Is every function with the intermediate value property a derivative?

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.

Solution 1:

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. :)

Solution 2:

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.