Let $f$ fail to be of bounded variation on [0,1]. Show that there is a point $x_0$ in [0,1] such that $f$ fails to be of bounded variation on each nondegenerate closed subinterval of [0,1] that contains $x_0$. I'm trying proving this directly, but I think maybe a proof by contradiction could work here. Couldn't we just assume that $f$ is of bounded variation on every closed, bounded subset of [0,1] to begin the proof by contradiction? Perhaps then we could split up the interval [0,1] so that we end up with a situation where $f$ is of bounded variation on a subinterval that contains $x_0$? I'd appreciate some help here.
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As Didier mentioned, the right statement probably is:
Here is an alternative proof: We use the standard notation $V_a^b(f))$ to denote the variation of $f$ on $[a,b]$. Let $A : =\{ x \in [0,1] | V_0^x(f) < \infty \}$. Then $0 \in A , 1 \notin A$ and it is very easy to see that $x_0=\sup A$ works. |
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I bet on compactness... You want to prove that if $f\notin BV([0,1])$, then the following holds: $$ \exists x_0\in [0,1] :\ \forall I \subseteq [0,1] \text{ nondegenerate closed subinterval containing } x_0,\ f\notin BV(I) $$ If you argue by contradiction, you assume that: $$\forall x\in [0,1],\ \exists I_x\subseteq [0,1] \text{ nondegenerate closed subinterval containing } x:\ f\in BV(I)$$ The family $\{\text{int }I_x\}_{x\in [0,1]}$ covers $[0,1]$ ($\text{int}$ denotes the interior w.r.t. the topology induced on $[0,1]$), hence by compactness you can find $x_1,\ldots ,x_N \in [0,1]$ s.t. $[0,1]=\bigcup_{n=1}^N \text{int }I_{x_n}$, so $[0,1]=\bigcup_{n=1}^N I_{x_n}$; now $f\in BV(I_{x_n})$ for $n=1,\ldots ,N$, therefore $f\in BV([0,1])$, which is a contradiction. I think it works... But you have to check the details. |
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I spent half an hour trying to prove this before I realised that it's obviously wrong. Take for instance: $f(x) = 0$ if $x \le \frac{1}{2}$, otherwise $f(x) = 1/(x-\frac{1}{2})$ Did you want $f$ to be continuous? I think it's still wrong. Something like: $f(x) = 0$ if $x \le \frac{1}{2}$, otherwise $f(x) = (x-\frac{1}{2})\sin(1/(x-\frac{1}{2}))$ |
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