Let $\{f_n\}$ be a decreasing sequence of continuous functions with $f_n:[0,1]\rightarrow\mathbb{R}$, with the property that there exists an $M\in(0,1)$ such that $|f_n|\leq M$. Furthermore, $f_n\rightarrow f$ pointwise, where $f$ is also continuous. Prove that the convergence is actually uniform.
This is what I tried: Let $\epsilon>0$ be given. Then as $f$ is continuous on $[0,1]$ it is uniformly continuous. Let $\delta$ be such that $\delta$-close points map to $\frac{\epsilon}{3}$-close points. Partition $[0,1]$ into intervals, all whose length are less than $\delta$. Let $I_1,...,I_k$ be such intervals. Let $y_i$ be the midpoint of $I_i$. Let $N_i$ be such that $|f_n(y_i)-f(y_i)|<\epsilon/3$. Let $N$ be the max of all $N_i$.
Let $x\in [0,1]$ be arbitrary. Then $x\in I_k$ for some $k$. Hence, if $n\geq N$, then: $|f_n(x)-f(x)|\leq |f_n(x)-f_n(y_k)|+|f_n(y_k)+f(y_k)|+|f(y_k)+f(x)|$By choice of $N$, we have that the middle term is less that $\epsilon/3$, by continuity of $f$, the last summand is also less than $\epsilon/3$, so I was wondering how can I finish it?
Thanks