The Bojanov--Naidenov inequality for quartics and second derivatives

We settle the case $n=4$, $k=2$ of the Bojanov–Naidenov problem for algebraic polynomials. Let $P$ be a real polynomial of degree at most four with ${\parallel P\parallel }_{C[-1,1]}\le 1$, and let $T_4(x)=8x^4-8x^2+1$. We prove that, for every $t\ge 0$, $${\int }_{-1}^1(\mathrm{\mid }{P}^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid }-t{)}_{+}dx\le {\int }_{-1}^1(\mathrm{\mid }{T}_4^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid }-t{)}_{+}dx\mathrm{.}$$ This tail estimate implies $${\int }_{-1}^1\phi (\mathrm{\mid }{P}^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid })dx\le {\int }_{-1}^1\phi (\mathrm{\mid }{T}_4^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid })dx$$ for every nondecreasing convex function $\phi :[0,\mathrm{\infty })\to \mathbb{R}$. If $\phi$ is strictly increasing and convex, equality can occur only for $P=\pm T_4$. The proof is elementary and finite. We interpolate at the five extremal points of $T_4$; convexity then reduces the problem to the $32$ sign choices at these nodes. At each vertex the second derivative is a quadratic polynomial, so the remaining work is an explicit comparison of level sets.