I want to prove the following result that I found in the literature:
Let $X,Y$ be normed vector spaces and $U$ an open set of X. If $f:U\rightarrow Y$ is a twice differentiable function in $a\in U$, then there exists $r>0$ such that $f$ is differentiable in $x$, for all $x\in B(a,r)$, where $B(a,r)$ is the open ball with center $a$ and radius $r$.
My attempt:
We want to show that $\exists r>0$ such that $\forall x\in B(a,r)$ and $\forall \varepsilon>0$, $\exists\delta>0$ such that if $\|h\|<\delta$, then $\|f(x+h)-f(x)-Df(x)h\|<\varepsilon\|h\|$.
Since $f$ is twice differentiable in $a$, then $f$ is differentiable in $a$, $f$ is continuous in $a$ and $Df$ is continuous in $a$. Let $\varepsilon>0$.
$\textbf{(1)}$ $f$ differentiable in $a\iff(\exists\eta>0$ such that if $\|h\|<\eta\implies\|f(a+h)-f(a)-Df(a)h\|<\varepsilon\|h\|$)
$\textbf{(2)}$ $f$ continuous in $a\iff(\exists\delta_1>0$ such that if $\|x-a\|<\delta_1\implies\|f(x)-f(a)\|<\varepsilon)$
$\textbf{(3)}$ $Df$ continuous in $a\iff(\exists\delta_2>0$ such that if $\|x-a\|<\delta_2\implies\|Df(x)-Df(a)\|<\varepsilon)$.
Taking $R=\min\{\delta_1,\delta_2\}$, $\delta=\min\{\frac{R}{2},\eta\}$ and $x\in B(a,\frac{R}{2})$, we have that
$\textbf{(**)}\|f(x+h)-f(x)-Df(x)h\|\leq\|f(x+h)-f(a+h)\|+\|f(a+h)-f(a)-Df(a)h\|+\|f(a)-f(x)\|+\|Df(a)h-Df(x)h\|$
By $\textbf{(2)}$, we have that $\|f(x+h)-f(a+h)\|\leq\|f(x+h)-f(a)\|+\|f(a)-f(a+h)\|<2\varepsilon$ since $\|x+h-a\|<\delta_1$ and $\|a+h-a\|<\delta_1$; and also, $\|f(a)-f(x)\|<\varepsilon$ since $\|x-a\|<\delta_1$.
By $\textbf{(1)}$, we have that $\|f(a+h)-f(a)-Df(a)h\|<\varepsilon\|h\|$ since $\|h\|<\delta<\eta$.
By $\textbf{(3)}$, we have that $\|Df(a)h-Df(x)h\|\leq\|Df(a)-Df(x)\|\|h\|<\varepsilon\|h\|$ since $\|x-a\|<\frac{R}{2}<\delta_2$.
Finally, from the above lines and $\textbf{(**)}$, we can conclude that
$\|f(x+h)-f(x)-Df(x)h\|<3\varepsilon+2\varepsilon\|h\|$.
Here is my question: as you can see, the first addend ($3\varepsilon$) does not depend on $\|h\|$, so I cannot conclude that $\|f(x+h)-f(x)-Df(x)h\|<\varepsilon\|h\|$ as it is required, and I think it is not possible to choose $\varepsilon$ as $\varepsilon\|h\|$ in order to get the desired result.
How can I fix this problem?