The sequence of function $\{f_n\}$ defined on $\mathbb{R}$, every function is decreasing function (if $x \geq y$ then $f_1(x)\geq f_1(y)$, $f_2(x)\geq f_2(y)$,.......) and sequences of function is decreasing function ($f_n\geq f_{n+1}$) and $\{f_n \}$ converges point wise.
Then $f_n$ converges to $f$ uniformly.
Example :
$f_n(x)=x+\frac{1}{n}$ , $n=1,2,3$..... defined on $(-\infty , 0]$. This sequence of function is uniformly convergence
My assumption is correct.
Perhaps one should first recall that one cannot prove that a statement holds simply exhibiting a case where it holds. Here, to exhibit a sequence $(f_n)$ such that all the hypotheses hold and such that $f_n\to f$ uniformly proves nothing about the validity of the general statement.
...Which happens to be false, as a single counterexample is enough to prove. Such as the following one.
Assume that $f_n(x)=1$ if $x\leqslant0$, $f_n(x)=1-nx$ if $0\leqslant x\leqslant\frac1n$ and $f_n(x)=0$ if $x\geqslant\frac1n$. Then, as desired, each function $f_n$ is nonincreasing, the sequence $(f_n(x))$ is nonincreasing for every fixed $x$, and $f_n\to f$ pointwise with $f(x)=1$ if $x\leqslant0$ and $f(x)=0$ if $x\gt0$.
But the convergence $f_n\to f$ is not uniform since $f_n(\frac1{n^2})=1-\frac1n$ and $f(\frac1{n^2})=0$ hence the difference $f_n(\frac1{n^2})-f(\frac1{n^2})$ does not converge to $0$.