Let $f_n\colon [0,1] \to \mathbb{R}$ be a non-negative function for each $n$ such that $$\lim_{x \to 0}f_n(x) = k$$ and $$\lim_{n \to \infty} f_n(x) = f(x).$$ Moreover, the first convergence above is non-increasing, and the second is non-decreasing.
Is there any way to conclude that $f_n(x) \to k$ as $x \to 0$ converges uniformly in $n$?
On
If by $$\lim_{x\to 0} f_n(x) = k$$ uniformly in $n$ you mean that for each $\epsilon >0$ there is a $\delta(\epsilon)$ such that, if $$|x| < \delta(\epsilon)$$ then $$|f_n(x)-k| < \epsilon$$ for all $n\in \Bbb N,$ then you might consider as a counterexample that satisfies your hypotheses $$f_n(x) = 1-e^{-nx^2}, \ \ n =1,2,\dots$$ (you can restrict it in $[0,1]$ if you prefer). We have $$\lim_{x\to 0} f_n(x) =0^+,$$ $$f_{n+1}(x) \geq f_n(x),$$ and $$\lim_{n\to +\infty}f_n(x) = f(x),$$ with $$f(x) = \begin{cases} 1 & (x\neq 0)\\ 0 & (x=0). \end{cases} $$
Definition of Uniform Convergence is state for converging a sequence of functions to specific function. It this case you can't conclude $\{f_n\}$ uniformly converges to $f$. And at just one point like $x=0$ as you try, there is no uniformly or nonuniformly convergency has meaning!