# name of uniformly convergence $f_n \to f$ when $|f_{n}(x) - f(x)|$ decreases monotonically

Suppose that $$(f_n)$$ is a sequence of real functions which $$f_n \to f$$ uniformly. Also suppose this condition is true for that sequence:

\begin{align} \forall n \in \mathbb{N} : |f_{n+1}(x) - f(x)| \leq |f_{n}(x) - f(x)| \tag{1} \end{align} Does adding condition(1) to uniformly convergence has specific convergence name?

Indeed, let us take $$f_n : \mathbb{R} \to \mathbb{R}$$ converging uniformly to $$f$$. Then, $$\sup_{x \in \mathbb{R}} | f_n(x) - f(x) | \to 0$$. Hence, up to extracting a subsequence, you can assume that $$\sup_{x \in \mathbb{R}} | f_n(x) - f(x) |$$ actually decreases to $$0$$ with $$\sup_{x \in \mathbb{R}} | f_{n+1}(x) - f(x) | \leq \sup_{x \in \mathbb{R}} | f_n(x) - f(x) |$$.
While these estimates don't exactly describe a pointwise-enhancing approximation, you can nevertheless "localize" them. Take $$[a,b] \subset \mathbb{R}$$. Up to extracting a subsequence, you can assume that $$\sup_{x \in [a,b]} | f_n(x) - f(x) |$$ actually decreases to $$0$$ with $$\sup_{x \in [a,b]} | f_{n+1}(x) - f(x) | \leq \sup_{x \in [a,b]} | f_n(x) - f(x) |$$.
Since you can do this for any rational interval, by a diagonal argument, you can even extract, from any uniformly converging sequence of real functions, a subsequence such that, $$\forall [a,b] \subset \mathbb{R}, \forall n \in \mathbb{N}, \quad \sup_{x \in [a,b]} | f_{n+1}(x) - f(x) | \leq \sup_{x \in [a,b]} | f_n(x) - f(x) |.$$ This corresponds to the intuition that the approximation error is monotonically decreasing locally (although indeed maybe not pointwise).