If $C^{1}(\mathbb{R}^{n})\ni g_{n}\to g$ uniformly and $\partial_{j}g_{n}\to g^{(j)}$ then $g \in C^{1}(\mathbb{R}^{n})$ and $\partial_{j}g=g^{(j)}$. I was studying Schwartz spaces and found this notes. On page 5, there is a Lemma which states:

If $\{g_{n}\}\subset C^{1}(\mathbb{R}^{n})$ converges uniformly to $g$ and $\partial_{j}g_{n}$ converges uniformly to $g_{(j)}$, then $g \in C^{1}(\mathbb{R}^{n})$ and $\partial_{j}g = g_{(j)}$.

I'd like to see the proof of the above lemma. Does anyone know how to prove or where to find such a proof?
 A: In essence, you're asking to verify that the space of bounded differentiable functions with bounded continuous derivative $(C^1(\mathbb R^N),\|\cdot\|_{C^1})$ is complete, where $\|f\|_{C^1} := \|f\|_{C^0} + \|f'\|_{C^0}$.
(Note that you used  $n$  to mean both a sequence index and the dimension. For me dimension will be $N$.)
(Also note that some people allow all unbounded continuously differentiable functions in the set $C^1$. These two sets are not the same)
The proof is by the FTC:  we have, for any $x\in\mathbb R^N$, $n\in\mathbb N$, $t\in\mathbb R$, and $i\in \{1,\dots,N\}$,
$$ g_n(x+te_i)=g_n(x)+\int_0^t g'_n(x+se_i)ds$$
Taking limits $n\to\infty$ we see, interchanging limit and integral (uniform convergence on compact, or DCT),
$$ g(x+te_i)=g(x)+\int_0^t g_{(i)}(x+se_i)ds$$
The FTC now implies $g\in C^1$ with $g'(x)=g_{(i)}$. Specifically if you set
$$ G(t) := g(x+te_i)-g(x), H(s) = g_{(i)}(x+se_i)$$
then the FTC says that since $G(t) = \int_0^t H(s)ds$ where $H\in C^0(\mathbb R)$, then $G'$ exists and $G'=H$, and of course $g_{(i)}(x)=H(0)=G'(0)=\lim_{t\to 0} \frac{g(x+te_i)-g(x)}t$. So $\partial_i g(x)$ exists for each $i,x$, and $\partial_i g$ is equal to the $C^0(\mathbb R^N)$ function $g_{(i)}$. Hence $g$ is differentiable in the sense of Frechét, with $\|g\|_{C^1}<\infty$, which proves the result.
