MHB Proving Differentiability of f at $x_0$

[mathmari]

Hey! :giggle:

I want to prove the following:

If $x_0$ is an inner point of $D$ ($x_0 \in \text{int } D$), so the differentiability of $f$ at $x_0$ is equivalent to each of the following two conditions.
(i) $\exists \alpha\in \mathbb{C}$ : $\forall \epsilon>0 \ \exists \delta>0\ \forall x\in B(x_0,\delta): \ |f(x)-f(x_0)-\alpha(x-x_0)|\leq \epsilon |x-x_0|$
(ii) There is a $\delta>0$, $\alpha\in \mathbb{C}$ and $r\in B(x_0, \delta)\rightarrow \mathbb{C}$ continuous at $x_0$, $r(x_0)=0$ such that $\forall x\in B(x_0, r): f(x)=f(x_0)+\alpha (x-x_0)+r(x)(x-x_0)$.
The definition is :

Let $D\subset \mathbb{C}$ and let $x_0\in D$ be an inner point of $D$, i.e. $D$ is a neighbourhood of $x_0$, i.e. there is $r>0$ with $B(x_0,r)\subset D$.
Let $f : D \rightarrow \mathbb{C}$.
$f$ is differentiable at $x_0$, if $\displaystyle{\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}}$ exists.
To show (i) to we define as $\alpha$ this limit? :unsure:

 

[I like Serena]

To show (i) to we define as $\alpha$ this limit?

Hey mathmari!

That seems as a good approach yes. (Nod)

 

[mathmari]

That seems as a good approach yes. (Nod)

So the direction $\Rightarrow$ is :

Suppose that $f$ is differentiable at $x_0$. Then $\displaystyle{\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}}$ exists, i.e. there is $\alpha\in \mathbb{C}$ such that $\displaystyle{\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\alpha}$, i.e. $\forall \epsilon>0 \ \exists \delta>0\ \forall x\in B(x_0,\delta)$ such that $\left |\frac{f(x)-f(x_0)}{x-x_0}-\alpha\right |\leq \epsilon \Rightarrow \left |f(x)-f(x_0)-\alpha (x-x_0)\right |\leq \epsilon |x-x_0|$.

So we have proven statement (i).

Could you give me a hint for (ii) ? :unsure:

 

[I like Serena]

Could you give me a hint for (ii) ?

We have $f(x)-f(x_0)-\alpha(x-x_0)=r(x)(x-x_0)$ don't we?

 

[mathmari]

We have $f(x)-f(x_0)-\alpha(x-x_0)=r(x)(x-x_0)$ don't we?

How do we get that $r(x)$ ? :unsure:

 

[I like Serena]

How do we get that $r(x)$ ?

We define it.
Let $r(x)=\frac{f(x)-f(x_0)-\alpha(x-x_0)}{x-x_0}$ and $r(x_0)=0$ on $B(x_0,\delta)$.
Can we prove that $r$ is continuous?
What did continuous mean again?

 

[mathmari]

We define it.
Let $r(x)=\frac{f(x)-f(x_0)-\alpha(x-x_0)}{x-x_0}$ and $r(x_0)=0$ on $B(x_0,\delta)$.
Can we prove that $r$ is continuous?
What did continuous mean again?

We have that $$\lim_{x\rightarrow x_0}r(x)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)-\alpha(x-x_0)}{x-x_0}\lim_{x\rightarrow x_0}=\frac{f(x)-f(x_0)}{x-x_0}-\alpha=\alpha-\alpha=0=r(x_0)$$ So $r$ is continuous at $x_0$. :unsure:

That means that we have proven in that way the statement (ii).

 

[mathmari]

As for the other direction, do we have to show that from (i) the differentiability follows and from (ii) the differentiability follows also?

:unsure:

 

[I like Serena]

As for the other direction, do we have to show that from (i) the differentiability follows and from (ii) the differentiability follows also?

Yep.

 

[mathmari]

Yep.

We suppose that (i) holds. Then $ |f(x)-f(x_0)-\alpha(x-x_0)|\leq \epsilon |x-x_0| \Rightarrow \left |\frac{f(x)-f(x_0)-\alpha(x-x_0)}{x-x_0}\right |\leq \epsilon \Rightarrow \left |\frac{f(x)-f(x_0)}{x-x_0}-\alpha\right |\leq \epsilon $.
This is the definition of $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\alpha$, which means that $f$ is differentiable in $x_0$.

We suppose that (ii) holds. Then $ f(x)=f(x_0)+\alpha (x-x_0)+r(x)(x-x_0) \Rightarrow \frac{f(x)-f(x_0)}{x-x_0}=\alpha +r(x)$. Taking the limit we get $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\rightarrow x_0}(\alpha +r(x))=\alpha$, which means that $f$ is differentiable in $x_0$. Is everything correct? :unsure:

 

[I like Serena]

Looks correct to me. (Nod)

 

[mathmari]

Looks correct to me. (Nod)

Great! Thank you very much! (Star)