Let be $f_n : \mathbb{R}^n \to \mathbb{R}$ differentiable functions such that $f_n$ converges uniformly to $f$.
Let us suppose that $f$ has a maximum strict local point in $x$.
I would like to clear my mind of hidden assumptions and understand what are the relations between $f$ and $\textrm{D}\; f$ when $f$ is a uniform limit but nothing is said about $\textrm{D}\; f_n$.
I know that most of the theorems ask for the uniform convergence of the derivative.
But is there any pointwise relations: it appears to me that $f_n$ has also maximum local points for some $n \geq N_0$ (due to uniform convergence), but I fail to show that such point exists.
And if such points exist, can we build a sequence of maximum local points which converges to $x$ such that $\textrm{D}_{x_n} f_n$ converges pointwise to $\textrm{D}_x f$ ?
Let us denote $r > 0$ some ray such that $x$ is a local strict maximum of $f$ in $B(x, r)$ the open ball.
Lemma 1: for all $n\in\Bbb N^*$, there exists $N(n) \in \Bbb N$ such as for all $m \geq N(n)$, $f_m$ admits a local maximum in $B_n = B_c(x,1/n)$ (the closed ball).
Proof of lemma 1:
Let $n \in \Bbb N^*$ be an integer. Let us take $\dfrac{1}{n} < r$. Consider $\alpha = \min_{y \in S(x, 1/n)} [f(x)-f(y)] > 0$.
Indeed, $\alpha$ cannot be null, let us suppose it would be null. That would mean there is some $y \in S(x, 1/n)$ such that $\lVert x - y \rVert = 1/n < r$, thus $x \neq y$.
But, that would mean that $x$ is not strict. Thus, $\alpha = 0$.
Because of the uniform convergence, there is $N(n) \in \Bbb N$ such as $|| f_m-f||_{\infty} <\alpha/2$ for $m\geq N(n)$. So we write, for $z\in S(x,1/n)$ (= $B_n \backslash \mathring B_n$) :
$$ \begin{align} f_m(x)-f_m(z) &= f(x) - f(z) + ( f(z) - f_m(z)) + (f_m(x) - f(x)) \\ & \geq f(x) - f(z) - 2\times\dfrac{\alpha}{2} \\ & > 0 \end{align} $$
Thus, the maximum is not in $S(x,1/n)$, it must be either $x$ or it is in $B(x, 1/n)$ the open ball.
Remark that due to the compacity of $B_n$, the image of $f$ on $B_n$ is compact, thus must have some maximum, but we just show that if maximum there is, it must be either $x$ or inside $B_n$ and not on the sphere.
Thus, there are $x^{(n)}_m \in \mathring B_n$ which is a local maximum for $f_m$ for all $m\geq N(n)$.
We build the sequence $(x_n)$ considering its values on each interval $[\![ N(n) ; N(n+1)[\![$ :
$$ \forall m \in [\![ N(n) ; N(n+1)[\![, \quad x_m = x^{N(n)}_{m} $$ By construction, $df_n(x_n) = 0$ for all $n \in \Bbb N$
Proof of convergence:
If we look what we proved in lemma 1, we see that:
\begin{equation*} \forall n \in \mathbb{N}^{*}, n \geq 1/r + 2 \implies \exists N(n) \in \mathbb{N}^*, \forall m \geq N(n), x_m^{(n)} \in \mathring B_n \end{equation*}
That is: $\forall n \in \mathbb{N}^{*}, n \geq 1/r + 2, \lvert x_n - x \rvert = \lvert x^{N(n)}_n - x \rvert < 1/n$
Thus, proving that $x_n \to x$ and we have that $df_n(x_n) = 0$ for all $n$, thus it converges to $df(x) = 0$ too trivially.