This question is concerned with a possible lemma which would be very useful in one of my current research projects, but which I am currently unable to prove. The project as a whole relates to the asymptotic growth rates of certain matrix products, but once all the dynamical parts are finished with, I need to obtain some quantitative information about the behaviour of the subderivatives of a certain convex function. This would be greatly simplified if the following lemma were true:

Lemma A. Let $f \colon [0,1] \to \mathbb{R}$ be a continuous convex function, and let $\varepsilon>0$. Suppose that for every $t \in (0,1)$, for every subderivative $f'(t)$ of $f$ at $t$ the inequality \[ f'(t)\leq \frac{f(t)-f(0)}{t} + \varepsilon\] holds. Then there exist subderivatives $f'(0)$ and $f'(1)$ of $f$ (at 0 and 1 respectively) such that $|f'(0)-f'(1)| \leq C\varepsilon$, where $C>0$ is a constant which does not depend on $f$ or on $\varepsilon$.

In visual terms, this means that for any $t \in (0,1)$ the slope of the straight line connecting $(0,f(0))$ and $(t,f(t))$ is constrained to be close to the "gradient" of $f$ at $t$, and I would like to deduce from this that the entire graph of $f$ does not admit very many different "slopes".

This problem can be re-stated in the following form, which I personally find somewhat easier to think about. Let us define $F \colon (0,1] \to \mathbb{R}$ to be the function which describes the left derivatives of the above function $f$. This $F$ is a well-defined monotone increasing function which is continuous on the left.

Lemma B. Fix $\varepsilon>0$. Let $F \colon (0,1] \to [0,\infty)$ be a monotone increasing function, continuous with respect to limits from the left, such that for every $t>0$ we have $F(t) \leq \frac{1}{t}\int_0^tF(s)ds + \varepsilon$. Is it necessarily the case that $\sup F - \inf F \leq C\varepsilon$ for some constant $C>0$ which does not depend on $F$ or on $\varepsilon$?

I've spent a couple of weeks thinking about this on-and-off without making very much headway, and I am beginning to get a little frustrated at not being able to solve a problem in one-dimensional real analysis! Of course, this does not preclude the possibility that the problem is actually very simple for someone equipped with the correct tools. So my question is: does anyone know whether either of the above two lemmas is true, or have any pointers which might be useful in establishing a proof?


1 Answer 1


Unfortunately the lemma is false. Given a candidate $C$, let $\varepsilon = 1$ and $F(t) = \ln(te^C+1)$. Then the hypotheses of Lemma B hold but the conclusion fails.

  • $\begingroup$ That certainly explains why I can't prove it! I will have a think about this and consider which extra hypotheses I might be able to introduce which apply to the particular function under consideration. $\endgroup$
    – Ian Morris
    Commented Aug 4, 2010 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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