Let $(X,d)$ be a compact metric space, and let $f_n : X \to \mathbb{R}$ define a sequence of functions $(f_n)$. Assume it to converge pointwise to a continuous function $f: X \to \mathbb{R}$, and $f_n(x)$ is a monotonic decreasing function for all $n$ and for all $x$ belonging to $X$. I eed to prove that $(f_n)$ converge to $f$ uniformly.
I thought that $(f_n)$ may continuous. because we know that if $f_n$ is continuous and converges uniformly to $f$, then $f$ is continuous. I try to prove like this: $$|f_n(x)-f_n(x_0)|\leq|f_n(x)-f(x)|+|f(x)-f(x_0)|+|f(x_0)-f_n(x_0)|<ε$$
Is this correct?
No. For a counterexample, take $X = [0,2]$, let $f_n$ be $\frac{1}{n}I_{[0,1]} + \frac{1}{2n}I_{[0,2]}$, where $I_A$ is the identifier function of $A$, so $$I_A(x) \ \left\{\array{1,&x\in A\\0,&x\not\in A.}\right.$$ Clearly $f_n \to 0$ pointwise, but none of the $f_n$ are continuous.