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Notes:

  1. 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 $x=0$

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

What I recall/understand:

  1. 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 $u,v: \mathbb R^2 \to \mathbb R$.

  2. 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 $(a,b)$.

  3. 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 $v$)

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  1. 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 $v$)

enter image description here

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 $(a,b)$.

  1. $f$ 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. $\frac{a+ib}{c+id} = e+if)$.

  2. 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 $(a,b)$.

  3. 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 $(a,b)$.

  4. $u_x(a,b)$ 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.

  5. $f_x(a,b)$ 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 $f_y$)

  6. $f_x(a,b)$ 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.

  7. $f_x(a,b)$ 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.

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

Questions:

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

  2. 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.)

BCLC
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    About 1. Yes. A function $f(z) = f(x+iy) = u(x,y) + i v(x,y)$ is complex differentiable if and only if the $\mathbb{R}^2 \to \mathbb{R}^2$ function $(u(x,y), v(x,y))$ is real differentiable and the derivatives satisfy the Cauchy-Riemann equations. So being complex differentiable is stronger than being differentiable as a real two-dimensional function (which is of course stronger than the real part being differentiable) – Zanzag Oct 27 '21 at 21:30
  • @Zanzag thanks. you can post as answer (if that's actually correct), but....well that's kind of what I'm talking about in Idea (1.1) and Idea (1.2). anyway how exactly do we prove your statement? i don't think either condition of your statement satisfies the sufficient conditions in (3) or (5) so...I guess it all comes down to thinking about the very definition of real differentiable? – BCLC Oct 27 '21 at 21:39
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    You can find the full proof in the section "Complex Differentiability" there: https://en.wikipedia.org/wiki/Cauchy–Riemann_equations – Zanzag Oct 27 '21 at 21:41
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    And yes, you need to work with the definition of real-differentiable in 3., not with the sufficient conditions. – Zanzag Oct 27 '21 at 21:42
  • @Zanzag thanks for the link! actually what i seem to be reading on wiki is something like (complex differentiable and real differentiable)-if and only if-(cauchy riemann and real differentiable). which of the following is what's going on? A - I'm correct and since complex differentiable implies real differentiable we get therefore your original statement (complex differentiable)-if and only if-(cauchy riemann and real differentiable). B - I'm incorrect since wiki really says the same thing as your original statement C - other. (see next comment) – BCLC Oct 27 '21 at 22:07
  • @Zanzag this is what wiki says: 'if f : C → C is a function which is differentiable when regarded as a function on R2, then f is complex differentiable if and only if the Cauchy–Riemann equations hold.' so i interpret as like $A \implies (B \iff C)$ and hence $(A \wedge B) \iff (A \wedge C)$ or something – BCLC Oct 27 '21 at 22:09
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    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. – Zanzag Oct 27 '21 at 23:35

2 Answers2

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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.

Stephen
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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.

Zanzag
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