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right now I'm reading a paper by Blower (Displacement convexity for generalized orthogonal ensemble) and I'm stuck at a rather delicate point. He uses the following equality, which is supposed to be deducable from the mean value theorem:

Lemma: Let $v$ be twice differentiable, then for any $x,y \in \mathbb{R}$ and $s \in [0,1]$ there is a $\bar{s} \in (0,1)$ such that $(1-s)v(x) + sv(y) - v((1-s)x+sy) = \frac{1}{2} s(1-s)(x-y)^2 v''((1-\bar{s})x+\bar{s}y)$.

I've tried to use the mvt (or rather Taylors theorem) on the LHS as a function of $s$, with little success, and with $x$, from which I get the following equality: (call the LHS $H$)

$$ H(x) = H(x) - H(y) = (x-y) \left( (1-s)v'(\xi) - (1-s) v'((1-s)\xi + sy) \right)$$

where $\xi \in (x,y),$ i.e. $\xi = (1-t)x + ty, t \in (0,1)$. Upon a second use the mvt this becomes

$$ H(x) = (x-y)^2(1-s)s \left( v''(\bar{\xi}) (1-t)\right). $$

I do not get any further equalities, as I do not know whether I can take the midpoint $t = \frac{1}{2}$. I can bound this from below by $(1-t) \ge \frac{1}{2}$ (as I can interchange $x$ and $y$ (and hence $s$ and $(1-s)$) if $t > 1/2$), which is sufficient for the proof. However: is there any way to show equality in the Lemma?

Thanks in advance for your responses, Arthur

Arthur Sinulis
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1 Answers1

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This looks formidable but becomes straightforward with suitable simplifying transformations.

First rescale to $x=0$, $y=1$, leading to

$$ (1-s)v(0) + sv(1) - v(s) = \frac{1}{2} s(1-s)v''(\bar s)\;. $$

Now subtract a linear function such that $v(0)=v(1)=0$, leading to

$$ - v(s) = \frac{1}{2} s(1-s)v''(\bar s)\;. $$

Now subtract a quadratic function such that $v(s)=0$, namely

$$ v(s)\frac{t(1-t)}{s(1-s)}\;, $$

leading to

$$ 0 = \frac{1}{2} s(1-s)v''(\bar s)\;, $$

and thus $v''(\bar s)=0$. Thus, the problem is reduced to showing that somewhere among three zeros the second derivative must vanish, which follows by applying the mean value theorem first to each subinterval, yielding two stationary points, and then to the interval between the stationary points.

(Of course you could subtract the quadratic function interpolating between $v(0)$, $v(s)$ and $v(1)$ all in one go, but then it's harder to see how that affects the equation under consideration.)

joriki
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    Wow, this is a clever solution. Did not really think of manipulating the equation before trying to use the mean value theorem. Thank you very much for your quick answer. – Arthur Sinulis Sep 10 '15 at 12:46
  • @ArthurSinulis: You're welcome. With 20/20 hindsight, the simplifications aren't all that necessary -- the lemma expresses the fact that the difference between $v$ and the quadratic interpolating polynomial through $v(x)$, $v(s)$ and $v(y)$ must have an inflection point somewhere between its three zeros. – joriki Sep 10 '15 at 16:37