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# Some questions about functions of bounded variation: Jordan's thoerem watch

1. I was trying to do some of the questions in the attachment. Here are my answers.

25) No, because f(x) might be a constant function. In this case, f(x) will not be the difference of two increasing functions...and so will not be of bounded variation.

26) No, because we cannot express it as the difference of two increasing functions.

27) f(x) = sinx = (sinx + 1) - sinx. So h(x)=sinx+1 and g(x)=sinx on [o 2pi]

28) If we partition [a,b] in such a way that for each (which is part of the partition), and each correspond to the midpoint of a step. I know this is not a formal answer, but I'm just trying to check my understanding.

Can anybody check my answers please? Or give me a hint if they are wrong?
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2. (Original post by Artus)
25) No, because f(x) might be a constant function. In this case, f(x) will not be the difference of two increasing functions...and so will not be of bounded variation.
This isn't right - think about it, the total variation of a constant function is 0, so it is definitely of bounded variation. For a counterexample to this claim you're going to need a fairly "complicated" function f - there's an example of one from probability which is the thing I thought of first.
3. (Original post by Mark13)
This isn't right - think about it, the total variation of a constant function is 0, so it is definitely of bounded variation. For a counterexample to this claim you're going to need a fairly "complicated" function f - there's an example of one from probability which is the thing I thought of first.
So my counterexample is wrong, but the statement (of question 25) is still false?
4. (Original post by Artus)
So my counterexample is wrong, but the statement (of question 25) is still false?
That's right - a continuous function need not have bounded variation on some closed interval. A sample path of Brownian motion would provide a good counterexample to this claim, but if you're not familiar with BM, then there are other examples that work too.
5. Yes.

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