Let $U\subset R^{n} $ and $f:U\to R$. Show that: $f$ is differetiable in $x_0\in U$ iff exist $A\in L(R^{n},R)$ and exist $r>0$:
$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$ with $\mathcal{S}^{n-1}$ is unitary sphere.
Attempt
$\Rightarrow$ Suppose that $f$ is differentiable in $x_0\in U$, then exist $A_{x_0}:R^{n}\to R$ linear transformation such that $f(x_0 +y)=f(x_0)+A_{x_0}(y)+||y||E_{x_0}(y)$ where $E_{x_0}(y)\rightarrow 0$ when $y\rightarrow 0$, equivalently
$\frac{f(x_{0} + y)-f(x_{0})}{||y||}-\frac{A_{x_0}(y)}{||y||}=E_{x_0}(y)$, now if i take $z=r\frac{y}{||y||}$ then $\frac{y}{||y||}\in \mathcal{S}^{n-1}$ and $y=\frac{||y||}{r}z$, if we will call $t=\frac{||y||}{r}$ imply
$\frac{f(x_{0} + tz)-f(x_{0})}{tr}-A_{x_0}(\frac{z}{r})=E_{x_0}(tz)$ and when i take limit when $t\rightarrow 0$ tehn
$\lim_{t\rightarrow 0}\left( \frac{f(x_{0} + tz)-f(x_{0})}{tr}-A_{x_0}(\frac{z}{r}) \right)=\frac{1}{r}\left[\lim_{t\rightarrow 0}\frac{f(x_{0} + tz)-f(x_{0})}{t}-A_{x_0}(z)\right]=\lim_{t\rightarrow 0}E_{x_0}(tz)=0$ hence
$\lim_{t\rightarrow 0}\frac{f(x_{0} + tz)-f(x_{0})}{t}=A_{x_0}(z)$ with $z\in r\mathcal{S}^{n-1}$.
$\Leftarrow$ Now by hypothesis $\lim_{t\rightarrow 0}\frac{f(x_{0} + tv)-f(x_{0})}{t}=A_{x_0}(v)$ with $v\in r\mathcal{S}^{n-1}$ uniformly, then $v=rw$ with $r>0$ and $w\in \mathcal{S}^{n-1}$ so
$\lim_{t\rightarrow 0}\frac{f(x_{0} + trw)-f(x_{0})}{t}=A_{x_0}(rw)=rA_{x_0}(w)$ or the same way
$\lim_{t\rightarrow 0}\frac{f(x_{0} + trw)-f(x_{0})}{rt}=A_{x_0}(w)$, but if we choose $h=tr$ when $t\rightarrow 0$ hence $h\rightarrow 0$ imply
$\lim_{h\rightarrow 0}\frac{f(x_{0} + hw)-f(x_{0})}{h}=A_{x_0}(w)$ and this is a definition of differentiablility in $x_0$.
My profesor say that i m not using the condition of uniformly no anywhere and my proof is confuse, therefore i have to proof that exist $r$, i just take one but i dont know how to show that exist, can somebody to help me please, thank you
Proof: $(\Rightarrow)$ Since $f$ is differentiable at $x_0\in U$, exist $A\in L(R^{n},R)$, such that
$$\lim_{h\to 0} \frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|}=0$$
So, for all $\varepsilon >0 $, there is $\delta_\varepsilon >0$ such that, for all $h\in \Bbb R^n$, if $\|h\|<\delta_\varepsilon$ then $$\frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|} < \varepsilon$$
It follows, immediately that, for any $r>0$, for all $\varepsilon >0 $, there is $\delta'= \frac{\delta_{\varepsilon/r}}{r} >0$ such that, for all $t>0$, if $t<\delta'$, then, for all $v\in r\mathcal{S}^{n-1}$, $\|tv\|=t\cdot r < \delta_{\varepsilon/r}$ and so
$$\frac{\|f(x_0+tv)-f(x_0)-tAv\|}{t \cdot r} < \varepsilon/r$$ So, $$\left \| \frac{f(x_0+tv)-f(x_0)}{t} - Av \right \| =\frac{\|f(x_0+tv)-f(x_0)-tAv\|}{t} < \varepsilon$$
Note that $\delta'$ independs on $v$. So, we have proved that $\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$.
$(\Leftarrow)$ Suppose exists $A\in L(\Bbb R^{n},\Bbb R)$ and exists $r>0$ such that $\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$.
Then, for all for all $\varepsilon >0 $, there is $\delta_\varepsilon >0$ such that, for all $t>0$, if $t<\delta_\varepsilon $, then, for all $v\in r\mathcal{S}^{n-1}$,
$$\left \| \frac{f(x_0+tv)-f(x_0)}{t} - Av \right \| < \varepsilon$$
So, for all $\varepsilon >0$, there is $\delta'=r \delta_{r\varepsilon}>0$, such that for all $h\in \Bbb R^n$, if $\|h\|<\delta'=r \delta_{r\varepsilon}$, then $t= \frac{\|h\|}{r} <\delta_{r\varepsilon}$ and $v=r \frac{h}{\|h\|} \in r\mathcal{S}^{n-1}$ and $h=tv$. So
\begin{align*} \frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|} &= \left \| \frac{f(x_0+tv)-f(x_0)- tAv }{t r} \right \| =\\ &= \frac{1}{r}\left \| \frac{f(x_0+tv)-f(x_0)}{t} - Av \right \| < \frac{1}{r} r\varepsilon = \varepsilon \end{align*} So, we have proved that $$\lim_{h\to 0} \frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|}=0$$ It means, $f$ is differentiable in $x_0\in U$.
Remark: In fact, we proved a stronger result. We proved that
In the first part of our proof, we proved that $(2 \Rightarrow 3)$. In the second part of our proof, we proved that $(1 \Rightarrow 2)$. The fact that $(3 \Rightarrow 1)$ is trivial.