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Is the product of two non-differentiable functions in $\mathbb{C}$ always non-differentiable ? similarly for differentiable functions ? and if $f$ is not differentiable in $\mathbb{C}$ does that imply that $f$ will always be differentiable in $\mathbb{R}$ ? If Yes , how should i approach this ?

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    Do you know the chain rule? – 2017-01-21
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    @ B. Pasternak Yes .How chain rule will be useful here? – 2017-01-21

2 Answers 2

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Choose:

$$f(x+iy)=\begin{cases} 1 & \text{if}& x\text{ rational}\\0 & \text{if}& x\text{ irrational}\end{cases}\quad g(x+iy)=\begin{cases} 0 & \text{if}& x\text{ rational}\\1 & \text{if}& x\text{ irrational}\end{cases}$$ Then, $\Re f$ and $\Re g$ are not continuous for all $z\in\mathbb{C}$, so $f$ and $g$ are not continuous, hence non differentiable for all $z\in\mathbb{C}.$ However, $fg=0$ is differentiable on $\mathbb{C}.$

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    Please correct me if i am wrong - i) as per your definition of $f$ and $g$ they both get mapped to $0$ and $1$ which are real numbers and a subset of complex numbers so its ok . 2) Here $f = \Re f $ and $g = \Re g$ 3) Similarly if my function $f = \Im f$ then i can also just change the definition of $f$ and $g$ with just $1$ replaced by $i$ then similar arguments continue as here also $\Im f = f$ and $\Im g = g$. – 2017-01-21
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    Yes, that is right. – 2017-01-21
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    I have just edited the question can you help me with this part -- if $f$ is not differentiable in $\mathbb{C}$ does that imply that $f$ will always be differentiable in $\mathbb{R}$ ? – 2017-01-21
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    For example, neither $f:\mathbb{C}\to \mathbb{C}$, $\;f(x+iy)=|x|$ is differenciable at $z=0+0i$, nor $f_{|\mathbb{R}}(x)=|x|$ at $x=0.$ – 2017-01-21
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    Yup that is correct. – 2017-01-21
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    And for the product of two differentiable functions case? just an example – 2017-01-21
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    The product of two differentiable functions is always differentiable. – 2017-01-21
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    Yes,yes product of two differentiable functions is always differentiable. – 2017-01-21
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No. Let $f$ be a non-differentiable function with $f(z)\ne 0$ for all $z\in\Bbb{C}$, then $1/f$ is also non-differentiable, but $f\cdot (1/f)=1$ is differentiable. For example, set $f(z)=\exp(\bar{z})$.

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    how we can show that if $f$ is non-differentiable then $\frac{1}{f}$ is also non-differentiable? – 2017-01-21
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    @BAYMAX Write out the definition of differentiability as a limit and show that [the limit does not exist](https://68.media.tumblr.com/tumblr_m8z1jigDLv1r5pqmto1_500.gif) – 2017-01-21
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    Is it like this --> $\lim_{h \to 0} \frac{\frac{1}{f(z_{0}+h)} - \frac{1}{f(z_{0})}}{h} = - \lim_{h \to 0}{\frac{f(z_{0} + h) - f(z_{0})}{h}} . \lim_{h \to 0}{\frac{1}{f(z_{0}).f(z_{0} + h )}}$ and finally as the former limit doesnot exist and hence limit of this will also not exist and hence deriviative of $\frac{1}{f}$ at $z_{0}$ will not exist too and as $z_{0}$ is arbitrary and hence $\frac{1}{f}$ will not be differentiable in whole $\mathbb{C}$ also. – 2017-01-21
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    Yes, that is the sort of argument I made to myself. To justify the existence of the limit $\lim_{h\to 0}\frac{1}{f(z_0)f(z_0+h)}$ I was assuming that $f$ was continuous – 2017-01-21
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    I am unable to find the assumption about where $f$ is continuous is taken? – 2017-01-21
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    If $f$ is continuous then $\lim_{h\to 0}\frac{1}{f(z_0+h)}=\frac{1}{z_0}$ exists. – 2017-01-21
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    Yes that is right.i think you missed $f$ in the denominator. – 2017-01-21