Proving properties of derivative of uniformly convergent functions Prove or disprove that if $\{f_n\}_{n=1}^\infty$ is a sequence in $C^2(\mathbb{R},\mathbb{R})$, and for some finite $M''$ and for all $n$, $\sup_{\mathbb{R}}|f_n''|\le M''$, and $f_n \to 0$ uniformly as $n \to \infty$, then for some $M'$ and all $n$, $\sup_{\mathbb{R}}|f_n'| \le M'$
I can't come up with any counter-examples, so I think it might be true. I was hoping someone could help me finish of the below proof, or give a better one.
My attempt: from Taylor's theorem: $$f_n'(x) = \frac{1}{2h}(f_n(x+2h)-f_n(x))-hf_n''(c)$$ for all $h>0$, and some $c \in [x,x+h]$.Thus, for all $h>0$, 
$$|f_n'(x)|\le \frac{1}{2h}|f_n(x+2h)-f_n(x)|+h|f_n''(c)|=\frac{1}{2h}|f_n(x+2h)-f_n (x)|+M''$$
My issue is that now I need to bind $f_n(x+2h)-f_n(x)$. Here's my attampt (which I know is wrong): Since $f''$ exists, $f$ is continuous, so $$\forall \epsilon>0, \forall x \in \mathbb{R} \exists \delta(\epsilon) >0 \text{such that} |x-y|<\delta(\epsilon) \implies |f(x)-f(y)|<\epsilon, \forall x,y \in \mathbb{R}$$
 Thus, choose $\epsilon =1, x\in \mathbb{R}$ choose $h<\delta(1)$, and then we have
$$|f_n'(x)|\le \frac{1}{2h}|f_n(x+2h)-f_n (x)|+M''\le \frac{1}{2\delta(1)}+M''$$
This is where I get lost, since this $\delta(1)$ depends on $x$, and may go to zero. I think I need to somehow use the fact that $f$ converges uniformly to $0$, but I'm not sure what to do. 
 A: The condition may not hold for a finitely many $f_n$.  For example, if we take $f_1(x) = x^2$ then $f_1''(x) = 2$ is bounded on $\mathbb{R}$ but $f_1'(x) = 2x$ is not.
However, for all sufficiently large $n$ there exists $M' >0$ such that $\sup_{\mathbb{R}}|f_n'| \leqslant M'.$
Since $f_n \to 0$ uniformly as $n \to \infty$, given some $M >0$ there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ and all $x \in \mathbb{R}$ we have $|f_n(x)| \leqslant M$.
By Taylor's theorem, for any $x$ and $h \neq 0$ there exists $c$ between $x$ and $x +h$ such that
$$f_n(x+h) = f_n(x) + f_n'(x) h + \frac{1}{2} f_n''(c)h^2.$$
Hence,
$$|f_n'(x)| = \left|\frac{f_n(x+h) - f_n(x)}{h} + f_n''(c) \frac{h}{2}\right| \\ \leqslant \frac{|f_n(x+h)| + |f_n(x)|}{|h|} + |f_n''(c)| \frac{|h|}{2}\\ \leqslant \frac{2M}{|h|} + \frac{M''}{2}|h|.$$
The function $x \mapsto 2M/x + M''x/2$ is minimized over $(0,\infty)$ at $x = 2 \sqrt{M/M''}$ with minimum value $2\sqrt{M M''}$.  
Since $h$ is arbitrary it follows that for all $n \geqslant N$ and for all $x \in \mathbb{R}$,
$$|f_n'(x)| \leqslant M' = 2\sqrt{M M''}, $$
which implies
$$ \sup_{\mathbb{R}}|f_n'| \leqslant M'$$
