Differentiability assumptions of Wirtinger derivatives

[Tendex]

TL;DR

Real differentiability with complex variables.

In defining the Wirtinger (aka Cauchy-Riemann) linear operators, often used in signal analysis and in proofs of complex derivatives and the Cauchy-Riemann equations, one assumes differentiability in the real sense. This assumption is usually seen as obvious in the complex analysis setting since holomorphic functions must be smooth but I wonder if there's any circumstance in complex analysis in which real differentiability is not automatic for a complex variable function. There are plenty of complex functions of a complex variable that are not holomorphic but I can't think of examples that are not real differentiable.

 

[Infrared]

I'm having trouble understanding what your question is. If you're just asking for a complex function that isn't (real) differentiable, then $f(z)=|z|$ works around $z=0$, but I don't know what this has to do with the operators $\partial,\overline{\partial}.$

 

[Tendex]

I'm having trouble understanding what your question is. If you're just asking for a complex function that isn't (real) differentiable, then $f(z)=|z|$ works around $z=0$, but I don't know what this has to do with the operators $\partial,\overline{\partial}.$

Thanks for the example. The relation was with defining complex conjugate operators. You need real differentiability to do it so the modulus function is indeed incompatible with them, then again is not C to C, which I guess is a requisite for defining the Wirtinger operators.

 

[Infrared]

The absolute value function is a map $\mathbb{C}\to\mathbb{C}$, since $\mathbb{R}\subset\mathbb{C}.$ You can look at $f(z)=z+|z|$ if you want an example with more interesting image.

I don't think you need real differentiability for the Cauchy-Riemann operators to make sense, only the partial derivative to exist.

 

[Tendex]

The absolute value function is a map $\mathbb{C}\to\mathbb{C}$, since $\mathbb{R}\subset\mathbb{C}.$

It is $\mathbb{R^2}\to\mathbb{R+}$, AFAICS. I was identifying $\mathbb{C}\to\mathbb{C}$ with $\mathbb{R^2}\to\mathbb{R^2}$ here.

I don't think you need real differentiability for the Cauchy-Riemann operators to make sense, only the partial derivative to exist.

Nope, you need the linear operator, the tangent space so $\mathbb{R^2}$-differentiability.

 

[Infrared]

It is $\mathbb{R^2}\to\mathbb{R+}$, AFAICS.

If $z=x+iy$ is a complex number, then $|z|=\sqrt{x^2+y^2}$ is also a complex number. That it is also a non-negative real number doesn't matter. Would you say that $f(z)=e^z$ is not a function $\mathbb{C}\to\mathbb{C}$ because it is also a map $\mathbb{C}\to\mathbb{C}\setminus\{0\}?$

Nope, you need the linear operator, the tangent space so $\mathbb{R^2}$-differentiability.

The expression $\overline{\partial}f=\frac{1}{2}(f_x+if_y)$ makes sense even if $f$ isn't real-differentiable, as long its partials exist. The operator is perfectly well-defined, even when you don't interpret it as $(0,1)$ tangent vector.

 

[Tendex]

If $z=x+iy$ is a complex number, then $|z|=\sqrt{x^2+y^2}$ is also a complex number. That it is also a non-negative real number doesn't matter. Would you say that $f(z)=e^z$ is not a function $\mathbb{C}\to\mathbb{C}$ because it is also a map $\mathbb{C}\to\mathbb{C}\setminus\{0\}?$

I edited while you answered.

The expression $\overline{\partial}f=\frac{1}{2}(f_x+if_y)$ makes sense even if $f$ isn't real-differentiable, as long its partials exist. The operator is perfectly well-defined, even when you don't interpret it as $(0,1)$ tangent vector.

You need to relate it to the non-conjugate operator.

 

[Infrared]

I edited while you answered.

If you want to view it as a map $\mathbb{R}^2\to\mathbb{R}^2$, then it is $(x,y)\mapsto (\sqrt{x^2+y^2},0).$ I don't see the issue.

You need to relate it to the non-conjugate operator.

I don't know what you mean by this. What does this have to do with differentiability of $f$?

 

[Tendex]

If you want to view it as a map $\mathbb{R}^2\to\mathbb{R}^2$, then it is $(x,y)\mapsto (\sqrt{x^2+y^2},0).$ I don't see the issue.

How about $(x,y)\mapsto (\sqrt{x^2+y^2},0, 0).$ if you prefer $\mathbb{R}^2\to\mathbb{R}^3$? Or any other dimension you like the codomain to be. Sorry but it doesn't work that way.

I don't know what you mean by this. What does this have to do with differentiability of $f$?

$f$ must be differentiable, how else do you relate the $x$ and $y$ directions in the partial derivatives of the two Wirtinger operators? They both must refer to the same axes to be operators of the same $f$. They are always defined for differentiable $f$ to make sense, check any textbook.

 

[Infrared]

How about $(x,y)\mapsto (\sqrt{x^2+y^2},0, 0).$ if you prefer $\mathbb{R}^2\to\mathbb{R}^3$? Or any other dimension you like the codomain to be. Sorry but it doesn't work that way.

I don't see your point. You wanted a non-differentiable map $\mathbb{C}\to\mathbb{C}.$ The map $x+iy\mapsto (\sqrt{x^2+y^2})+0i$ is such a map. When you identify $\mathbb{C}$ as $\mathbb{R}^2$ in the usual way, then you get the map $(x,y)\mapsto (\sqrt{x^2+y^2},0).$

They are always defined for differentiable $f$ to make sense, check any textbook.

This is not true. One common problem is to find distributional solutions to $\overline{\partial}u=f.$ Depending on the regularity of $f$, you cannot hope for a differentiable solution $u$.

 

[Tendex]

I don't see your point. You wanted a non-differentiable map $\mathbb{C}\to\mathbb{C}.$ The map $x+iy\mapsto (\sqrt{x^2+y^2})+0i$ is such a map. When you identify $\mathbb{C}$ as $\mathbb{R}^2$ in the usual way, then you get the map $(x,y)\mapsto (\sqrt{x^2+y^2},0).$

No, I am saying the modulus function is real-valued, not $\mathbb{C}\to\mathbb{C}$.

This is not true. One common problem is to find distributional solutions to $\overline{\partial}u=f.$ Depending on the regularity of $f$, you cannot hope for a differentiable solution $u$.

I don't know what you are talking about here. Certainly not about the Wirtinger operators defined for differentiable f.

 

[Infrared]

No, I am saying the modulus function is real-valued, not $\mathbb{C}\to\mathbb{C}$.

A real-valued function is also complex-valued since $\mathbb{R}\subset\mathbb{C}$. If you want to be ultra formal, there is an inclusion map $\mathbb{R}\to\mathbb{C},$ and you can compose a map $f:\mathbb{C}\to\mathbb{R}$ with this inclusion to get a map $\tilde{f}:\mathbb{C}\to\mathbb{C}.$

I also gave you the example of $f(z)=z+|z|$ if you wanted something that wasn't real-valued.

I don't know what you are talking about here. Certainly not about the Wirtinger operators defined for differentiable f.

I am talking about the differential operators $\partial=\frac{1}{2}\left(\partial_x-i\partial_y\right)$ and $\overline{\partial}=\frac{1}{2}\left(\partial_x+i\partial_y\right).$

 

[WWGD]

A real-valued function is also complex-valued since $\mathbb{R}\subset\mathbb{C}$. If you want to be ultra formal, there is an inclusion map $\mathbb{R}\to\mathbb{C},$ and you can compose a map $f:\mathbb{C}\to\mathbb{R}$ with this inclusion to get a map $\tilde{f}:\mathbb{C}\to\mathbb{C}.$

I also gave you the example of $f(z)=z+|z|$ if you wanted something that wasn't real-valued.I am talking about the differential operators $\partial=\frac{1}{2}\left(\partial_x-i\partial_y\right)$ and $\overline{\partial}=\frac{1}{2}\left(\partial_x+i\partial_y\right).$

Not sure if this is what you're asking but the bar differential operator must be 0 for f to be Complex-Differentiable.

 

[Tendex]

A real-valued function is also complex-valued since $\mathbb{R}\subset\mathbb{C}$.

But I'm insisting on the real identification because the thread centers on real differentiability, So the same way the domain can be seen as $\mathbb{R^2}$, the codomain can be $\mathbb{R}$

I am talking about the differential operators $\partial=\frac{1}{2}\left(\partial_x-i\partial_y\right)$ and $\overline{\partial}=\frac{1}{2}\left(\partial_x+i\partial_y\right).$

So when $\overline{\partial f}=0$ which defines holomorphicity how is $f$ not differentiable?

 

[Infrared]

But I'm insisting on the real identification because the thread centers on real differentiability, So the same way the domain can be seen as $\mathbb{R^2}$, the codomain can be $\mathbb{R}$

So what is wrong with the map $(x,y)\mapsto(\sqrt{x^2+y^2},0)$ when we identify $\mathbb{R}^2=\mathbb{C}$? It's certainly not real differentiable.

You still haven't commented on whether you're satisfied with the function $f(z)=z+|z|.$

So when $\overline{\partial f}=0$ which defines holomorphicity how is $f$ not differentiable?

A theorem is that any (even weak) solution to $\overline{\partial f}=0$ is in fact analytic. That doesn't mean you can only apply $\overline{\partial}$ to real-differentiable functions. If you're familiar with distributional (weak) derivatives, you don't even need to assume that the partials of $f$ exist in the strict sense. Finding weak solutions to PDEs like $\overline{\partial}u=f$ is often useful.

 

[Tendex]

So what is wrong with the map $(x,y)\mapsto(\sqrt{x^2+y^2},0)$ when we identify $\mathbb{R}^2=\mathbb{C}$? It's certainly not real differentiable.

You still haven't commented on whether you're satisfied with the function $f(z)=z+|z|.$A theorem is that any (even weak) solution to $\overline{\partial f}=0$ is in fact analytic. That doesn't mean you can only apply $\overline{\partial}$ to real-differentiable functions. If you're familiar with distributional (weak) derivatives, you don't even need to assume that the partials of $f$ exist in the strict sense. Finding weak solutions to PDEs like $\overline{\partial}u=f$ is often useful.

Thanks for your interest but I don't think we are understanding each other. One can define operators with all kind of purposes and set requirements for the functions they apply to accordingly, in the context of complex analysis the Wirtinger derivatives are defined assuming f smooth, that is all I'm saying, I thought this was understood all along. The modulus function is usually considered real-valued independently of the fact it could be seen as complex-valued, but not all complex valued functions are real-valued and since this adds nothing to the fact it is not real-differentiable it can be safely ignored.

 

[Infrared]

Can you rephrase your question because I don't know how $|z|$ or $z+|z|$ are not examples of

There are plenty of complex functions of a complex variable that are not holomorphic but I can't think of examples that are not real differentiable.

If you don't like using $|z|$, then please respond to whether $z+|z|$ is an acceptable function for you.

 

[Tendex]

Can you rephrase your question because I don't know how $|z|$ or $z+|z|$ are not examples of
If you don't like using $|z|$, then please respond to whether $z+|z|$ is an acceptable function for you.

No,no , these examples were fine and answer my question and I said it then, I'm talking about the subsequent posts.

 

[Infrared]

Oh okay, I couldn't tell whether that was resolved or not.

Anyway I can find references for the $\overline{\partial}$ operator being applied to non-smooth functions in complex analysis/geometry if you're interested. One big theorem about solving the $\overline{\partial}$ problem in general is Hormander's $L^2$ estimates.

 

[Tendex]

Oh okay, I couldn't tell whether that was resolved or not.

Anyway I can find references for the $\overline{\partial}$ operator being applied to non-smooth functions in complex analysis/geometry if you're interested. One big theorem about solving the $\overline{\partial}$ problem in general is Hormander's $L^2$ estimates.

I guess they are, hadn't heard about it. I wonder if they are applied just to several complex variables or also to one complex variable.

 

[Tendex]

I am talking about the differential operators $\partial=\frac{1}{2}\left(\partial_x-i\partial_y\right)$ and $\overline{\partial}=\frac{1}{2}\left(\partial_x+i\partial_y\right).$

So if you apply that definition of a change of real basis for $f$ where you define z and zbar real variables from x and y, it means $f$ is differentiable. I actually still fail to see why you insist that $f$ is not differentiable in this definition.
A PDE like $\overline{\partial}u=f$ has nothing to do with the above definition.