A differential inclusion relating to the slope of a convex function - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:34:55Z http://mathoverflow.net/feeds/question/34527 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34527/a-differential-inclusion-relating-to-the-slope-of-a-convex-function A differential inclusion relating to the slope of a convex function Ian Morris 2010-08-04T16:31:22Z 2010-08-04T18:13:01Z <p>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:</p> <blockquote> <p><strong>Lemma A</strong>. 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 $\varepsilon$.</p> </blockquote> <p>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".</p> <p>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. </p> <blockquote> <p><strong>Lemma B.</strong> 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 $\varepsilon$?</p> </blockquote> <p>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?</p> http://mathoverflow.net/questions/34527/a-differential-inclusion-relating-to-the-slope-of-a-convex-function/34537#34537 Answer by Tracy Hall for A differential inclusion relating to the slope of a convex function Tracy Hall 2010-08-04T18:13:01Z 2010-08-04T18:13:01Z <p>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.</p>