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The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is a perfectly valid statement (i.e. not in any sense merely formal--no quotation marks required). All you need to know is that if an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid--if $\int f(x,y) dy$ converges in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula for the F.T. of the constant function. First consider the classical fact that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., I think that this is just kicking the ball further down the park since one then has to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden (and has been since the 60´s of the previous century). There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many examplesW:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous;

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher distributional derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never vanishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $(\frac 1{\phi´} D_x)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.

The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is a perfectly valid statement (i.e. not in any sense merely formal--no quotation marks required). All you need to know is that if an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid--if $\int f(x,y) dy$ converges in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula for the F.T. of the constant function. First consider the classical fact that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., I think that this is just kicking the ball further down the park since one then has to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden (and has been since the 60´s of the previous century). There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many examples:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous;

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher distributional derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never vanishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $(\frac 1{\phi´} D_x)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.

The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is the Dirac distribution is a perfectly valid statement (i.e. not in any sense just formal). All you need to know is that if a an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid  -- if $\int f(x,y) dy$ in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula. First consider the classical case that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., then I think that you are just kicking the ball down the park since you then have to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden. There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many situations:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous. (we can even do this when $f$ is a measure, as is the delta distribution);

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never anishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $\big(\frac 1{\phi´} D_x\big)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.

The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is a perfectly valid statement (i.e. not in any sense merely formal--no quotation marks required). All you need to know is that if an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid--if $\int f(x,y) dy$ converges in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula for the F.T. of the constant function. First consider the classical fact that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., I think that this is just kicking the ball further down the park since one then has to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden (and has been since the 60´s of the previous century). There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many examplesW:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous;

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher distributional derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never vanishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $(\frac 1{\phi´} D_x)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.

The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is the dirac distribution is a perfectly valid statement (i.e. not in any sense just formal). All you need to know is that if a an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid--if $\int f(x,y) dy$ in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula. First consider the classical case that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., then I think that you are just kicking the ball down the park since you then have to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden. There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many situations:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous. (we can even do this when $f$ is a measure, as is the delta distribution);

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never anishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $(\frac 1{\phi´} D_x)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.

The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is the Dirac distribution is a perfectly valid statement (i.e. not in any sense just formal). All you need to know is that if a an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid  -- if $\int f(x,y) dy$ in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$

This allows a simple proof of the required formula. First consider the classical case that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.

Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., then I think that you are just kicking the ball down the park since you then have to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.

However, there is another path which is already well-trodden. There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many situations:

We can define $f\circ \phi$

1. when $f$ and $g$ are functions, say continuous. (we can even do this when $f$ is a measure, as is the delta distribution);

2. when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never anishes;

3. by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.

In 2. the composition is defined as $\big(\frac 1{\phi´} D_x\big)^n (F\circ \phi)$.

Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.