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Suppose$f_x$ and $f_y$ exist in a neighborhood of z. Then if $f_x$ and $f_y$ are continuous at z and $f_x$ =$if_y$ there, $f$ is differentiable (or holomorphic) at z.

Proof: Let $f$ = $u$ + i $v$, h = $\xi$ + i$\eta$. By the Mean-Value Theorem (for real functions of a real variable).

$\frac{u(z+h)-u(z)}{h}$= $\frac{u(x+\xi, y + \eta)-u(x,y)}{\xi + i\eta}$ = $\frac{u(x+\xi, y + \eta)-u(x+\xi,y)}{\xi + i\eta}$ + $\frac{u(x+\xi, y)-u(x,y)}{\xi + i\eta}$
=$\frac{\eta}{\xi+i\eta}u_y(x+\xi, y+\theta_1\eta) + \frac{\xi}{\xi+i\eta}u_x(x+\theta_2\xi,y) $

My professor said that I need $f_x$ and $f_y$ continuous at z in order to apply Mean-Value Theorem. Why do I need derivatives to be continuous and their existence is not enough? Thanks for your help.

Mrnobody
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  • Have you looked up the statement and proof of the mean value theorem to see that continuity is required and where it is used in the proof? Do you understand that the mean value theorem is used in the equation you have recited? – Eric Towers Mar 26 '21 at 22:14
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    Yes I looked but I couldn't connect the dots. In mean value theorem f is differentiable is enough it doesn't state that $f'$ should be continuous as well. I am a bit confused – Mrnobody Mar 26 '21 at 22:19
  • @Mrnobody Are you sure the theorem is not applied to the derivative? – Dole Mar 26 '21 at 22:51

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The version of the mean value theorem you are using establishes the existence of $c_1 \in (y,y+\eta)$ such that $$ \frac{u(x+\xi, y+\eta) - u(x+\xi,y)}{\xi + \mathrm{i}\eta} = \frac{\eta}{\xi + \mathrm{i}\eta}u_y(x+\xi, c_1) \text{.} $$ This is not the form given. To get $c_1 = y + \theta_1 \eta$ we must know that $u_y(x+\xi, y + \theta_1 \eta)$ takes every value in the set $$ \left\{ u_y(x+\xi, c) \mid c \in (y,y+\eta) \right\} \text{.} $$ An easy way to do so is to apply the intermediate value theorem, which here requires that $u_y$ is continuous in its second argument. That is, $u$ is continuously differentiable with respect to $y$.

The other term in your long equation give that $u$ is continuously differentiable with respect to $x$. The same development for $v$ gives the requirement that it is continuously differentiable in each of $x$ and $y$.

One hypothesis that gives all four of these is that $f_x = u_x + \mathrm{i} v_x$ and $f_y = u_y + \mathrm{i} v_y$ are continuous, so $f$ is continuously differentiable with respect to both $x$ and $y$.

Maybe a weaker hypothesis would work. $f$ merely differentiable in $x$ and $y$ is insufficient because there is no analog of the intermediate value theorem that can give the $\theta_1$ and $\theta_2$ for potentially nowhere continuous derivatives. See this answer for a reminder of how awful the derivative of a differentiable but not continuous differentiable function can be -- and an easy visual example of why $\theta_1$ (and $\theta_2$) need not exist for such a function.

Eric Towers
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