If $L: V\rightarrow V$ is a linear operator on a finite dimensional vector spaces, then by choosing a basis, you can always write it as a matrix, and write the relation $w = Lv$ as $w_i = \sum_{j} L_{ij}v_j$.

Notably, this does not hold for operators between infinite dimensional spaces in general, at least not like this.

In the physics literature, this is however not an obstacle, and authors often essentially assume that every operator between function spaces has a kernel, i.e. if $L$ is a linear map between suitable (and usually not explicitly defined) function spaces, the relation $g = Lf$ can be written as an integral operator relation $$g(x) = \int L(x, y)\, dy,$$ which is really just the continuum analogue of the matrix equation above. The bivariate function $L(x, y)$ above is called the kernel of the operator $L$.

As mentioned, most linear operators don't have a kernel, but this doesn't stop people from treating them as if they had kernels. Sometimes using function notation for distributions can make it appear that an operator has a kernel when in fact it doesn't and still retain at least some semblance of rigour.

So with that in mind, if $D$ is a linear differential operator between some (implicit) function spaces, then a bivariate function $G(x, y)$ is called a Green function for the operator if $G$ is the kernel of an inverse of $D$ (I say "an" instead of "the" because most differential operators are not invertible, and for example elliptic operators require appropriate boundary conditions for unique solution to exist, so an elliptic operator is invertible only if th function space is restricted by a choice of boundary condition).

Let's for a moment suppose that $D$ can be written as an integral operator (it usually can't but whatever), then we have $$f(x)=\int\int D(x, z)G(z, y)f(y)\,dz\,dy,$$ which means that the inner integral just produces the Dirac delta, $\int D(x, z)G(z,y)\, dz = \delta(x-y)$. Which is really just the continuum analogue of $\delta_{ij=}\sum_k A_{ik}B_{kj}$.

But if we recall that $D(x,y)$ is just the fictious kernel of $D$, then this van be written as $$D_x G(x,y) = \delta(x-y), $$ where the index $x$ in $D_x$ signifies that the operator acts on the $x$ variable of $G(x,y)$ only.

So $G(x, y)$ is bivariate by design because it is supposed to be a "continuum matrix" representation of a linear operator.

However is the operator $D$ is translation invariant, then so is $G$ and in this case one can define a single-variable function $\hat G(z)$ by $\hat G(x-y) = G(x, y)$. It is then a very common abuse of notation to use the same symbol for $G$ and $\hat G$.

If $L: V\rightarrow V$ is a linear operator on a finite dimensional vector spaces, then by choosing a basis, you can always write it as a matrix, and write the relation $w = Lv$ as $w_i = \sum_{j} L_{ij}v_j$.

Notably, this does not hold for operators between infinite dimensional spaces in general, at least not like this.

In the physics literature, this is however not an obstacle, and authors often essentially assume that every operator between function spaces has a kernel, i.e. if $L$ is a linear map between suitable (and usually not explicitly defined) function spaces, the relation $g = Lf$ can be written as an integral operator relation $$g(x) = \int L(x, y)f(y)\, dy,$$ which is really just the continuum analogue of the matrix equation above. The bivariate function $L(x, y)$ above is called the kernel of the operator $L$.

As mentioned, most linear operators don't have a kernel, but this doesn't stop people from treating them as if they had kernels. Sometimes using function notation for distributions can make it appear that an operator has a kernel when in fact it doesn't and still retain at least some semblance of rigour.

So with that in mind, if $D$ is a linear differential operator between some (implicit) function spaces, then a bivariate function $G(x, y)$ is called a Green function for the operator if $G$ is the kernel of an inverse of $D$ (I say "an" instead of "the" because most differential operators are not invertible, and for example elliptic operators require appropriate boundary conditions for unique solution to exist, so an elliptic operator is invertible only if th function space is restricted by a choice of boundary condition).

Let's for a moment suppose that $D$ can be written as an integral operator (it usually can't but whatever), then we have $$f(x)=\iint D(x, z)G(z, y)f(y)\,dz\,dy,$$ which means that the inner integral just produces the Dirac delta, $\int D(x, z)G(z,y)\, dz = \delta(x-y)$. Which is really just the continuum analogue of $\delta_{ij=}\sum_k A_{ik}B_{kj}$.

But if we recall that $D(x,y)$ is just the fictious kernel of $D$, then this can be written as $$D_x G(x,y) = \delta(x-y), $$ where the index $x$ in $D_x$ signifies that the operator acts on the $x$ variable of $G(x,y)$ only.

So $G(x, y)$ is bivariate by design because it is supposed to be a "continuum matrix" representation of a linear operator.

However if the operator $D$ is translation invariant, then so is $G$ and in this case one can define a single-variable function $\hat G(z)$ by $\hat G(x-y) = G(x, y)$. It is then a very common abuse of notation to use the same symbol for $G$ and $\hat G$.

If $L: V\rightarrow V$ is a linear operator on a finite dimensional vector spaces, then by choosing a basis, you can always write it as a matrix, and write the relation $w = Lv$ as $w_i = \sum_{j} L_{ij}v_j$.

Notably, this does not hold for operators between infinite dimensional spaces in general, at least not like this.

In the physics literature, this is however not an obstacle, and authors often essentially assume that every operator between function spaces has a kernel, i.e. if $L$ is a linear map between suitable (and usually not explicitly defined) function spaces, the relation $g = Lf$ can be written as an integral operator relation $$g(x) = \int L(x, y)\, dy,$$ which is really just the continuum analogue of the matrix equation above. The bivariate function $L(x, y)$ above is called the kernel of the operator $L$.

As mentioned, most linear operators don't have a kernel, but this doesn't stop people from treating them as if they had kernels. Sometimes using function notation for distributions can make it appear that an operator has a kernel when in fact it doesn't and still retain at least some semblance of rigour.

So with that in mind, if $D$ is a linear differential operator between some (implicit) function spaces, then a bivariate function $G(x, y)$ is called a Green function for the operator if $G$ is the kernel of an inverse of $D$ (I say "an" instead of "the" because most differential operators are not invertible, and for example elliptic operators require appropriate boundary conditions for unique solution to exist, so an elliptic operator is invertible only if th function space is restricted by a choice of boundary condition).

Let's for a moment suppose that $D$ can be written as an integral operator (it usually can't but whatever), then we have $$f(x)=\int\int D(x, z)G(z, y)f(z)\,dz\,dy,$$ which means that the inner integral just produces the Dirac delta, $\int D(x, z)G(z,y)\, dz = \delta(x-y)$. Which is really just the continuum analogue of $\delta_{ij=}\sum_k A_{ik}B_{kj}$.

But if we recall that $D(x,y)$ is just the fictious kernel of $D$, then this van be written as $$D_x G(x,y) = \delta(x-y), $$ where the index $x$ in $D_x$ signifies that the operator acts on the $x$ variable of $G(x,y)$ only.

So $G(x, y)$ is bivariate by design because it is supposed to be a "continuum matrix" representation of a linear operator.

However is the operator $D$ is translation invariant, then so is $G$ and in this case one can define a single-variable function $\hat G(z)$ by $\hat G(x-y) = G(x, y)$. It is then a very common abuse of notation to use the same symbol for $G$ and $\hat G$.

If $L: V\rightarrow V$ is a linear operator on a finite dimensional vector spaces, then by choosing a basis, you can always write it as a matrix, and write the relation $w = Lv$ as $w_i = \sum_{j} L_{ij}v_j$.

Notably, this does not hold for operators between infinite dimensional spaces in general, at least not like this.

In the physics literature, this is however not an obstacle, and authors often essentially assume that every operator between function spaces has a kernel, i.e. if $L$ is a linear map between suitable (and usually not explicitly defined) function spaces, the relation $g = Lf$ can be written as an integral operator relation $$g(x) = \int L(x, y)\, dy,$$ which is really just the continuum analogue of the matrix equation above. The bivariate function $L(x, y)$ above is called the kernel of the operator $L$.

As mentioned, most linear operators don't have a kernel, but this doesn't stop people from treating them as if they had kernels. Sometimes using function notation for distributions can make it appear that an operator has a kernel when in fact it doesn't and still retain at least some semblance of rigour.

So with that in mind, if $D$ is a linear differential operator between some (implicit) function spaces, then a bivariate function $G(x, y)$ is called a Green function for the operator if $G$ is the kernel of an inverse of $D$ (I say "an" instead of "the" because most differential operators are not invertible, and for example elliptic operators require appropriate boundary conditions for unique solution to exist, so an elliptic operator is invertible only if th function space is restricted by a choice of boundary condition).

Let's for a moment suppose that $D$ can be written as an integral operator (it usually can't but whatever), then we have $$f(x)=\int\int D(x, z)G(z, y)f(y)\,dz\,dy,$$ which means that the inner integral just produces the Dirac delta, $\int D(x, z)G(z,y)\, dz = \delta(x-y)$. Which is really just the continuum analogue of $\delta_{ij=}\sum_k A_{ik}B_{kj}$.

But if we recall that $D(x,y)$ is just the fictious kernel of $D$, then this van be written as $$D_x G(x,y) = \delta(x-y), $$ where the index $x$ in $D_x$ signifies that the operator acts on the $x$ variable of $G(x,y)$ only.

So $G(x, y)$ is bivariate by design because it is supposed to be a "continuum matrix" representation of a linear operator.

However is the operator $D$ is translation invariant, then so is $G$ and in this case one can define a single-variable function $\hat G(z)$ by $\hat G(x-y) = G(x, y)$. It is then a very common abuse of notation to use the same symbol for $G$ and $\hat G$.