Difference/comparison/relation of real-differentiable vs complex-differentiable Notes:

*

*I don't mean like this where you just convert $x$ to $z$: $\mathrm{e}^{-1/x^2}$ real differentiable but not complex differentiable at $x=0$


*I also don't mean holomorphic instead of complex-differentiable.

What I recall/understand:


*If I have $f: \mathbb C \to \mathbb C$, $f=u+iv$ with $u,v: \mathbb C \to \mathbb R$, then we may view $f$ alternatively as $f=[u,v]$ for $f: \mathbb R^2 \to \mathbb R^2$ with $u,v$ viewed as $u,v: \mathbb R^2 \to \mathbb R$.


*The vector function $f: \mathbb R^2 \to \mathbb R^2$ is defined as real-differentiable at a point $(a,b)$ if each component $u,v$ is real differentiable at $(a,b)$.


*That $u: \mathbb R^2 \to \mathbb R$ is real-differentiable at a point $(a,b)$ is defined as this weird thing from James Stewart, Calculus (similar for $v$)



*Sufficient condition: That $u: \mathbb R^2 \to \mathbb R$ is real-differentiable a point $(a,b)$ has a sufficient condition from this theorem also from James Stewart, Calculus (similar for $v$)


which I read for $u$ as

If $u_x$ and $u_y$ exist in an open disc containing $(a,b)$ and are continuous at $(a,b)$, then $u$ is real-differentiable at $(a,b)$.



*$f$ is defined complex-differentiable at $(a,b)$ if $\lim_{(x,y) \to (a,b)} \frac{f(x,y)-f(a,b)}{(x,y)-(a,b)}$ exists (where of course we have $\frac{(a,b)}{(c,d)} = (e,f)$ s.t. $\frac{a+ib}{c+id} = e+if)$.


*Sufficient condition: If $f_x$ and $f_y$ exist in an open disc containing $(a,b)$ and are continuous at $(a,b)$ and satisfy Cauchy-Riemann at $(a,b)$, then $f$ is complex-differentiable at $(a,b)$.


*Necessary condition: If $f$ is complex-differentiable at $(a,b)$, then $f_x$ and $f_y$ exist at $(a,b)$ and satisfy Cauchy-Riemann at $(a,b)$.


*$u_x(a,b)$ is defined as $u_x(a,b) := \lim_{x \to a} \frac{u(x,b)-u(a,b)}{x-a}$ (similar for $u_y$ and then similar for $v$), if this exists.


*$f_x(a,b)$ is defined as $f_x(a,b) := \lim_{x \to a} \frac{f(x,b)-f(a,b)}{x-a}$, if this exists. (similar for $f_y$)


*$f_x(a,b)$ exists if and only if $u_x(a,b)$ and $v_x(a,b)$ exist. (similar for $f_y$)

*

*9.1. This is because $\frac{f(x,b)-f(a,b)}{x-a} = [\frac{u(x,b)-u(a,b)}{x-a},\frac{v(x,b)-v(a,b)}{x-a}]$: take limit of both sides.



*$f_x(a,b)$ is equivalently defined as $f_x(a,b) := u_x(a,b) + iv_x(a,b) = [u_x(a,b), v_x(a,b)]$, if $u_x(a,b)$ and $v_x(a,b)$ exist. (similar for $f_y$)

*

*10.1. This is also because $\frac{f(x,b)-f(a,b)}{x-a} = [\frac{u(x,b)-u(a,b)}{x-a},\frac{v(x,b)-v(a,b)}{x-a}]$: again, take limit of both sides.



*$u$ differentiable at $(a,b)$ implies $u_x$ and $u_y$ exist at $(a,b)$. Confirmed here: Questions about real differentiable

Questions:

*

*Does $f$ complex-differentiable imply $f$ real-differentiable but not conversely? Why/why not?

*

*1.1. Idea 1.1: That's what it looks like from the sufficient conditions (3) and (5).


*1.2. Idea 1.2: If we pretend the sufficient conditions (3) and (5) are also necessary conditions, then obviously real doesn't imply complex because real is missing Cauchy-Riemann. Seems like this is what's going on there: Do $u$ and $v$ real-differentiable and $u$ and $v$ satisfy Cauchy-Riemann equations hold if and only if $f$ is complex-differentiable?.




*Much of elementary complex analysis talks about how (complex-)holomorphic is much stronger in complex as compared to real-holomorphic (real differentiable in an open neighbourhood of $p$). What about complex analogues of real properties like differentiable: How much stronger is complex-differentiable compared to real-differentiable? (I also asked smooth and analytic.)
 A: If $f$ is not real-differentiable, then it can't be complex differentiable. If it is real-differentiable, then it is complex-differentiable if and only if it satisfies the Cauchy-Riemann equation. So, in summary, $f$ is $C$-differentiable if and only if it is $\mathbb{R}^2$-differentiable and it satisfies Cauchy-Riemann equations.
A: Perhaps the following will clarify somewhat: given a function $f:V \to W$, where $V$ and $W$ are finite-dimensional $\mathbf{R}$-vector spaces, we say $f$ is differentiable at $v \in V$ if there is a linear map $L:V \to W$ such that
$$\lim_{h \to 0} \frac{f(v+h)-f(v)-L(h)}{|h|}=0,$$ and we call $L$ (which is unique if it exists) the derivative of $f$ at $v \in V$.
Exactly the same definition works if one simply replaces $\mathbf{R}$ by $\mathbf{C}$ (or indeed by any local field): supposing $V$ and $W$ are finite-dimensional $\mathbf{C}$-vector spaces and $f:V \to W$ is a function, we say $f$ is $\mathbf{C}$-differentiable at $v \in V$ if there is a $\mathbf{C}$-linear map $L:V \to W$ such that
$$\lim_{h \to 0} \frac{f(v+h)-f(v)-L(h)}{|h|}=0,$$ and we call $L$ the derivative of $f$ at $v$ in this case as well. Now a $\mathbf{C}$-linear map is in particular $\mathbf{R}$-linear, and so if we regard complex vector spaces $V$ and $W$ as $\mathbf{R}$-vector spaces by forgetting their complex structure, such a map is differentiable with derivative $L$. Conversely, if it is differentiable (in the real sense) then it is $\mathbf{C}$-differentiable exactly if its derivative $L$ is $\mathbf{C}$-linear. If you unwind this in case $V=W=\mathbf{C}$ you get precisely the Cauchy-Riemann equations.
