Holomorphic functions and real functions: continuity of partial derivatives Given $f: A \subset \mathbb{C} \rightarrow \mathbb{C}$ a holomorphic function, I can represent the function as $f=u+iv, \  u,v:A \rightarrow \mathbb{R}$ and so $f$ can be seen as a function from $A' \subset \mathbb{R}^2$ to $\mathbb{R}^2$ under the identification of $\mathbb{C} \ni z=x+iy$ with $(x,y) \in \mathbb{R}^2$ and of $f$ with $F=(u,v)$. 
Now, the problem arise (for me) in two different direction, the first: 
A theorem in [Markushevich] ["Theory of functions of a complex variable"]1 says that if we look at $f$ as $F$ (notation above) then $f$ is holomorphic if and only if $F$ is real differentiable and solve Cauchy-Riemann conditions. 
Now, this gives us a good method to say when $f$ is holomorphic (at least when we can manage somehow the differentiability of $F$), but (and so the book continues) there is a nice sufficent condition on $F$ which is the continuity of partial derivatives of $u(x,y)$ and $v(x,y)$ (plus C-R equations) to imply holomorphicity of $f$. And that's ok, but this is only a sufficent condition. Wikipedia's page on holomorphic functions seems to do confusion: it says (in "Properties")

If one identifies $\mathbb{C}$ with $\mathbb{R}^2$, then the holomorphic functions coincide with those functions of two real variables with continuous first derivatives which solve the Cauchy–Riemann equations, a set of two partial differential equations.

Is it me, or this is wrong?
Second question: how the fact that $f$ is holomorphic, and so analytical, relates with $u(x,y)$ and $v(x,y)$? The problem, to me, arises when I try to show that a holomorphic function $f=u+iv$ solve the equation $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$. 
i) How can I say that the second (partial) derivatives of $u$ exists?
ii) What about the mixed (partial) derivatives of $u$ which are supposed to cancel? How the symmetry requirements on $u$ are satisfied?
In summary: which are some good characterisations of holomorphicity in terms of $u,v$?
Thank you in advance
 A: I'll take $A\subseteq\mathbb{C}$ to be an open set. A function $f:A\rightarrow
\mathbb{C}$ is holomorphic in $A$ if for every $z_{0}\in A$ there exists the
limit
$$
\lim_{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\ell\in\mathbb{C},
$$
which just says that the derivative $f^{\prime}(z_{0})$ exists in $\mathbb{C}%
$. Some books unfortunately require $f^{\prime}$ to be continuous, which just
muddles the waters. It is not needed, you get it for free. 
Once you know that $f$ is holomorphic, you can prove that $f^{\prime}$ is
continuous and it is itself holomorphic, and then you can prove that there
exist derivatives of any order and are all holomorphic. To prove this you fix
an open ball $B$ in $A$ and prove that for every rectifiable close curve
$\gamma$ contained in $B$ you have that $\int_{\gamma}f(z)\,dz=0$. Note that
to make sense of the integral you only need $f$ to be continuous and not even
differentiability. Cauchy proved this theorem assuming that $f^{\prime}$ was
continuous (and this is why some books add this to the definition) but later
Goursat proved that the result continues to hold assuming that $f$ is only differentiable.
Once you have $\int_{\gamma}f(z)\,dz=0$ for every rectifiable close curve
$\gamma$ contained in $B$ you prove Cauchy's formula, that is, that for every
$z_{0}\in B$
\begin{equation}
f(z_{0})=\frac{1}{2\pi i}\int_{\partial B(z_{0},r)}\frac{f(z)}{z-z_{0}%
}dz\label{cauchy formula}%
\end{equation}
where the closed ball $\overline{B(z_{0},r)}$ is contained in $B$. The point
now is that the right-hand side is a function $h(z_{0})$ which is given by an
integral depending on a parameter and as a function of $z_{0}$ you have that
$\frac{1}{z-z_{0}}$ is as regular as you want. So you can use theorems on
differentiation under the integral sign to conclude that that the right-hand
side has derivatives of any order and they are all continuous. In turn the
same is true for the left-hand side. So now you know that $f$ has derivatives
of any order and they are all continuous. At this point, if you define
$u(x,y):=$ real part of $f(x+iy)$ and $v(x,y):=$immaginary part of $f(x+iy)$,
then $u$ and $v$ are $C^{\infty}$ because they are given by the composition of
the function $(x,y)\mapsto x+iy$ which is $C^{\infty}$ with the function $f$
which is also $C^{\infty}$. In particular you can use the chain rule and take
second derivatives of $u$ and $v$ and prove that the Cauchy-Riemann equations
are satisfied and that $u$ and $v$ is harmonic (the mixed derivatives cancel
out because $u$ and $v$ are $C^{\infty}$). Hope this answers your second questions.
Now the converse implication is not perfect. There is a nice review article of
Gray and Morris 
. One good result is the following. Take $U\subseteq
\mathbb{R}^{2}$ be an open set and let $u:U\rightarrow\mathbb{R}$ and
$v:U\rightarrow\mathbb{R}$ be such that there exist the partial derivatives
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$.
Note that the existence of partial derivatives DOES not imply that $u$ and $v$
are differentiable in $U$. You also assume that $u$ and $v$ satisfy the
Cauchy-Riemann equations. 
These hypotheses are not enough, since they do not even imply that $u$ and $v$
are continuous.  The function
$$
f(z):=\left\{
\begin{array}
[c]{ll}%
\exp(-z^{-4}) & \text{if }z\neq0\\
0 & \text{if }z=0,
\end{array}
\right.
$$
does not have a derivative at $z=0$ but the corresponding functions $u$ and
$v$ satisfy the Cauchy-Riemann equations in $\mathbb{R}^{2}$.
So you need an extra hypothesis. There are several variants. One that I like
is that $u:U\rightarrow\mathbb{R}$ and $v:U\rightarrow\mathbb{R}$ admit
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$ in
$U$, $u$ and $v$ satisfy the Cauchy-Riemann equations, and that the function
$f(z):=u(x,y)+iv(x,y)$, where $z=x+iy$, is continuous in the domain
$A=\{z=x+iy:\,(x,y)\in U\}$. The idea of the proof is to to mollify the
functions $u$ and $v$, taking $u_{\varepsilon}=\varphi_{\varepsilon}\ast u$
and $v_{\varepsilon}=\varphi_{\varepsilon}\ast v$, where $\varphi
_{\varepsilon}$ is a nice kernel, and prove that the $C^{\infty}$ functions
$u_{\varepsilon}$ and $v_{\varepsilon}$ satisfy the Cauchy-Riemann equations.
Using that, you can prove that $\int_{\gamma}f_{\varepsilon}(z)\,dz=0$ for
every rectifiable close curve $\gamma$ contained in $B$, where $f_{\varepsilon
}(z):=u_{\varepsilon}(x,y)+iv_{\varepsilon}(x,y)$, where $z=x+iy$, and then
continue as before to prove that $f_{\varepsilon}$ is analytic. In turn you
get the Cauchy formula (1) for $f_{\varepsilon}$ and then since $f$ is
continuous, you can let $\varepsilon\rightarrow0$ to prove that $f$ satisfies
the Cauchy formula and so is holomorphic.
Hope this answers your first question. Read the paper, it is well-written.
A: Let $\Omega\subseteq\mathbb{C}$ open, $z_{0}\in\Omega,\ f:\Omega\rightarrow\mathbb{C}$.


*

*Theorem: $f$ is complex-differentiable at $z_{0}$ iff it is real-differentiable at $z_{0}$ and satisfies the Cauchy-Riemann equations.


Since being holomorphic in $\Omega$ implies $f\in C^{\infty}(\Omega)$ (but being holomorphic at a single point does not implies to be $C^{\infty}$ at the point), we also have:


*

*Corolary: $f$ is complex-differentiable in $\Omega$ iff in $\Omega$ it is real-differentiable, satisfies the Cauchy-Riemann equations and the patial derivatives are continous.


EDIT: Proof of the theorem: $f$ is complex-differentiable at $z_{0}$ iff $\exists a,b\in\mathbb{R}$:
$$ \mathbb{lim}_{z\rightarrow z_{0}} \frac{|f(z)-f(z_{0})-(a+ib)(z-z_{0})|}{|z-z_{0}|}=0$$
iff $\exists a,b\in\mathbb{R}$:
$$ \mathbb{lim}_{(x,y)\rightarrow (x_{0},y_{0})}\frac{\left | f(x,y)-f(x_{0},y_{0})-\begin{pmatrix}
a & -b\\ 
b & a
\end{pmatrix}\left ( \begin{pmatrix}
x\\ 
y
\end{pmatrix}-\begin{pmatrix}
x_{0}\\ 
y_{0}
\end{pmatrix}\right )\right |}{\left | \begin{pmatrix}
x\\ 
y
\end{pmatrix}-\begin{pmatrix}
x_{0}\\ 
y_{0}
\end{pmatrix} \right |} =0$$
iff $f$ is real-differentiable and satisfies the Cauchy-Riemann equations.
