, 2018 at 14:55
• It is true that the identities you gave cannot be true, for they fail with $f(x, y)=1$, as Bertram points out. Commented Feb 26, 2018 at 20:47
• @Giuseppe Negro, thanks. What if the function has compact support? Commented Feb 27, 2018 at 9:09

Your proof shows that $\int \Big(f(x,y) + x\partial_xf(x,y) + y\partial_yf(x,y)\Big)\delta(x+y)\mathrm dx\mathrm dy = 0$. However you can't just leave away the integral - think about the difference between $f=0$ and $\int_{\mathbb{R}} f(x)\mathrm dx = 0$. Indeed, for $f=1$ we have $\Big(f(x,y) + x\partial_xf(x,y) + y\partial_yf(x,y)\Big)\delta(x+y) = \delta(x+y)\neq 0$. Here one actually has to be quite careful, since the integral is not defined for a general $f$, as the distribution $\delta(x+y)$ does not have compact support, so we need some restriction (e.g. compact support) for the restriction of $f$ to its support $\{y = -x\}$. But choosing some compactly supported function $f$ which is constant $1$ on a neighbourhood of some point $(x_0,-x_0)$, one still sees that $\Big(f(x,y) + x\partial_xf(x,y) + y\partial_yf(x,y)\Big)\delta(x+y)\neq 0$ by integrating the left-hand side against a test function with support in this neighbourhood.

(Here I had some computation, which was wrong as was pointed out by Giuseppe in the comments.)

Some care must be taken when applying coordinate transformations to $\delta$ functions since they transform as densities. One way to remember the correct formula is that $\delta(x+y)\mathrm dx\mathrm dy$ is invariant under automorphisms. Since I am bad at remembering where all the relevant signs go, let me just do the calculation in the original coordinate system. (In your case, it should be OK since the linear transformation $(x,y)\mapsto (u,v)$ has determinant $1$.)

We can write $x\partial_x + y\partial_y = \frac{1}{2}(x+y)(\partial_x + \partial_y) + \frac{1}{2}(x-y)(\partial_x-\partial_y)$, which is essentially your substitution $(x,y)\mapsto (u,v)$. Now $(x+y)\delta(x+y) = 0$, and deriving this we get \begin{align*} \frac{1}{2}(\partial_x + \partial_y)(x+y)\delta(x+y) &= \delta(x+y) + \frac{1}{2}(x+y)(\partial_x +\partial_y)\delta(x+y) \end{align*} Similarly we have $(\partial_x-\partial_y)\delta(x+y) = 0$, which one can check by integrating against a smooth compactly supported test function $\phi$: \begin{align*} \int \phi(x,y)(\partial_x-\partial_y)\delta(x+y)\mathrm dx\mathrm d y &:= -\int (\partial_x\phi(x,y) - \partial_y\phi(x,y))\delta(x+y)\mathrm dx\mathrm dy\\ &:= -\int (\partial_x\phi(x,y) - \partial_y\phi(x,y))|_{y = -x}\mathrm dx\\ &= -\int \frac{d}{dx} \phi(x,-x) \mathrm dx\\ &= 0 \end{align*}

These two calculations combine to give your integral formula.

All in all, one sees that $\delta(x+y)$ does not have a particularly nice transformation under infinitesimal scaling, which shouldn't be too surprising since in the coordinates $u,v$ it is written as $C\cdot \delta(u)$, and the $\delta$ function has different scaling behaviour than the constant function.

• I suspect there's an error in this answer; please see my answer. The passage in which you set y=-x in the integral (after "they can also be done by integrating against a smooth compactly supported test function “) look suspicious. Commented Feb 26, 2018 at 15:00
• Sorry about my abuse of language. My equal sign means that two distributions are the same. However, your result contradicts with mine using change of coordinate and that of @ Giuseppe Negro by a factor of 2. Commented Feb 26, 2018 at 15:01
• @GiuseppeNegro Thank you, I did indeed have an error. I have taken out that section. user260822, I don't understand what you mean by abuse of language. Equality of distributions has a very specific meaning, namely that integration against any test function gives the same result. Using this definition, the equality of distributions is wrong, while the integral equality is true (it's the special case where the test function is constant $1$, which actually needs some restriction on the behaviour of both sides at infinity, such as compact support). Commented Feb 26, 2018 at 16:39
• @Bertram, thanks for the update. Could you please explain the last few lines of your equations? Commented Feb 27, 2018 at 10:02
• The last one holds by partial integration: The integral of a (compactly supported) total derivative vanishes. The one before it is the chain rule applied to the composition $\phi\circ\Delta$, where $\Delta:x\to (x,x)$ is the diagonal. I had a typo in the first line (forgot $-\partial_y$) which I now fixed, the remaining equalities are the definition of the derivative of a distribution and the delta-function, respectively. Commented Feb 27, 2018 at 12:34

I think that the correct result is $$(x\partial_x+y\partial_y)\delta(x+y)=-\delta(x+y).$$ Proof: the operator $x\partial_x+y\partial_y$ is the generator of dilations, in the sense that for any distribution $F$ we have $$(x\partial_x+y\partial_y)F(x,y)=\partial_{\epsilon=0} F(e^\epsilon x, e^\epsilon y).$$ Applying this formula with $F(x, y)=\delta(x+y)$ and using the property $$\delta(\lambda t)=\lambda^{-1}\delta(t),$$ which is proved by changing variable in the formal integral $\int_{\mathbb R}\delta(t)\phi(t)\, dt$, we see that $$(x\partial_x+y\partial_y)\delta(x+y)=\partial_{\epsilon=0}\delta(e^\epsilon(x+y))=\partial_{\epsilon=0} e^{-\epsilon}\delta(x+y)=-\delta(x+y).$$

• Thanks for the comment. Your statement is correct, which I have also shown using the change of coordinate. However, it does not tell anything about the identity I am questioning about. Could you please take a look at my last identity? Commented Feb 26, 2018 at 14:58
• Sorry for not answering, I wanted to add weight to your first formula, since there is an extra factor of 2 in Bertrand answer. I will see your last identity as soon as I can. Commented Feb 26, 2018 at 15:04
• Hey Giuseppe, thanks for pointing out my mistake! I think in your application of the scaling transformation you are neglecting that $\delta(x+y)$ is constant and not "weight $-1$" in the direction $x-y$. Commented Feb 26, 2018 at 16:29
• @Bertram: sorry for distorting your name to "Bertrand" in my previous comment. Apart from that, I think that the result is correct, and the fact that $\delta(x+y)$ is "weight $-1$ " only in one direction (whatever that means) is responsible for the coefficient $-1$. I mean that the distribution $F(x, y)=\delta(x)\delta(y)$, which is "weight $-2$", would have given the result $(x\partial_x+y\partial_y)F(x, y)=-2F(x, y)$. Commented Feb 26, 2018 at 20:45