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 $$\pi_n : t_0^n < ... < t_n^n$$

Since $$Cos(ct)$$

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

$$V_f (t) = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } <br /> \sum_{k=1}^n \|{(f(t_k^n)-f(t_{k-1}^n)}| = \lim_{n\rightarrow \infty} Sup_{ t_k^n,t_{k-1}^n \in \pi_n } \sum_{k=1}^n |f^{'} (G)| (t_k^n - t_{k-1}^n)$$

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

$$\int_{a}^{b} |f^{'} (x)| dx$$

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

$$Sup V_f (t) = Sup |c|(b-a) <\infty$$

 

[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

$$V_f ((0,t)) = \int_{0}^{t} |f'(t)|dt = \int_{0}^{t} |ccos(ct)|dt \leq \int_{0}^{t} |c|dt=|c|t<\infty$$ 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?)

which 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?)
$$|\alpha| \lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(f(t_k^n)-f(t_{k-1})|+ |\beta|\lim_{n\rightarrow \infty}Sup_{ t_k^n,t_{k-1}^n \in \pi_n }\sum_{k=1}^n |(g(t_k^n)-g(t_{k-1})| =$$

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