# Function fails to be of bounded variation

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.

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.

• Indeed, compactness is going to have to come in somehow. Note that the theorem is false if you replace $[0,1]$ by $(0,1)$. Apr 8, 2011 at 15:09
• Good job.
– Did
Apr 8, 2011 at 15:10
• @Didier, @Pacciu: Something's wrong. See my answer. Apr 8, 2011 at 17:39
• (Just came back to a computer after two days without any.) The faulty step is to say that the family of the interiors covers $[0,1]$. But this step holds again if you modify things as suggested in a comment on TonyK's answer.
– Did
Apr 10, 2011 at 20:39
• @Didier: So the problem is that $x_0$ may be contained in the boundary of the closed interval... Fair! But I bet Sachin's statement has to be interpreted as you wrote in your comment on TonyK's answer. Thank you Didier. ;-D Apr 10, 2011 at 22:39

As Didier mentioned, the right statement probably is:

Show that there is a point x0 in [0,1] such that f fails to be of bounded variation on each nondegenerate closed subinterval of [0,1] which is a neighborhood of x0 in [0,1].

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.

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 $f(x) = (x-\frac{1}{2})\sin(1/(x-\frac{1}{2}))$

• It seems to me both of your $f$s fail to be of bounded variation in $[0,1]$ and in any neighbourhood of $1/2$; hence Sachin's assertion holds with $x_0=1/2$. But maybe I can't see the point... Apr 9, 2011 at 12:27
• (Just came back to a computer after two days without any.) If you replace on each nondegenerate closed subinterval of $[0,1]$ that contains $x_0$ by on each nondegenerate closed subinterval of $[0,1]$ which is a neighborhood of $x_0$ in $[0,1]$, it sems the assertion holds. (But I do not understand your second example: the function is BV.)
– Did
Apr 10, 2011 at 20:37
• @Pacciu: OP's conjecture didn't say 'any neighbourhood of $x$', it said 'each nondegenerate closed subinterval that contains $x$'. So it's false. In my counterexample, if $x=1/2$ just take the interval $[0,1/2]$. Apr 12, 2011 at 5:27
• @Didier: So we are agreed that OP's statement is false as it stands. And you were right about my second example function: I have corrected it. Apr 12, 2011 at 5:41