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Assume $f$ is Lebesgue measurable and bounded on $[0,1]$ and that $$\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\to0\;\text{as}\;h\to0.$$ Show that $f$ is constant for almost every $x$ in $[0,1]$.

My question relates to my proposed proof below. At no point did I rely on the boundedness of $f$ on $[0,1]$, only its measurability and so I suspect my reasoning is faulty and would appreciate somebody that could (in their answer) scrutinize and correct my proof.

(Proposed) Proof. Assume first that $h\to0^{+}$ and suppose that we may interchange the two limits. Then since the difference quotients are non-negative, we conclude $f'(x)$ exists and is equal to $0$ for almost every $x$ in $[0,1]$, and the claim follows. (As pointed out, we don't necessarily know the limit of the integrands exists even when we assume the interchange is valid, although this is probably not difficult to reason one's way out of -- more damning is below.)

The goal quickly turns to justifying this interchange. To that end, the presumed existence of the above limit implies that for any $0<M<\infty$ there is some positive real number $h(M)$ such that for any $h<h(M)$ we have $$\left|\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\right|<M.$$ Since $m([0,1])=1$ (in particular, we are integrating over a set of finite measure) and the difference quotients are non-negative, we conclude that $$\frac{|f(x+h)-f(x)|}{h}<M\;\text{a.e.}\;x.$$ Since this bound is independent of $h$ once $h<h(M)$, it follows that at some point along the sequence each of the difference quotients is pointwise a.e. dominated by $M$, which is integrable on any set of finite measure. Hence the interchange of limits discussed above is valid, and with the case of $h\to0^{-}$ being handled similarly, the claim is proved.

(As the comment points out, we cannot conclude anything about the upper (or lower) bound of a function based off the value of its integral, even if the integration is taken over a set of finite measure and the integrand in question is nonnegative; approximations to the identity serve or otherwise spikey functions on sets of small measure as counter-examples.)

Edit. Fatou's lemma implies $$\int_{0}^{1}\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}\;dx\leq\lim_{h\to0}\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx=0.$$ Hence $$\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}=0\;\text{a.e.}\;x.$$ (The $\liminf$ of course exists without further justification.)

We now need a way to obtain the same for $\limsup_{h\to0}\frac{|f(x+h)-f(x)|}{h}.$ This could be accomplished by showing the limit exists under the assumptions given, in which case the $\liminf$ is equal to the $\limsup$. Any ideas how to execute this, or otherwise approach the problem?

I'm wondering too if my original argument above could be modified by taking into account the assumed boundedness of $f$ in order to exclude the possibility that its difference quotients form a "spikey" sequence of integrands.

Edit. Here is another approach I am considering by posting this question Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$.

Peter Tamaroff mentioned the type-writer sequence being a counter-example to the post linked.

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Sargera
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  • http://www.math.ucla.edu/grad/handbook/hbquals.shtml -- See problem #1 from Analysis Spring 2013. I don't understand your point though; there are no singularities or otherwise obstructions to speak of that would prevent us from integrating along $[0,1]$... – Sargera Sep 05 '13 at 22:23
  • Regarding the posing of the question, the original takes $f$ which is defined on $\mathbb{R}$, which solves the problem. However, note that $\int_E gd\mu<M$ and $\mu(E)\leq 1$ don't imply that $\mu({x\in E\mid g(x)>M})=0$. – Jonathan Y. Sep 05 '13 at 22:27
  • Isn't this true though when $g\geq0$? How can you have $\int_{0}^{1}g(x);dx<M$ with $g(x)>M$ on a set of positive measure? – Sargera Sep 05 '13 at 22:35
  • Thank you for making me realize my naivety. Approximations to the identity much... (: – Sargera Sep 05 '13 at 22:46
  • Sure thing; sometimes outside perspective is the only way to avoid these things. Another thing that's bothering me is that you're implicitly assuming the existence of a pointwise limit (a.e.), so that we can contend ourselves to trying to switch the order. – Jonathan Y. Sep 05 '13 at 22:51
  • @Jonathan I incorporated your comments in the most recent edit. – Sargera Sep 06 '13 at 00:29

1 Answers1

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If one is allowed to use Lebesgue's differentiation theorem, one can prove the result as follows.

Since the integrand is nonnegative, we have $\lim_{h\to 0^+} \int_0^\alpha \frac{\vert f(x+h)-f(x)\vert}{h}\, dx=0$ for every $\alpha\in[0,1)$, and hence $\lim_{h\to 0^+} \int_0^\alpha \frac{f(x+h)-f(x)}{h}\, dx=0$.

Putting $F(u):=\int_0^u f(x)dx$, one may write \begin{eqnarray}\int_0^\alpha (f(x+h)-f(x))\, dx&=&\int_h^{\alpha+h} f(u)du-\int_0^\alpha f(x)dx\\&=&F(\alpha+h)-F(\alpha)-F(h)\, . \end{eqnarray} So the above limit becomes $$\lim_{h\to 0^+}\left(\frac{F(\alpha+h)-F(\alpha)}{h}-\frac{F(h)}h\right)=0\, . \tag{$*$}$$ Now, since $f$ is integrable on $[0,1]$ (being bounded), we know from Lebesgue's differentiation theorem that $\lim_{h\to 0} \frac{F(\alpha+h)-F(\alpha)}{h}=f(\alpha)$ for almost every $\alpha\in [0,1)$. By $(*)$, it follows that $l=\lim_{h\to 0^+} \frac{F(h)}h$ exists, and that $f(\alpha)=l$ almost everywhere. Hence, $f$ is a.e. constant.

Etienne
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  • Can I ask what motivated this proof for you? In other words, what trial and error did you go through before you thought of attacking the problem by defining such $F$? – Sargera Sep 09 '13 at 02:52
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    I think it was "clear" from the beginning that Lebesgue's differentiation theorem should be helpful, because of the combination "integral+difference quotient+almost everywhere". So it was natural to introduce the function $F$. – Etienne Sep 09 '13 at 04:57