Are these proofs correct(bounded and finite variation).

[Azael]

First of all if you read this and the latex is all messed upp I am probably working on getting it right so please be patient till I get it right. No need to post a comment that it doesn't work. Thanks

I haven't taken a pure maths class in over 2,5 years so I can hardly remember how to write proofs

Problem 1.

Let C \in \mathbb{R} be a arbitrary number. Show that the function
f:[a,b]\rightarrow \mathbb{R}

given by f(t)=cos(ct)

is of bounded variation. i.e it satisifies the condition

Sup V_f (t) < \infty

Proof.

V_f (t) = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } <br /> \sum_{k=1}^n |{(f(t_k^n)-f(t_{k-1}^n)}|

with \pi_n : t_0^n &lt; ... &lt; t_n^n

Since Cos(ct)

is differentiable we can rewrite V_f (t) with the mean value theorem

V_f (t) = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } <br /> \sum_{k=1}^n \|{(f(t_k^n)-f(t_{k-1}^n)}| = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } \sum_{k=1}^n |f^{&#039;} (G)| (t_k^n - t_{k-1}^n)

wich is equal to(according to the definition of the riemann integral)

\int_{a}^{b} |f^{&#039;} (x)| dx

with f(x)=f(t)=cos(ct) we get

\int_{a}^{b} |-csin(ct)| dx \leq \int_{a}^{b} |c| dx = |c|(b-a)

So

Sup V_f (t) = Sup |c|(b-a) &lt;\infty

 

[Azael]

problem 2
show that the function g:[0,\infty) \rightarrow \mathbb{R}
given by g(t)=sin(ct) is of finite variation.

In the same manner as in problem one we get

V_f ((0,t)) = \int_{0}^{t} |f&#039;(t)|dt = \int_{0}^{t} |ccos(ct)|dt \leq \int_{0}^{t} |c|dt=|c|t&lt;\infty for all t equal to or larger than zero. (how do I get the equal to or larger than symbol in latex? and the "for all" symbol?)

wich shows that the function is of finite variation

 

[Azael]

problem 3
If the functions f,g:[0,\infty) \rightarrow \mathbb{R} are of finite variation, show that any linear combination of them,
\alpha f + \beta g:[0,\infty) \rightarrow \mathbb{R} \alpha,\beta \in \mathbb{R} are of finite variation.

V_{f,g} ((0,t))=\lim_{n\rightarrow \infty} \Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |{\alpha f(t_k^n)+\beta g(t_k^n)-\alpha f(t_{k-1}^n)-\beta g(t_{k-1}^n)|=
\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n| \alpha (f(t_k^n)-f(t_{k-1})+\beta (g(t_k^n)-g(t_{k-1})| \leq

(using triangle inequality)

\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |\alpha (f(t_k^n)-f(t_{k-1})|+|\beta (g(t_k^n)-g(t_{k-1})| =
(can I just move the alpha and beta outside of the lim,sup and sum like this?)
|\alpha| \lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(f(t_k^n)-f(t_{k-1})|+ |\beta|\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(g(t_k^n)-g(t_{k-1})| =

|\alpha| V_f((0,t))+|\beta| V_f((0,t)) &lt;\infty for all t equal to or larger than zero since V_f((0,t))&lt;\infty and V_g((0,t)) &lt;\infty for all t equal to or larger than zero.

 

[Azael]

I hope I haven't made any error that makes the whole thing incomprehensible :(

 

[StatusX]

For the first question, the basic idea of the proof seems correct, although I'm not sure how V_f is a function of t, it seems more a function from intervals into the real numbers. And I'm not sure what you're taking the sup of when you write sup V_f(t) or sup |c|(b-a). Besides these details it looks correct. For the second, do they mean "of bounded variation on the set [0,t], for all t in R"? If so, this is just a special case of the last problem. The third proof seems correct as well.

 

[Azael]

yes you are quite correct I just made some errors :)

Thanks a lot for checking, appreciate it.