Real Analysis: Prove BV function bounded and integrable

• Jan 17th 2010, 12:34 AM
kfdleb
Real Analysis: Prove BV function bounded and integrable
1. The problem statement, all variables and given/known data

f is of bounded variation on [a;b] if there exist a number K such that

$\displaystyle \sum$$\displaystyle ^{n}_{k=1}|f(ak)-f(ak-1)| \displaystyle \leqK a=a_0<a_1<...<a_n=b; the smallest K is the total variation of f I need to prove that 1) if f is of bounded variation on [a;b] then it is bounded on [a;b] 2) if f is of bounded variation on [a;b], then it is integrable on [a;b] 2. The attempt at a solution i thought of using triangle inequation such that 0<=|f(b)-f(a)|<=\displaystyle \sum$$\displaystyle ^{n}_{k=1}$|f(ak)-f(ak-1)| $\displaystyle \leq$K

but im not really sure how to prove the two statements
any help is really appreciated
thanks
• Jan 17th 2010, 05:14 AM
Plato
Quote:

Originally Posted by kfdleb
1. The problem statement, all variables and given/known data

f is of bounded variation on [a;b] if there exist a number K such that

$\displaystyle \sum$$\displaystyle ^{n}_{k=1}|f(ak)-f(ak-1)| \displaystyle \leqK a=a_0<a_1<...<a_n=b; the smallest K is the total variation of f I need to prove that 1) if f is of bounded variation on [a;b] then it is bounded on [a;b] 2) if f is of bounded variation on [a;b], then it is integrable on [a;b] 2. The attempt at a solution i thought of using triangle inequation such that 0<=|f(b)-f(a)|<=\displaystyle \sum$$\displaystyle ^{n}_{k=1}$|f(ak)-f(ak-1)| $\displaystyle \leq$K

but im not really sure how to prove the two statements
any help is really appreciated
thanks

Please learn how to use LaTeX at this site.
Then repair this post so it is readable.
• Jan 17th 2010, 08:13 AM
Dinkydoe
On other sites they use $\displaystyle code$ but here tex = math
• Jan 17th 2010, 02:31 PM
Drexel28
Quote:

Originally Posted by kfdleb
1. The problem statement, all variables and given/known data

f is of bounded variation on [a;b] if there exist a number K such that

$\displaystyle \sum$$\displaystyle ^{n}_{k=1}|f(ak)-f(ak-1)| \displaystyle \leqK a=a_0<a_1<...<a_n=b; the smallest K is the total variation of f I need to prove that 1) if f is of bounded variation on [a;b] then it is bounded on [a;b] 2) if f is of bounded variation on [a;b], then it is integrable on [a;b] 2. The attempt at a solution i thought of using triangle inequation such that 0<=|f(b)-f(a)|<=\displaystyle \sum$$\displaystyle ^{n}_{k=1}$|f(ak)-f(ak-1)| $\displaystyle \leq$K

but im not really sure how to prove the two statements
any help is really appreciated
thanks

Just so our notation is straight. We have that $\displaystyle V\left(\phi,P\right)=\sum_{j=1}^{n}\left|\phi\left (x_j\right)-\phi\left(x_{j-1}\right)\right|$. Then the variation of $\displaystyle \phi$ on $\displaystyle [a,b]$ is $\displaystyle \sup_{P\in\mathcal{P}} V\left(\phi,P\right)$ where $\displaystyle \mathcal{P}$ is the set of all partitions of $\displaystyle [a,b]$.

Then to prove that $\displaystyle \phi$ is of bounded variation is simple since. We have that that $\displaystyle \left|\phi(x)-\phi(k)\right|\leqslant \left|phi(x)-\phi(k)\right|+\left|\phi(x)-\phi(b)\right|\leqslant \text{Var}(\phi)$ and since $\displaystyle \phi$ is of bounded variation we have that $\displaystyle \text{Var}(\phi)<\infty$. From where it follows that $\displaystyle |\phi(x)|\leqslant \text{Var}(\phi)+|\phi(a)|<\infty$, thus it is bounded.

What do you have for the second one? Let me know.