Rolle's Theorem states that $f(1/2)=f(-1/2)+f'(x)$ has a root in the open real
interval $(-1/2,1/2)$ if $f$ is continuous and differentiable. How large can the absolute value of such a root
$\xi$
be if $f$ is a polynomial of degree $d$? (We choose $\xi$ to be the real root of minimal norm in the case of several solutions.)
A few comments:
$\xi$ can be arbitrarily close to the boundary if there is no condition on the degree.
$\xi$ is always $0$ in degree $2$.
I have examples (using random polynomials) with $\xi=.28867…$ for $d=3$
(the correct value is perhaps $1/(2\sqrt{3})$ given for example with $f(x)=x^3$),
and $\xi=.324…$ for $d=5$.
Considering for $f$ a complex polynomial, the equation $f(1/2)=f(-1/2)+f'(x)$
seems always to have a complex solution of norm at most 1/2. Is this true?
If yes, does the real constant working for real polynomials of degree $d$
also work for complex polynomials of degree $d$? (I have no counterexample.)
A positive answer for complex polynomials would imply the statement for entire
functions (where the smallest root could perhaps be on the boundary of the
complex disc of radius $1/2$).
Best Answer
For real polynomials $\xi$ can be $1/2 - O(1/d^2)$ but no closer to $1/2$ than that. The value $1/\sqrt{12} = 0.288675\ldots$ is correct for $d=3$, and also for $d=4$; for larger $d=2n-1$ or $d=2n$, the supremum occurs at half the largest root of the $n$-th Legendre polynomial $P_n$, and is not attained because at the extremal points $f$ also has points of inflection satisfying the Rolle condition, as at $x=0$ for the quintic polynomial $f(x) = x^3-4x^5$ which shows the supremum $\sqrt{3/20} = .387298\ldots$. (The reason $n=2$ is different is that $P_2(x) = (3x^2-1)/2$, and indeed $1/\sqrt{12}$ is half the largest root $1\sqrt{3}$, but there are no inflection points because $P_2$ has no zeros in the interior of $[-1/\sqrt{3},+1/\sqrt{3}]$. For $n=1$ the Legendre polynomial vanishes only at zero, which we know must be a Rolle point when $f$ is quadratic.)
Here's a Wolfram Alpha plot for $d=11$, for which $\xi = .4662\ldots$:
Given $n$, let $p(x)=P_n(2x)$, so that $p$ is a degree-$n$ orthogonal polynomial on $[-1/2,+1/2]$; denote its roots by $x_i$ in increasing order, $x_1 < x_2 < x_3 < \cdots < x_n$, with each $x_i = -x_{n+1-i}$ by symmetry. In particular $x_1 = -x_n$. For any polynomial $f$ of degree $d \leq 2n$, let $g(x) = f'(x) - cx$ where $c = f(1/2) - f(-1/2)$. I claim $g$ cannot be positive on $[x_1,x_n]$. Indeed $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = 0$, but by Gaussian quadrature $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx$ is a positive linear combination of the $g(x_i)$ — contradiction. On the other hand, $g$ can be nonnegative but not identically zero on $[x_1,x_n]$, which happens iff $$ g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2) $$ for some $b>0$ and $a \in [-b/x_n,b/x_n]$. (Note that $a=0$ deals with $d=2n-1$.) Indeed the same Gaussian-quadrature formula shows that $g$ must vanish on each $x_i$, with double roots except possibly at $x_1$ and $x_n$; this gives $g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2)$, and then the inequalities on $a,b$ follow from nonnegativity on $[x_1,x_n]$. Conversely, if $g$ is of that form then it is indeed nonnegative on $[x_1,x_n]$, and $$ \int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = \int_{-1/2}^{1/2} \phantom. p(x) \frac{(ax+b)\phantom. p(x)}{x_n^2-x^2} \phantom. dx = 0 $$ because the integral is the inner product on $[-1/2,+1/2]$ of $p$ with a polynomial of degree less than $n$.
We can then take $f_0(x)$ to be an indefinite integral of $g(x)$, which satisfies $f_0(-1/2)=f_0(+1/2)$ and is strictly increasing on $[x_1,x_n]$. For each $\epsilon>0$ we can then perturb $f_0$ to get a polynomial $f$ of the same degree, still taking the same values at $-1/2$ and $+1/2$, but with a strictly positive derivative on $[x_1+\epsilon,\phantom, x_n-\epsilon]$. For instance we may take $f(x) = f_0(\theta x) + \delta\cdot x$ with $\theta$ slightly larger than $1$ and $\delta>0$ chosen so that $f(-1/2)=f(+1/2)$. This completes the proof.