Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $\Vert f(t)\Vert\leq A$ and $\Vert f''(t)\Vert \leq B$ for all $t\in (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0\in (t_1,t_2)$ and $\alpha>0$ be such that $(t_0-\alpha,t_0+\alpha)\subset (t_1,t_2)$.
I would like to prove that $\Vert f'(t_0)\Vert\leq 2A/\alpha + B\alpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
First, it's enough to consider the case $n=1$: Say $v\in\Bbb R^n$ with $||v||=1$ and let $g(t)=f(t)\cdot v$. Then $|g'|\le A$ and $|g''(t)|\le B$; if we show that $|g'(t)|\le C$ then $|f'(t)\cdot v|\le C$ for all $v$ with $||v||=1$, hence $|f'(t)|\le C$.
Taylor's Theorem shows that for $t\in (t_0-\alpha,t_0+\alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|\le\frac12 B(t-t_0)^2<\frac12 B\alpha^2.$$
So $$2A\ge|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)| \ge|t-t_0||f'(t_0)|-\frac12 B\alpha^2$$Chooing $t$ close to $t_0\pm\alpha$ now gives $$2A\ge\alpha|f'(t_0)|-\frac12B\alpha^2,$$or$$|f'(t_0)|\le2A/\alpha+\frac12 B\alpha.$$