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.

As it is well known every continuous function has the intermediate value property, but even some discontinuous functions like $$f(x)=\left\{ \begin{array}{cl} \sin\left(\frac{1}{x}\right) & x\neq 0 \\ 0 & x=0 \end{array} \right.$$ are having this property.
In fact we know that a derivative always have this property.

My question is, if for every function $f$ with the intermediate value property exists a function $F$ such that $F'=f$. And if so, is $F$ unique (up to a constant you may add)?

My attempts till now: (Just skip them if you feel so)

The most natural way of finding those functions would be integrating, so I guess the question can be reduced to, if functions with the intermediate value property can be integrated.
This one depends heavy on how you define when a functions is integrable, we (my analysis class) said that we call a function integrable when it is a regulated function (the limits $x\to x_0^+ f(x)$ and $x\to x_0^- f(x)$ exists ) .
As my example above shows, not every function with the intermediate value property is a regulated function. But if we say a function is integrabel when every riemann sum converges the above function is integrable, so it seems like this would be a better definition for my problem.

Edit: As Kevin Carlson points out in a commentar that being a derivative is different from being riemann integrable, he even gave an example for a function which is differentiable but it's derivative is not riemann integrable. So we can't show that those functions are riemann integrable as they are not riemann integrable in general. Now I have no clue how to find an answer.

share|improve this question
4  
Though it's a deep point, it's important to point out that being a derivative is not the same thing as being integrable. The main example is the derivative of Volterra's function, which you can read about on Wikipedia. This is why a precise statement of the fundamental theorem of calculus supposes that $f$ is integrable, or often that it's even continuous. (Here by "integrable" I mean the Riemann sums converge.) –  Kevin Carlson Mar 11 '13 at 7:48
    
@KevinCarlson ok that helps a lot so so my idea how to reduce the problem doesn't work –  Dominic Michaelis Mar 11 '13 at 7:57
1  
what about "Conway base 13 function"? –  Laura Mar 11 '13 at 8:06
2  
@Tai yeah that sounds good, but how to proof that conway base 13 function is not a derivative? –  Dominic Michaelis Mar 11 '13 at 8:12
2  
FYI, derivatives are also Baire $1$ functions‌​. Bruckner's book Differentiation of Real Functions discusses in detail many of the attractive properties that functions simultaneously Darboux (have the Intermediate Value Property) and Baire $1$ have, along with how far in many other ways such functions can be from being a derivative. –  Dave L. Renfro Mar 11 '13 at 19:05

3 Answers 3

up vote 20 down vote accepted

If you compose $ \tan^{-1} $ with Conway’s Base-$ 13 $ Function, then you get a bounded real-valued function on the open interval $ (0,1) $ that satisfies the Intermediate Value Property but is discontinuous at every point in $ (0,1) $. Therefore, by Lebesgue’s theorem on the necessary and sufficient conditions for Riemann-integrability, this function is not Riemann-integrable on any non-degenerate closed sub-interval of $ (0,1) $.

Now, it cannot be the derivative of any function either, because by the Baire Category Theorem, if a function defined on an open interval has an antiderivative, then the function must be continuous on a dense set of points. This thread may be of interest to you. :)

share|improve this answer
2  
Denote Conway’s Base-$ 13 $ Function by $ f $. Then $ \tan^{-1} \circ f $ is bounded between $ - \dfrac{\pi}{2} $ and $ \dfrac{\pi}{2} $. Next, $ \tan^{-1} \circ f $ cannot be continuous, otherwise $ f = \tan \circ \tan^{-1} \circ f $ would be continuous, which is a contradiction. –  Haskell Curry Mar 11 '13 at 21:54
    
Is Conway's Base-13 Function measurable? –  JSchlather Mar 11 '13 at 22:15
1  
@JSchlather: According to this MathOverflow thread, Conway’s Base-$ 13 $ Function is Borel-measurable. –  Haskell Curry Mar 11 '13 at 23:18
    
Thanks the link not only answered my question but led me to an interesting discussion about whether or not the absence of AC in constructing functions implies measurability. –  JSchlather Mar 12 '13 at 0:11

Another conterexample is this: let $(a_n, b_n), n = 1, 2, \ldots$ be the sequence of all open intervals in $\mathbb{R}$ with rational endpoints. Let $C_1$ be some Cantor set inside $(a_1, b_1)$. Because $C_1$ is closed and has no interior, $(a_2, b_2) - C_1$ contains some open subinterval. Construct Cantor set $C_2$ inside this subinterval. We can continue this procedure (constructing new Cantor set $C_n$ in $(a_n, b_n)$ that does not intersect previously created $C_1, C_2, \ldots$) -- this is essentialy the same argument as in proof of Baire's theorem.

Now, we have a sequence of Cantor sets $C_1, C_2, ...$ such that 1) $C_i \cap C_j = \emptyset$ for $i \ne j$ and 2) $\bigcup_{i \geq 1} C_i$ is dense in $\mathbb{R}$. For each $n$, pick a bijection $f_n: C_n \to \mathbb{R}$ (there exists one, because $C_n$ has cardinality of continuum). Then construct a function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = f_n(x)$ if $x \in C_n$ and $f(x) = 0$ if $x \not \in C_n$ for any $n$. It's easy to see that $f$ has intermediate value property: if we have some $x < y$ and $f(p) < f(q)$, there's some interval with rational endpoints $(a_n, b_n) \subset (x, y)$, there's $C_n \subset (a_n, b_n)$, so for any $z \in (f(p), f(q))$ there's $w \in C_n$ such that $f(w) = z$: $w$ is just $f_n^{-1}(z)$ (remember that $f_n$ was a bijection).

More interestingly, the set $\bigcup_{n \geq 2} C_n$ is also dense in $\mathbb{R}$, so we can repeat above construction with any function $f_1: C_1 \to \mathbb{R}$, not necessarily a bijection, and we still will obtain a function with the intermediate value property. Now, there are $2^{\mathfrak{c}}$ different functions $C_1 \to \mathbb{R}$, so there are $2^{\mathfrak{c}}$ functions $\mathbb{R} \to \mathbb{R}$ with the intermediate value property. Since there are only $\mathfrak{c}$ continuous functions $\mathbb{R} \to \mathbb{R}$, there are also $\mathfrak{c}$ derivatives, and since $\mathfrak{c} < 2^{\mathfrak{c}}$, there are lots of functions with intermediate value property that are not derivatives.

share|improve this answer
    
+1 I like this one, looks nice to me will read it tomorrow more extrensive it is late now –  Dominic Michaelis Mar 30 '13 at 21:18

Here is a (bounded, Baire class 1) "natural" example of a function with the intermediate value property that is not a derivative. I mentioned it also in this answer.

Consider first the function you mentioned, $$f(x)=\left\{\begin{array}{cl}\sin\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 0&\mbox{ if }x=0.\end{array}\right.$$ This function is a derivative, because, letting $g$ be the function $$ g(x)=\left\{\begin{array}{cl}x^2\cos\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 0&\mbox{ if }x=0,\end{array}\right. $$ and setting $$h(x)=\left\{\begin{array}{cl}2x\cos\left(\frac1x\right)&\mbox{ if }x\ne 0,\\ 0&\mbox{ if }x=0,\end{array}\right.$$ then $h$ is continuous, and $f(x)=g'(x)-h(x)$ for all $x$. But continuous functions are derivatives, so $h$ is also a derivative. Now take $$j(x)=\left\{\begin{array}{cl}\sin\left(\frac1x\right)&\mbox{ if }x\ne0,\\ 1&\mbox{ if }x=0.\end{array}\right.$$ This function still has the intermediate value property, but $j$ is not a derivative. Otherwise, $j-f$ would also be a derivative, but $j-f$ does not have the intermediate value property (it has a jump discontinuity at $0$).

The reference

A.C.M. van Rooij, and W.H. Schikhof, A second course on real functions, Cambridge University Press, 1982,

discusses this example in great detail, showing that instead of sine, one can use any periodic derivative:

If $j:[0,\infty)\to\mathbb R$ is a derivative, and $j(x+1)=j(x)$ for all $x\ge 0$, one can set $J$ be an antiderivative of $j$, let $A=J(1)-J(0)$, and define $h:[0,1]\to\mathbb R$ by $$ h(x)=\left\{\begin{array}{cl} j\left(\frac1x\right)&\mbox{ if } 0<x\le 1,\\ A&\mbox{ if }x=0.\end{array}\right. $$ One can then argue that $h$ is a derivative and, letting $i$ be any function that coincides with $h$ except at $0$, where it takes a value different from $A$ but close, we have an example of an $i$ with the intermediate value property, bounded, and of Baire class 1, that is not a derivative.

To see that $h$ is indeed a derivative, notice first that $A=0$ without loss of generality (replacing $j$ by $j-A$, $h$ by $h-A$, and $J$ by $\hat J(x)=J(x)-Ax$). Now set $$ H(x)=\left\{\begin{array}{cl}-x^2J\left(\frac1x\right)+2\int_{\frac1x}^\infty \frac{J(s)}{s^3}\,ds&\mbox{ if }0<x\le 1,\\ 0&\mbox{ if }x=0.\end{array}\right. $$ One can then verify that $H'=h$ (using that our choice of $A=0$ makes $J$ periodic and therefore bounded, to ensure that $H'(0)=0$).

Van Rooij and W.H. Schikhof then proceed to consider specific examples of functions $j$ that they use to verify the following:

  • There is a derivative $f$ such that $|f|$ is not a derivative.
  • There is a derivative $f$ such that $f^2$ is not a derivative.
  • There is derivative $f$ with, say, $1\le f\le 2$, such that $1/f$ is not a derivative.
share|improve this answer

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.