Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let a real-valued function $f$ be continuous on $[0,1].$ Then there exists a number $a$ such that the integral $$\int_0^1\frac 1 {|f(x)-a|}\, dx $$ diverges. How to prove that statement?

share|improve this question
    
: whats source? –  Maisam Hedyelloo Jun 1 '13 at 10:13
7  
@Maisam Hedyelloo: The human mind. –  user64494 Jun 1 '13 at 10:25
1  
There might be two cases of interest. 1) when $f$ vanishes on $(0,1)$; and 2) when it does not. If it does vanish, pick $a=0$ and the integral should diverge. If it does not, then pick $a=f(x_0)$ for some $x_0$ in $(0,1)$? –  Cameron Williams Jun 1 '13 at 21:01
1  
Interesting question! I do not know the answer, but a counterexample must be quite pathological. –  André Nicolas Jun 2 '13 at 2:43
    
f(0) and f(1) seem to work. –  dimensio1n0 Jun 7 '13 at 2:21
show 4 more comments

4 Answers

up vote 11 down vote accepted
+150

Take a nondecreasing rearrangement $r(x)$ of the function $f(x)$ (some discussion of this may be found at http://en.wikipedia.org/wiki/Convex_conjugate). This involves finding a measure-preserving transformation of the interval $[0,1]$ that transforms $f$ into $r$. In particular, your integrals $\int_0^1\frac 1 {|f(x)-a|}\, dx$ are all preserved (for every $a$). Now apply the result that every monotone function is a.e. differentiable (see http://en.wikipedia.org/wiki/Monotonic_function). Take a point $p$ where the function $r$ is differentiable. Then $a=r(p)$ does the trick, because $r(x)-a$ can be bounded in terms of a linear expression.

Note that the existence of a nondecreasing rearrangement of a function $f$ admits an elegant proof in the context of its hyperreal extension $^\star f$, which we will continue to denote by $f$. Namely, take an infinite hypernatural $H$ and consider a partition of the hyperreal interval $[0,1]$ into $H$ segments, by means of partition points $0, \frac{1}{H}, \frac{2}{H}, \frac{3}{H}, \ldots, \frac{H-1}{H}, 1$. Now rearrange the values $f(\frac{i}{H})$ of the function at partition points in increasing order, and permute the $H$ segments accordingly. The standard part of the resulting function is the desired monotone function $r$.

Note 1. I should point out that one does not really need to use the result that monotone functions are a.e. differentiable. Consider the convex hull of the graph of $r(x)$, and take a point where the graph touches the boundary of the convex hull (other than the endpoints 0 and 1). Setting $a$ equal to the $x$-coordinate of the point does the job.

share|improve this answer
    
What if the derivative is 0 at every differentiator point, as in the Lebesgue function of a cantor set? –  Brian Rushton Jun 6 '13 at 14:14
1  
The constant functions are obvious counterexamples, so to make the problem meaningful in this case one could declare that the integral $\int_0^1 \frac{1}{0}dx$ "diverges". –  user72694 Jun 6 '13 at 14:32
    
There are strictly monotonic functions whose derivative is 0 at every point where it is defined. –  Brian Rushton Jun 6 '13 at 14:33
    
If one lets $a$ be the value of the function at that point, the integral $\int \frac{1}{f(x)-a}dx$ will indeed diverge. –  user72694 Jun 6 '13 at 14:38
    
True! That's a good point... –  Brian Rushton Jun 6 '13 at 14:45
show 9 more comments

Here's a proof that $\int_0^1\frac1{\lvert f(x)-a\rvert}dx = \infty$ for every $a$ in the image of $f$ and outside of a meagre set. In particular, if $f$ is not constant, then there are uncountably many such $a$ in every neighborhood of the image of $f$. [Note: I also agree that user72694's proof works fine, and is completely independent of the proof I'll give here.]

First, define $g(a)$ to be the given integral, and let $[a_0,a_1]$ be the image of $f$. Assuming that $f$ is non-constant, we have $a_0 < a_1$. Letting $b_0 < b_1$ be in $[a_0,a_1]$ then Fubini's theorem gives, $$ \int_{b_0}^{b_1}g(a)\,da= \int_{b_0}^{b_1}\!\!\!\int_0^1\frac1{\lvert f(x)-a\rvert}\,dxda =\int_0^1\!\!\!\int_{b_0}^{b_1}\frac1{\lvert f(x)-a\rvert}\,dadx=\infty. $$ Here, the integral of $1/\lvert f(x)-a\rvert$ wrt $a$ is infinite whenever $f(x)$ is in $[b_0,b_1]$ (because $1/x$ is not integrable at the origin), which happens for $x$ in a nontrivial interval, so the double integral is infinite. This means that $g$ is not integrable (and, hence, is unbounded) in any nontrivial interval $[b_0,b_1]$ in the image of $f$.

Next, for each $K > 0$, set $S_K=\lbrace a\in[a_0,a_1]\colon g(a) > K\rbrace$. As $g$ is unbounded in the neighborhood of any point, the set $S_K$ is dense in $[a_0,a_1]$. Furthermore, $S_K$ is open for each positive $K$. To see this, note that $g_n(a)\equiv\int_0^11_{\lbrace\lvert f(x)-a\rvert > 1/n\rbrace}/\lvert f(x)-a\rvert dx$ is a sequence of continuous functions increasing to $g$, so $g$ is lower semicontinuous.

Hence, we have $$ \left\lbrace a\in[a_0,a_1]\colon g(a)=\infty\right\rbrace=\bigcap_{n=1}^\infty S_n, $$ which is a countable intersection of dense open sets in $[a_0,a_1]$ so, by definition, its complement is meagre.

share|improve this answer
    
Very nice!${}{}$ –  Nick Strehlke Jun 6 '13 at 20:35
    
Both answers to the question under consideration, which belongs to mathematical folklore and is authored by S. Konyagin, are right and nice. Unfortunately, I cannot share bounty between the two answers. –  user64494 Jun 7 '13 at 4:55
2  
That's interesting. His wiki page that you linked does not say anything about the question. Can you elaborate? –  user72694 Jun 7 '13 at 7:22
add comment

I don't think this statemnt is correct if $a\in[0,1]$. Maybe I'm wrong but you can take the counter example $f(x)=\frac 1 {|sin(x)+cos(x)|}$. So let's make two points: 1.looking at $|f(x)-a|$,notice that $f(x)=sin(x)+cos(x)$ is continious but moreover is nonegative. if $a\in[0,1]$ the integral is well defined. 2.Then the antriderivative of the function is wel defined according to maple and for all a in [0,1] the claim is correct.


EDIT: Let's try reductio ad absurdum and assume $\int_0^1 \frac 1 {|f(x)-a|} dx$ converges to $L\in\mathbb R$. Also let $a\in \mathbb R,f:[0,1]\rightarrow\mathbb R$. since the function is defined on a closed interval $\exists x_0 \in \mathbb R$ s.t $x_0$ is a maximum (concluding from weirestrass first lemma for continious function on closed intervals) . since $|f(x)-a|\leq |f(x_0)-a| \Longrightarrow \frac {1}{|f(x)-a|}\geq \frac {1}{|f(x_0)-a|}$. and taking $a\to f(x_0)$, $|f(x_0)-a|\to 0$ but then $ \frac 1 {|f(x_0)-a|} \to \infty$ and finally $L=\int_0^1 \frac 1 {|f(x)-a|}dx\geq \int_0^1 \frac 1 {|f(x_0)-a|}\to \infty$ which is undifined in contradiction for that we assumed that $L\in\mathbb R$.

share|improve this answer
2  
Thank you for the interest to the question. Unfortunately, you did miss the point. The statement does not assert that $a \in [0,1].$ –  user64494 Jun 1 '13 at 10:04
    
I edited the post. –  Coargu Aliquis Jun 1 '13 at 10:32
    
Mmh i think there are some flaws in the proof (hope,that I'm not wrong) 1) You can't assert that exists $x_0$ s.t. $f'(x_0)=0$ only because f is defined on a close, maxima can be reach on the extremis of the interval and therefore the derivate is always $\neq 0$. Just take as counterexample $f(x)=x-\frac{2}{3}$. 2) the disequality about absolute value is not true in general, consider the previous $f$, if you take the maxima $=\frac{1}{3}$ and $a>0$ and a random point in $[0,\frac{2}{3}]$ you obtain a contradiction. –  Riccardo Jun 1 '13 at 21:11
    
a. the derivative does not matter for the question. The important thing is you have a maximum (extreme value therom which is also explained here en.wikipedia.org/wiki/Extreme_value_theorem). b. I don't think you can take the maximum from $[0, \frac 2 3]$ since you choose $x_0$ from $[\frac 2 3,1]$. if f is non-negtive the claim is correct. for the counter-example you wrote since the claim is correct in [\frac 2 3, 1],since the $\int_{\frac 2 3}^{1}$ diverges the whole integral divereges. For negitve function take $x_1$, the minimum of the function in almost a same way. –  Coargu Aliquis Jun 2 '13 at 2:28
    
If $x_0$ is a maximum for $|f(x)-a|$ then $a$ is (by implicit assumption) fixed, so you can't make $a\to f(x_0)$ and expect the inequality $|f(x) - a|\leq |f(x_0)-a|$ to continue to hold for all $x\in [0,1]$. –  Nick Strehlke Jun 2 '13 at 2:42
show 4 more comments

The Devil's staircase is a counterexample. When the integral is defined, it converges. For instance, at 0, the Devil's staircase is approximately between $(C_1 x)^{\frac{\ln 2}{\ln 3}}$ and $(C_2 x)^{\frac{\ln 2}{\ln 3}}$ (the curves connecting the left endpoints of intervals and right endpoints, respectively), and so your integral converges if $a=0$. Near any other point where the integral is defined, the integral converges since the set is self-similar, so all such points are like 0.

All of this is in Chapter four of Frank Jones' Lebesgue integration book, as well as pages 521.

Devil's staircase from Wikipedia

share|improve this answer
    
Good point! I'll have to think of a way to address that. –  Brian Rushton Jun 2 '13 at 2:47
    
I believe this works now. –  Brian Rushton Jun 6 '13 at 14:30
3  
The constant functions are obvious counterexamples; so to make the problem meaningful in this case one could declare that the integral $\int_E \frac 1 0 dx$ over the set $E$ of positive measure diverges. This is a usual convention in real analysis. –  user64494 Jun 6 '13 at 15:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.