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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?

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This is rather a long comment: I have not seen this notion before, and don't know how I would call it. Where did you encounter this notion and what do you expect that it adds to the (already quite strong) notion of uniform convergence?

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).

Hence, my gut feeling is that your enhanced notion is not that different from usual uniform convergence.

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