Locally uniformly convergent sequence I tried to prove the following:
1) $f\in C^{k}(\Bbb{R}^n)$
2) $f\in C_b^{\alpha}(\Bbb{R}^n)$
3) $\|f\|_{C^{\alpha}_{b}(\Bbb{R}^n)}\le M$
Here we assume that 1) and 2) hold. Then, 
$$
\|f\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}=\|f+f_n-f_n\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}\le\|f_n\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}+\|f-f_n\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}\le K + \|f-f_n\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}.
$$
Now,  we let $n\rightarrow \infty$ and see that $\|f\|_{C^{k,\alpha}_{b}(\Bbb{R}^n)}\le K$.
I dont think that my approach is correct since I do not use the locally uniform property.  I am also really struggling with 2). How can I prove this proposition? Any hints are appreciated thank you.
 A: First of all, you need $\alpha>0$ here; the statement fails for $k\in\mathbb{N}$ and $\alpha=0$, because a uniformly convergent sequence of differentiable functions without uniform control on the modulus of continuity of the derivative may lose differentiability in the limit. For example, $\sqrt{x^2+1/n} \to |x|$.
You are right that the locally uniform convergence is not needed; pointwise is enough.
Assume $\alpha>0$. Since the sequences $\sup|D^\beta f_n|$ and $[D^\beta f_n]_\alpha$  are  bounded, we can assume they converge, by passing to a subsequence of $\{f_n\}$. That is, as $n\to\infty $, we have $\sup|D^\beta f_n| \to M_\beta$ for $|\beta|\le k$ and $[D^\beta f_n]_\alpha \to L_\beta$ for $|\beta|=k$. The goal is to show that $\sup|D^\beta f| \to M_\beta$ for $|\beta|\le k$ and $[D^\beta f]_\alpha \le L_\beta$ for $|\beta|=k$.
Let's deal with the case $\beta=0$ first. The fact that $|f|\le M_0$ everywhere follows from the pointwise convergence $f_n\to f$, since this is a pointwise inequality. Same goes for the inequality $|f(x)-f(y)| \le L_0 |x-y|^\alpha $.  Done.
The above will also work for other values of $\beta$, as soon as we get convergence of derivatives $D^\beta f_n$. This requires passing to a subsequence since in general they won't be convergent (consider for example $f_n(x) = n^{-1}\sin nx$ which is bounded in $C^{0,\alpha}$ for any $\alpha\in (0,1]$, and converges to $0$ uniformly, yet $f'_n(x) = \cos nx$ does not converge to $0$).
On the ball $\{x: |x|\le N\}$ the functions $D^\beta f_n$, $|\beta|\le k$, are equicontinuous and uniformly bounded; thus, we can pick a subsequence along which they all converge uniformly on this ball. Do the same for $N=1,2,3,\dots$ and pick a diagonal subsequence to get $f_{n_j}$ with all derivatives of orders up to $k$ converging locally uniformly on $\mathbb{R}^n$. Note that the limit of $f_{n_j}$ will be $f$, since we have pointwise convergence to $f$. And since uniform convergence of derivatives allows us to interchange derivative and limit, the derivatives of $f_{n_j}$ converge to the corresponding derivatives of $f$.  The proof completes by applying the same reasoning as in the case $\beta=0$. 
