I'm having trouble with the following :
Let $f$ be : $$f(x) = \max a_ix + b_i$$ For i in $[|1,n|]$ such as : $$a_1<a_2<...<n$$ In addition, we suppose that for each k in $[|1,n|]$, there is at least a $x$ such as $f(x)=a_kx +b_k$. (*)
Show that in this case, on each segment $[\frac{b_{k-1}-b_k}{a_k- a_{k-1}},\frac{b_{k}-b_{k+1}}{a_{k+1}- a_{k}}]$, $f(x)= a_kx +b_k$.
I have trouble proving this without any assumption on the $b_i$. When I translate (*), I get $n-1$ inequalities that should hold true and, according to the result, two of those inequalities are actually enough and I can't quite see why.
Could anybody clarify please ?
If $j>i$, consider the point $x_0$ where the lines $y=a_ix+b_i$ and $y=a_jx+b_j$. Assume $i<k<j$. Now if $a_kx_0+b_k<a_ix_0+b_i=a_jx_0+b_j$, it follows that $a_kx+b_k<a_jx+b_j$ for $x>x_0$ and $a_kx+b_k<a_ix+b_i$ for $x<x_0$, contradicting (*). Conclude that the individual $a_ix+b_i$ become "visible" fro left to right in their given order.