Is the pointwise limit of monotonically increasing functions a uniform limit? For a real-valued sequence of functions $\{f_n\}_{n=1}^\infty$ on the closed interval $[0,1]$, each $f_n$ is monotonocally increasing and suppose that the sequence of functions $\{f_n\}_{n=1}^\infty$ converges to the function $f$ by $n\xrightarrow{}{}\infty$ at each point

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*Show that the inequality holds\begin{align*}\sup_{z\in[x,y]}|f_n(z)-f(z)|\leq \max\{|f_n(x)-f(y)|,|f_n(y)-f(x)|\}\end{align*}

*Show that, when the function $f$ is continuous, the sequence of functions $\{f_n\}_{n=1}^\infty$ converges to $f$ uniformly on $[0,1]$
I know that a sequence of functions $\{f_n\}$ is bounded on some interval that is convergent uniformly to $f$ on that interval iff $||f_n-f||\xrightarrow{}{}0$. But this will not help to solve this problem, how do we get this inequality? Any hints would be appreciated.
 A: $f_n(z)-f(z) \leq f_n(y)-f(x)$ because $f_n(z) \leq f_n(y)$ (by monotonicity) and $f(z) \geq f(x)$ (so $-f(z) \leq -f(x)$). Similarly you get $f(z)-f_n(z) \leq f(y)-f_n(x)$. From these you conclude that  $f_n(z)-f(z) \leq \max \{|f_n(y)-f(x)|,|f(y)-f_n(x)|\} $ and $f(z)-f_n(z) \leq \max \{f_n(y)-f(x),f(y)-f_n(x)\} $. If $a \leq b$ and $-a \leq c$ then $|a| \leq c$ so we now have $|f(z)-f_n(z)| \leq \max \{|f_n(y)-f(x)|,|f(y)-f_n(x)|\} $. Taking sup over all $ z\in [x,y]$ we get the inequality in the first part.
Let $\epsilon >0$ and use uniform continuity of $f$ to choose $\delta >0$ such that $|f(y)-f(x)|<\epsilon$ whenver $|x-y| \leq \delta$.  Divide $[0,1]$ into subintervals $[x_i,x_{i+1}]$ with $x_{i+1}-x_i <\delta$ for each $i$.  Apply the inequality in the first part to each of the intervals $[x_i,x_{i+1}]$. Choose $n_1, n_2$ such that $n \geq n_1$ implies $|f_n(x_i)-f(x_i)|<\epsilon$ for $n \geq n_1$ and $n \geq n_2$ implies $|f_n(x_{i+1})-f(x_{i+1})|<\epsilon$ for $n \geq n_1$. Now you see that $\sup_{x_i \leq z \leq  x_{i+1}} |f_n(z)-f(z)|<2\epsilon$ and this is true for all $i$. Hence $\sup_{0\leq z \leq  1} |f_n(z)-f(z)|<2\epsilon$ finishing the proof.
