- #1
jdz86
- 21
- 0
Homework Statement
A.) Prove: [tex]m_{i}[/tex](f) + [tex]m_{i}[/tex](g) [tex]\leq[/tex] [tex]m_{i}[/tex](f+g) [tex]\leq[/tex] [tex]M_{i}[/tex](f+g) [tex]\leq[/tex] [tex]M_{i}[/tex](f) + [tex]M_{i}[/tex](g)
B.) Prove: [tex]M_{i}[/tex]([tex]\alpha[/tex]f) = {[tex]\alpha[/tex][tex]M_{i}[/tex](f) [tex]\alpha[/tex] [tex]\succ[/tex] 0 or [tex]\alpha[/tex][tex]m_{i}[/tex](f) [tex]\alpha[/tex] [tex]\prec[/tex] 0
and
[tex]m_{i}[/tex]([tex]\alpha[/tex]f) = {[tex]\alpha[/tex][tex]m_{i}[/tex](f) [tex]\alpha[/tex] [tex]\succ[/tex] 0 or [tex]\alpha[/tex][tex]M_{i}[/tex](f) [tex]\alpha[/tex] [tex]\prec[/tex] 0
C.) Take P = [tex]P_{1}[/tex] = {a,b}, and write [tex]m_{i}[/tex] = m, [tex]M_{i}[/tex] = M (i=1). Give an example for each of the following:
(a) m(f) + m (g) [tex]\prec[/tex] m(f+g)
(b) M(f+g) [tex]\prec[/tex] M(f) + M(g)
(c) both (a) and (b) hold
Homework Equations
P = {[tex]x_{0}[/tex],...,[tex]x_{n}[/tex]} and is a partition of [a,b]
[tex]M_{i}[/tex] = {sup {f(x) : x [tex]\in[/tex] [[tex]x_{i-1}[/tex], [tex]x_{i}[/tex]]}}
[tex]m_{i}[/tex] = {inf {f(x) : x [tex]\in[/tex] [[tex]x_{i-1}[/tex], [tex]x_{i}[/tex]]}}
[tex]\alpha[/tex] is just any multiplier
The Attempt at a Solution
For A.) and B.) i was going to take different sets and show it that way, but it will be a long drawn out process, i can't put my finger on the "shortcut/abbreviated" version. Any suggestions on what method to use??
And for C.) (a) f = [-2,-1) g = [-1,0]
(b) f = [0,1) g = [1,2]
(c) ??