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

Question

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?

Answer 1

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.

Answer 2

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 $$(x\partial_x+y\partial_y)\delta(x+y)=\partial_{\epsilon=0}\delta(e^\epsilon(x+y))=\partial_{\epsilon=0} e^{-\epsilon}\delta(x+y)=-\delta(x+y).$$