Approximation of convex functions

[Rigel1]

Let $f:[0,1]\to RR$ be a continuous convex function.
Given $n\in NN$, let $x_i = i/n$, $i=0,...,n$ be a partition of $[0,1]$, and let $g_n$ be the piecewise affine functions such that
$g_n(x_i) = f(x_i)$ for every $i=0,...,n$, and $g_n$ is affine on each interval $[x_{i-1}, x_{i}]$, $i=1,...,n$.

1. (Easy) Prove that for every $\epsilon>0$ there exists $N\in NN$ such that
$|f(x)-g_n(x)| < \epsilon$ for every $x\in [0,1]$, for every $n\ge N$.

2. (Not so easy) Show that there exist numbers $a, b_0\in RR$, $b_1,..., b_n\ge 0$ such that
$g_n(x) = a + b_0 x + \sum_{i=1}^n b_i |x-x_i|$, for every $x\in [0,1]$.

[gugo82]

OK Leo!

[Leonardo891]

My try for 1bis.

From 1 we know that
$AA \epsilon > 0, EE N \in NN : AA n >= N, AA x \in [0,1], |f(x)-g_n(x)|<\epsilon $
Since f is convex than is convex also f restricted to $[x_{i-1},x_i], i \in {1,n}$ but in $[x_{i-1},x_i]$ $g_n(x)$ is a line for $(x_{i-1},f(x_{i-1}))$ and $(x_{i},f(x_{i}))$ so by definition of convexity we have $0 <= f(x)-g_n(x)<\epsilon $ in $[x_{i-1},x_i]$.
Since this is true $AA i \in {1,n}$ we have the thesis.

[gugo82]

... Just to use convexity in 1. too.

1 bis. Show that for each [tex]$\varepsilon >0$[/tex] there exists [tex]$N\in \mathbb{N}$[/tex] s.t. for all [tex]$n\geq N$[/tex] and [tex]$x\in [0,1]$[/tex] we have [tex]$0\leq f(x)-g_n(x)<\varepsilon$[/tex].

[Rigel1]

"Leonardo89":
Why is the convexity necessary? There are some mistakes in my proof?

Well done.
For this first point convexity is not necessary.
Only (uniform) continuity is used.

[Leonardo891]

Hello!
Here there's my try for 1).

Since $f$ is a continuous function on a compact it's uniformly continuous then
$AA \epsilon > 0, EE \delta > 0 : AA x,y \in [0,1], |x-y|< \delta => |f(x)-f(y)|<\epsilon$
Let be $ N \in NN : frac{1}{N}< \delta $.
Let be $x \in [0,1]$, $n>=N$ and let be $ x_i, i \in {0,...,n} $ the partition like the above description.

$EE i \in {0,...,n-1} : x\in[x_i,x_{i+1}]$ then
$|f(x)-g_n(x)|<=|f(x)-g_n(x_i)|+|g_n(x_i)-g_n(x)|<=|f(x)-f(x_i)|+|g_n(x_i)-g_n(x_{i+1})|=|f(x)-f(x_i)|+|f(x_i)-f(x_{i+1})|<2 \epsilon $

Why is the convexity necessary? There are some mistakes in my proof?