Show that, for $x$ close enough to $a$, $||f(x)|| \leqslant e^{k||x-a||} \;\;\,||f(a)||.$ Exercice : Let $E$ be a Banach space, $U$ an open of $E$, and $f$ a differentiable application from $U$ to $\mathbb{R}$ such that, for all $x\in U$, $||df(x)|| \leqslant k||f(x)||$.
Show that for $x$ close enough to $a$, $||f(x)|| \leqslant e^{k||x-a||} \;\;\,||f(a)||$.
My attempt :
Let \begin{align*}g : [0,1&] \longrightarrow \mathbb{R} \\ &t \longmapsto g(t)=f(a+t(x-a))\end{align*}
By MVT, $||g(1)-g(0)|| \leqslant  \underset{t\in[0,1]}{\sup} ||dg(t)||  \leqslant \underset{t\in[0,1]}{\sup} ||df(a+t(x-a)|| \,||x-a|| \leqslant k||x- a|| \,||f(x)||$, or $g(1)=f(x)$ and $g(0)=f(a),$ we get $$||f(x)-f(a)||\leqslant k||x-a|| \,||f(x)||.$$
Or $||f(x)|| \leqslant ||f(x)-f(a)||+||f(a)||, $ thus $||f(x)|| \leqslant k||x-a||\,||f(x)||+||f(a)||.$
$||f||$ is continuous so for $x$ close enough to $a$, we have $||f(x) \sim ||f(a)|| $, and and $k||x-a||+1\sim e^{k||x-a||} \;\;$, then for $x$ close enough to $a$, $$||f(x)|| \leqslant e^{k||x-a||} \;\;\,||f(a)||.$$

I think I should use an $\varepsilon-\delta$ approch . Any help is appreciated.
 A: First
suppose $f(a) >0$, let $\eta(x) = \log f(x)$. Then we have
$\|D \eta (x) \| \le k$ for $x$ close to $a$ (so that $f(x) > 0$) and
so $|\eta(x)-\eta(a)| = |\log {f(x) \over f(a)}|\le k \|x-a\|$ from which we get
$\log {f(x) \over f(a)}\le k \|x-a\|$ and so
$f(x) \le f(a) e^{k \|x-a\|}$.
If $f(a) \neq 0$, we see that same condition applies to $t f$, where $t$ is chosen so that $tf(a)=|f(a)|$.
Finally, if $f(a) = 0$, then choose $\delta>0$ such that for $x \in C=\overline{B}(a,\delta)$ we have
$\|Df(x)\| \le {1 \over 2}$ and $k \delta < 1$. Then using the mean value theorem we have $|f(x)-f(a)| \le {1 \over 2} \|x-a\|$ for $x \in C$.
Now suppose $|f(x)-f(a)| \le {1 \over 2} k^n \|x-a\|^{n+1}$ for $x \in C$.
Then
\begin{eqnarray}
|f(x)-f(a)| &\le& \|Df(\zeta)\|\|x -a\| \\
&\le& k |f(\zeta)-f(a)| \|\|x-a\| \\
&\le& {1 \over 2} k^{n+1} \|\zeta-a\|^{n+1} \|x-a\| \\
&=& {1 \over 2} k^{n+1} \|x-a\|^{n+2}
\end{eqnarray}
In particular, letting $ n \to \infty$ we see that $f(x) = f(a)$ and hence the
result holds close to $a$.
