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1. Homework Statement [/b]
If f has a continuous derivative on [a,b], and if P is any partition of [a,b], show that V(f,P)\leq \intablf'(t)l dt. Hence, Vba\leq\intablf'(t)ldt.
Monotone function \subset BV[a,b]
\sumf(ti+1)-f(ti) = lf(b) - f(a)l
Let P = {a=t0 < t1 < ... < tn}. So if we divide our function into monotone segments we have:
V1(f,P) = \sumf(ti+1)-f(ti) = lf(a1) - f(a)l
V2(f,P) = \sumf(ti+1)-f(ti) = lf(a2) - f(a1)l
.
.
.
Vn(f,P) = \sumf(ti+1)-f(ti) = lf(b)- f(an-1)l
Then, treating this segments independently of the whole, we see that
v1(f,P)= lf(a)-f(a1)l=\intablf'(t)ldt = lf(a1) -f(a)l
etc.
Adding them all up V(f,P)= \intablf'(t)l dt, which satisfies our prompt.
If f has a continuous derivative on [a,b], and if P is any partition of [a,b], show that V(f,P)\leq \intablf'(t)l dt. Hence, Vba\leq\intablf'(t)ldt.
Homework Equations
Monotone function \subset BV[a,b]
\sumf(ti+1)-f(ti) = lf(b) - f(a)l
The Attempt at a Solution
Let P = {a=t0 < t1 < ... < tn}. So if we divide our function into monotone segments we have:
V1(f,P) = \sumf(ti+1)-f(ti) = lf(a1) - f(a)l
V2(f,P) = \sumf(ti+1)-f(ti) = lf(a2) - f(a1)l
.
.
.
Vn(f,P) = \sumf(ti+1)-f(ti) = lf(b)- f(an-1)l
Then, treating this segments independently of the whole, we see that
v1(f,P)= lf(a)-f(a1)l=\intablf'(t)ldt = lf(a1) -f(a)l
etc.
Adding them all up V(f,P)= \intablf'(t)l dt, which satisfies our prompt.
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