You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$

$$\langle T,\phi \rangle =0$$

Let's take a closer look. First,note that for any $\varepsilon>0$, we have the distributional equality

$$\frac{\partial}{\partial y}\delta(|x-y|-\varepsilon)=\delta'(y-(x+\varepsilon))+\delta'(y-(x-\varepsilon))$$

Second, note that both the Dirac Delta and its derivative have supprt $\{0\}$. Hence, $\operatorname{supp}\delta(y-x)=\{y|y=x\}$ and $\operatorname{supp} \delta'(y-(x\pm \varepsilon))=\{y|y=x\pm \varepsilon\}$. Inasmuch as these sets are disjoint for all $\varepsilon>0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$.

Explantion Using Sheafs:

Here, we provide a rigorous way to define the product of distributions that have disjoint supports. To that end, we proceed.

Distributions are defined as sheafs. As such, a distribution is zero if it is zero in a neighborhood of every point.

Suppose we have two distributions, $T_1$ with $\operatorname{supp}\{T_1\}=S_1$ and $T_2$ with $\operatorname{supp}\{T_2\}=S_2$, where $S_1 \cap S_2=\emptyset$. We seek to define their product, $T_1T_2$.

Let $x$ be any point in the domain (e.g., $\mathbb{R}^2$). Obviously, it is impossible for $x$ to be in both $S_1$ and $S_2$. Suppose

So, suppose $x\notin S_1$. Then, there exists an open neighborhood $N_1$ of $x$ where $T_1$ is the zero distribution. To have a consistent definition of a product, we must have $T_1T_2=0\cdot T_2=0$ on $N_1$.

Similarly, suppose $x\notin S_2$, Then, there exists an open neighborhood $N_2$ of $x$ where $T_2$ is the zero distribution. Hence $T_1T_2=0\cdot T_1=0$ on $N_2$.

Finally, the entire space is covered by the open sets $\{N_1$} and $\{N_2\}$ and the distribution $T_1T_2$ is, therefore, zero everywhere.

You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$

$$\langle T,\phi \rangle =0$$

Let's take a closer look. First,note that for any $\varepsilon>0$, we have the distributional equality

$$\frac{\partial}{\partial y}\delta(|x-y|-\varepsilon)=\delta'(y-(x+\varepsilon))+\delta'(y-(x-\varepsilon))$$

Second, note that both the Dirac Delta and its derivative have supprt $\{0\}$. Hence, $\operatorname{supp}\delta(y-x)=\{y|y=x\}$ and $\operatorname{supp} \delta'(y-(x\pm \varepsilon))=\{y|y=x\pm \varepsilon\}$. Inasmuch as these sets are disjoint for all $\varepsilon>0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$.

Explantion Using Sheafs:

Here, we provide a rigorous way to define the product of distributions that have disjoint supports. To that end, we proceed.

Distributions are defined as sheafs. As such, a distribution is zero if it is zero in a neighborhood of every point.

Suppose we have two distributions, $T_1$ with $\operatorname{supp}\{T_1\}=S_1$ and $T_2$ with $\operatorname{supp}\{T_2\}=S_2$, where $S_1 \cap S_2=\emptyset$. We seek to define their product, $T_1T_2$.

Let $x$ be any point in the domain (e.g., $\mathbb{R}^2$). Obviously, it is impossible for $x$ to be in both $S_1$ and $S_2$.

So, suppose $x\notin S_1$. Then, there exists an open neighborhood $N_1$ of $x$ where $T_1$ is the zero distribution. To have a consistent definition of a product, we must have $T_1T_2=0\cdot T_2=0$ on $N_1$.

Similarly, suppose $x\notin S_2$, Then, there exists an open neighborhood $N_2$ of $x$ where $T_2$ is the zero distribution. Hence $T_1T_2=0\cdot T_1=0$ on $N_2$.

Finally, the entire space is covered by the open sets $\{N_1$} and $\{N_2\}$ and the distribution $T_1T_2$ is, therefore, zero everywhere.

You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$

$$\langle T,\phi \rangle =0$$

Let's take a closer look. First,note that for any $\varepsilon>0$, we have the distributional equality

$$\frac{\partial}{\partial y}\delta(|x-y|-\varepsilon)=\delta'(y-(x+\varepsilon))+\delta'(y-(x-\varepsilon))$$

Second, note that both the Dirac Delta and its derivative have supprt $\{0\}$. Hence, $\operatorname{supp}\delta(y-x)=\{y|y=x\}$ and $\operatorname{supp} \delta'(y-(x\pm \varepsilon))=\{y|y=x\pm \varepsilon\}$. Inasmuch as these sets are disjoint for all $\varepsilon>0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$.

Explantion Using Sheafs:

Distributions are defined as sheafs. As such, a distribution is zero if it is zero in a neighborhood of every point.

Suppose we have two distributions, $T_1$ with $\operatorname{supp}\{T_1\}=S_1$ and $T_2$ with $\operatorname{supp}\{T_2\}=S_2$, where $S_1 \cap S_2=\emptyset$. We seek to define their product, $T_1T_2$.

Let $x$ be any point in the domain (e.g., $\mathbb{R}^2$). Obviously, it is impossible for $x$ to be in both $S_1$ and $S_2$. Suppose

So, suppose $x\notin S_1$. Then, there exists an open neighborhood $N_1$ of $x$ where $T_1$ is the zero distribution. To have a consistent definition of a product, we must have $T_1T_2=0\cdot T_2=0$ on $N_1$.

Similarly, suppose $x\notin S_2$, Then, there exists an open neighborhood $N_2$ of $x$ where $T_2$ is the zero distribution. Hence $T_1T_2=0\cdot T_1=0$ on $N_2$.

Finally, the entire space is covered by $N_1$ and $N_2$ and the distribution $T_1T_2$ is zero everywhere.

You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$

$$\langle T,\phi \rangle =0$$

Let's take a closer look. First,note that for any $\varepsilon>0$, we have the distributional equality

$$\frac{\partial}{\partial y}\delta(|x-y|-\varepsilon)=\delta'(y-(x+\varepsilon))+\delta'(y-(x-\varepsilon))$$

Second, note that both the Dirac Delta and its derivative have supprt $\{0\}$. Hence, $\operatorname{supp}\delta(y-x)=\{y|y=x\}$ and $\operatorname{supp} \delta'(y-(x\pm \varepsilon))=\{y|y=x\pm \varepsilon\}$. Inasmuch as these sets are disjoint for all $\varepsilon>0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$.

Explantion Using Sheafs:

Here, we provide a rigorous way to define the product of distributions that have disjoint supports. To that end, we proceed.

Distributions are defined as sheafs. As such, a distribution is zero if it is zero in a neighborhood of every point.

Suppose we have two distributions, $T_1$ with $\operatorname{supp}\{T_1\}=S_1$ and $T_2$ with $\operatorname{supp}\{T_2\}=S_2$, where $S_1 \cap S_2=\emptyset$. We seek to define their product, $T_1T_2$.

Let $x$ be any point in the domain (e.g., $\mathbb{R}^2$). Obviously, it is impossible for $x$ to be in both $S_1$ and $S_2$. Suppose

So, suppose $x\notin S_1$. Then, there exists an open neighborhood $N_1$ of $x$ where $T_1$ is the zero distribution. To have a consistent definition of a product, we must have $T_1T_2=0\cdot T_2=0$ on $N_1$.

Similarly, suppose $x\notin S_2$, Then, there exists an open neighborhood $N_2$ of $x$ where $T_2$ is the zero distribution. Hence $T_1T_2=0\cdot T_1=0$ on $N_2$.

Finally, the entire space is covered by the open sets $\{N_1$} and $\{N_2\}$ and the distribution $T_1T_2$ is, therefore, zero everywhere.