The problem goes like: Suppose that $f\in B[a,b]$. If $V^b_{a+\epsilon}f\leq M$ for all $\epsilon >0$, does it follow that f is of bounded variation on $[a,b]$?

I think the answer is yes. Since $V^b_{a+\epsilon}f$ is uniformly bounded and f is also bounded, also since $\epsilon$ is arbitrary, we have f is of bounded variation of $[a,b]$. Is this correct?

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    $\begingroup$ The key point is of course the uniform bound, i.e., the fact that $M$ does not depend on $\epsilon$ in the precise formulation in the body of this post (but this is not the case in the question in the title of this post) $\endgroup$ – Hagen von Eitzen Jan 24 '16 at 0:01
  • $\begingroup$ Just curious as to why the assumption that $f$ is bounded? If $V^b_{a+\epsilon}f\leq M$ for all $\epsilon>0$ then $f$ must be bounded anyway on $(a,b]$ by $M + |f(b)|$ and so $f$ is bounded on $[a,b]$ by $M+|f(a)|+|f(b)|$. $\endgroup$ – B. S. Thomson Jan 24 '16 at 0:37
  • $\begingroup$ "...as it is it makes no sense" is sometimes more commonly expressed on this website by the phrase "I don't understand the argument." Fix $a<x<b$ then $$|f(b)-f(x)| \leq V^b_{a+\epsilon}f\leq M$$ for sufficiently small $\epsilon$. Hence $|f(x)|\leq M + |f(b)|$. Thus $M + |f(b)|$ is an upper bound for $f$ on $(a,b]$. $\endgroup$ – B. S. Thomson Jan 24 '16 at 16:20

Indeed, if $a$ is a point of a given partition, then it is sufficient to take $\epsilon$ sufficiently small to make sure that the next point is already in $[a+\epsilon,b]$. Boundedness of $f$ takes care of the first interval.

  • $\begingroup$ One thing to add perhaps. What is your best estimate for $V_a^b f$? $\endgroup$ – B. S. Thomson Jan 24 '16 at 0:01
  • $\begingroup$ It would be the same bound, right? $\endgroup$ – metricspace Jan 24 '16 at 0:09
  • $\begingroup$ Almost the same. (Function need not be continuous here.) $\endgroup$ – B. S. Thomson Jan 24 '16 at 0:11
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    $\begingroup$ $V_a^b f \leq M+\lim_{c\to a^+} \sup\{|f(a)-f(x)| : x\in (a,c]\}\leq M+\sup_{x\in (a,b]}|f(a)-f(x)|.$ E.g. if $f(x)=0$ for $x\in (a,b]$ and $f(a)=1$, with $M=0$; and $V_a^bf=1.$ $\endgroup$ – DanielWainfleet Jan 24 '16 at 0:38
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    $\begingroup$ How about $M+\limsup_{x\to a}|f(x)-f(a)|$? $\endgroup$ – John B Jan 24 '16 at 0:45

Since the other answer is more a wave of the hand than an answer one would submit to your instructor I thought a proper, economical, write-up might be welcome. (Not that there is anything wrong with "hand-waving" we all do it: the intention is "Here is the idea -- You write it up yourself now.")

Also the other poster was less than pleased by the discussion and asked (one presumes) for details.

Problem. Let $f:[a,b]\to \mathbb R$ have the property that the variation $V_c^b f \leq M$ for all $a<c<b$. Show that $f$ has bounded variation on $[a,b]$ and that $$V_a^b f \leq M + |f(a+)-f(a)|.$$ In particular, if $f$ is continuous on the right at $a$, then $V_a^b f \leq M$.

First observe that, for any $a<x<b$ $$ |f(b) -f(x)| \leq V_x^b f \leq M.$$ Hence $|f|$ is bounded on $(a,b]$ by $M+|f(b)|$.

Now take any subdivision $a=a_0<a_1 <a_2< \dots < a_n=b$ and any $a<t<a_1$. Observe that $$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq |f(t)-f(a)| + |f(a_1)-f(t)| + \dots +|f(a_n)-f(a_{n-1})| $$ $$ \leq |f(t)-f(a)| + M .$$ This is true for all such $t$ and $f$ is bounded, so $$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq \liminf_{t\to a+} |f(t)-f(a)| + M< \infty.$$

Since this is true for all such subdivisions of $[a,b]$ we have $$ V_a^b f \leq \liminf_{t\to a+} |f(t)-f(a)| + M$$ and so $f$ has bounded variation on $[a,b]$.

All functions of bounded variation are regulated (i.e. they have finite right and left hand limits at each point) and so $$\liminf_{t\to a+} |f(t)-f(a)| = \lim_{t\to a+} |f(t)-f(a)| = |f(a+)-f(a)|$$ completing the proof.

  • $\begingroup$ Just a question, what if the partition is inifinite? like in the case of $x^{-2}sin(1/x)$ $\endgroup$ – metricspace Jan 27 '16 at 1:34
  • $\begingroup$ The definition for variation of a function on an interval uses finite partitions. So for all purposes (proofs/examples/applications) that is all you need. If you really want to use an infinite sequence, say $0 < ...< a_{n+1}< a_n < ... < 1$ with $a_n\to 0$ to estimate the variation for $x^{-2}\sin x^{-1}$ on $[0,1]$ you would fix an integer $m$ and use the finite partition $\{0, a_m,a_{m-1}, ..., a_1, 1\}$. So, while it appears you are using an infinite partition you would write it up using this finite partition. That help? $\endgroup$ – B. S. Thomson Jan 27 '16 at 3:58
  • $\begingroup$ Downvoted perhaps out of spite? This answer is intended more as a elaboration since the problem as posed had a redundancy (assuming $f$ is bounded) and did not ask, as it should, for a best estimate on $V_a^b f$, If you don't like the elaboration, please do downvote. $\endgroup$ – B. S. Thomson Jan 28 '16 at 2:40

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