January 01, 1970
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 \(\left\lVert P\right\rVert_{C[-1,1]}\leq 1\), and let \(T_4(x)=8x^4-8x^2+1\). We prove that, for every \(t\geq0\), \[\int_{-1}^{1} \bigl(|P''(x)|-t\bigr)_+\,dx \leq \int_{-1}^{1} \bigl(|T_4''(x)|-t\bigr)_+\,dx .\] This tail estimate implies \[\int_{-1}^{1}\varphi(|P''(x)|)\,dx \leq \int_{-1}^{1}\varphi(|T_4''(x)|)\,dx\] for every nondecreasing convex function \(\varphi:[0,\infty)\to\mathbb{R}\). If \(\varphi\) 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.
Let \(\pi_n\) be the space of real algebraic polynomials of degree at most \(n\), and put \[\pi_n^\circ:=\{P\in\pi_n:\left\lVert P\right\rVert_{C[-1,1]}\leq 1\}.\] We write \[T_n(x)=\cos(n\arccos x),\qquad -1\leq x\leq 1,\] for the Chebyshev polynomial of the first kind. Naidenov [1] recorded the following problem of Bojanov. Is it true that \[\label{eq:BN-general} \int_{-1}^{1}\varphi(|P^{(k)}(x)|)\,dx < \int_{-1}^{1}\varphi(|T_n^{(k)}(x)|)\,dx\tag{1}\] for every strictly increasing convex function \(\varphi\) on \([0,\infty)\), every \(P\in\pi_n^\circ\setminus\{\pm T_n\}\), and every \(k=2,\ldots,n-1\)?
The endpoint cases have a different nature. The case \(k=n\) follows from a classical extremal property of the Chebyshev polynomial, while the case \(k=1\) was proved by Bojanov [2]; related Markov-type results had appeared earlier in [3], [4]. The problem is also known in some special classes. Bojanov and Rahman [5] treated polynomials whose zeros lie in \([-1,1]\), and Avvakumova [6] proved the case \(k=2\) for a larger intermediate class. More recent work studies Bojanov–Naidenov-type questions in other settings, including differentiable functions on the real line, trigonometric polynomials, and splines, together with links to sharp Kolmogorov- and Bernstein-type inequalities [7]–[9]. These papers deal with different spaces or metrics from 1 . The question considered here is the original algebraic-polynomial problem on \([-1,1]\).
This paper proves the full quartic case, namely \(n=4\) and \(k=2\). In fact, we prove a little more than the convex integral inequality: the integrated tails of \(|P''|\) are dominated by those of \(|T_4''|\). This is an increasing-convex-order statement, and it gives 1 in the quartic case.
The argument is quite specific to degree four. A quartic is determined by its values at the five extremal points of \(T_4\). After this interpolation, a simple convexity argument allows us to enlarge the feasible set to the full cube \([-1,1]^5\) and then pass to its vertices. There are only \(32\) of them, and their second derivatives reduce, up to sign and reflection, to a short list of quadratic polynomials. The rest of the proof compares the level sets of these quadratics with the level sets of \(T_4''\). Thus the finite character of the proof is also its limitation: it gives a complete solution of this concrete intermediate case, but it is not a direct argument for higher degrees.
Theorem 1. Let \(P\in\pi_4\) satisfy \(\left\lVert P\right\rVert_{C[-1,1]}\leq 1\), and let \[T_4(x)=8x^4-8x^2+1.\] Then, for every \(t\geq 0\), \[\label{eq:tail-main} \int_{-1}^{1}\bigl(|P''(x)|-t\bigr)_+\,dx \leq \int_{-1}^{1}\bigl(|T_4''(x)|-t\bigr)_+\,dx .\qquad{(1)}\] Consequently, for every nondecreasing convex function \(\varphi:[0,\infty)\to\mathbb{R}\), \[\label{eq:convex-main} \int_{-1}^{1}\varphi(|P''(x)|)\,dx \leq \int_{-1}^{1}\varphi(|T_4''(x)|)\,dx .\qquad{(2)}\] If \(P\neq\pm T_4\) and \(\varphi\) is strictly increasing and convex, then ?? is strict.
We briefly outline the proof. For a fixed tail parameter \(t\), the map from the five interpolating values to the integrated tail of \(|P''|\) is convex. Its maximum on \([-1,1]^5\) is therefore attained at a vertex. At the vertices we only have to handle finitely many quadratic polynomials, up to sign and reflection; these are compared with \[T_4''(x)=96x^2-16.\]
We start with a standard tail representation for increasing convex functionals. The lemma is stated with the strict part included, because this will be needed at the end of the proof.
Lemma 1. Let \(f,g\) be measurable functions on a finite measure space, with values in \([0,M]\). Suppose \[\label{eq:tail-order-general} \int (f-t)_+\,d\nu\leq \int (g-t)_+\,d\nu, \qquad 0\leq t\leq M.\qquad{(3)}\] Then \[\int \varphi(f)\,d\nu\leq \int \varphi(g)\,d\nu\] for every nondecreasing convex function \(\varphi:[0,M]\to\mathbb{R}\). If, in addition, \[\label{eq:strict-at-zero} \int f\,d\nu<\int g\,d\nu,\qquad{(4)}\] then the last inequality is strict for every strictly increasing convex function \(\varphi\).
Proof. Every convex function on \([0,M]\) can be written in the form \[\label{eq:hinge-representation} \varphi(u)=\varphi(0)+mu+ \int_{[0,M]}(u-s)_+\,d\mu(s),\tag{2}\] where \(m\in\mathbb{R}\) and \(\mu\) is a nonnegative finite Borel measure. When \(\varphi\) is nondecreasing, we may choose the linear coefficient with \(m\geq0\), after absorbing a possible atom at \(0\) into \(m\). Integrating 2 and applying ?? proves the non-strict inequality.
Now assume ?? . Put \[D(t):=\int(g-t)_+\,d\nu-\int(f-t)_+\,d\nu .\] Then \(D\) is continuous and \(D(0)>0\). Hence \(D(t)>0\) on some interval \(0\leq t\leq\delta\). If \(m>0\), the strict inequality already follows from the linear term. If \(m=0\) and \(\varphi\) is strictly increasing, then \(\mu((0,\delta))>0\); otherwise the representation 2 would make \(\varphi\) constant on \([0,\delta]\). Therefore \[\int \varphi(g)\,d\nu-\int \varphi(f)\,d\nu = mD(0)+\int_{[0,M]}D(s)\,d\mu(s)>0.\] ◻
Let \[x_j=\cos\frac{j\pi}{4},\qquad j=0,1,2,3,4.\] Thus \[x_0=1,\qquad x_1=\frac{\sqrt2}{2},\qquad x_2=0, \qquad x_3=-\frac{\sqrt2}{2},\qquad x_4=-1.\] For \(y=(y_0,\ldots,y_4)\in\mathbb{R}^5\), let \(P_y\) be the unique polynomial of degree at most four satisfying \[P_y(x_j)=y_j,\qquad j=0,\ldots,4.\] For fixed \(t\geq 0\), define \[\label{eq:Ft} F_t(y):=\int_{-1}^{1}\bigl(|P_y''(x)|-t\bigr)_+\,dx .\tag{3}\]
Lemma 2. For each \(t\geq 0\), the map \(F_t:\mathbb{R}^5\to\mathbb{R}\) is convex. Consequently, \[\max_{y\in[-1,1]^5}F_t(y)\] is attained at a vertex of the cube \([-1,1]^5\).
Proof. Interpolation depends linearly on the nodal values. Thus, for fixed \(x\), the map \(y\mapsto P_y''(x)\) is linear. The scalar function \[u\mapsto (|u|-t)_+=\max\{u-t,-u-t,0\}\] is convex, so \(y\mapsto (|P_y''(x)|-t)_+\) is convex as well. Integrating over \([-1,1]\) keeps convexity, which proves the first assertion.
For the second assertion, write an arbitrary point of the cube as a convex combination of vertices, say \(y=\sum_v\lambda_v v\). Then \[F_t(y)\leq \sum_v\lambda_vF_t(v)\leq \max_vF_t(v).\] Hence a maximum is attained at some vertex. ◻
The nodal vectors coming from polynomials in \(\pi_4^\circ\) form only a subset of \([-1,1]^5\). We shall nevertheless work on the full cube. This enlargement is harmless, since we only need an upper bound.
Let \(\ell_j\) be the Lagrange basis polynomial associated with \(x_j\). A direct calculation gives \[\begin{align} \ell_0''(x)&=12x^2+6x-1,\\ \ell_1''(x)&=-24x^2-6\sqrt2\,x+4,\\ \ell_2''(x)&=24x^2-6,\\ \ell_3''(x)&=-24x^2+6\sqrt2\,x+4,\\ \ell_4''(x)&=12x^2-6x-1. \end{align}\] Therefore \[P_y''(x)=\sum_{j=0}^{4}y_j\ell_j''(x)=A(y)x^2+B(y)x+C(y),\] where \[\begin{align} A(y)&=12y_0-24y_1+24y_2-24y_3+12y_4,\tag{4}\\ B(y)&=6y_0-6\sqrt2\,y_1+6\sqrt2\,y_3-6y_4,\tag{5}\\ C(y)&=-y_0+4y_1-6y_2+4y_3-y_4.\tag{6} \end{align}\]
At the vertices we have \(y_j\in\{-1,1\}\), hence \(32\) possible second derivatives. The transformations \(q(x)\mapsto -q(x)\) and \(q(x)\mapsto q(-x)\) leave all level-set and tail quantities unchanged. After using these symmetries and removing duplicates, the list is: \[\label{eq:vertex-list} \begin{array}{c|l} \text{type} & q(x)=P_y''(x) \\ \hline 0 & 0 \\ L & 12\sqrt2\,x+4 \\ Q_1 & 24x^2+12x-2 \\ Q_2 & 48x^2+12\sqrt2\,x-8 \\ Q_3 & 24x^2+12(\sqrt2-1)x-6 \\ Q_4 & 48x^2-12 \\ Q_5 & 72x^2+12x-14 \\ Q_6 & 24x^2+12(\sqrt2+1)x-6 \\ Q_7 & 48x^2-4 \\ Q_* & 96x^2-16. \end{array}\tag{7}\] The last row is \(T_4''\) itself. It comes from the Chebyshev nodal vector \[(1,-1,1,-1,1),\] and its negative corresponds to \(-T_4\).
We shall need the following elementary data. For a quadratic row we write \[q(x)=a(x-h)^2-v.\]
| Type | \(a\) | \(h\) | \(v\) | \(q(-1)\) | \(q(1)\) | \(\norm{q}_{C[-1,1]}\) |
|---|---|---|---|---|---|---|
| \(0\) | – | – | – | \(0\) | \(0\) | \(0\) |
| \(L\) | – | – | – | \(4-12\sqrt2\) | \(4+12\sqrt2\) | \(4+12\sqrt2\) |
| \(Q_1\) | \(24\) | \(-\frac14\) | \(\frac72\) | \(10\) | \(34\) | \(34\) |
| \(Q_2\) | \(48\) | \(-\frac{\sqrt2}{8}\) | \(\frac{19}{2}\) | \(40-12\sqrt2\) | \(40+12\sqrt2\) | \(40+12\sqrt2\) |
| \(Q_3\) | \(24\) | \(\frac{1-\sqrt2}{4}\) | \(\frac{21}{2}-3\sqrt2\) | \(30-12\sqrt2\) | \(6+12\sqrt2\) | \(6+12\sqrt2\) |
| \(Q_4\) | \(48\) | \(0\) | \(12\) | \(36\) | \(36\) | \(36\) |
| \(Q_5\) | \(72\) | \(-\frac1{12}\) | \(\frac{29}{2}\) | \(46\) | \(70\) | \(70\) |
| \(Q_6\) | \(24\) | \(-\frac{1+\sqrt2}{4}\) | \(\frac{21}{2}+3\sqrt2\) | \(6-12\sqrt2\) | \(30+12\sqrt2\) | \(30+12\sqrt2\) |
| \(Q_7\) | \(48\) | \(0\) | \(4\) | \(44\) | \(44\) | \(44\) |
| \(Q_*\) | \(96\) | \(0\) | \(16\) | \(80\) | \(80\) | \(80\) |
Lemma 3. If \(P\in\pi_4^\circ\), then \[\left\lVert P''\right\rVert_{C[-1,1]}\leq 80=\left\lVert T_4''\right\rVert_{C[-1,1]}.\] Equality is possible only for \(P=\pm T_4\).
Proof. Let \(y_j=P(x_j)\). Then \(y\in[-1,1]^5\) and \(P=P_y\). The map \[y\mapsto \left\lVert P_y''\right\rVert_{C[-1,1]}\] is convex, since it is the supremum, over \(x\in[-1,1]\), of the convex functions \(y\mapsto |P_y''(x)|\). Its maximum on the cube is therefore attained at a vertex. Table 1 shows that the two Chebyshev vertices have norm \(80\), while every other vertex has norm strictly smaller than \(80\).
Suppose equality holds for some \(y\in[-1,1]^5\). Write \(y=\sum_v\lambda_v v\) as a convex combination of vertices. Convexity gives \[80=\left\lVert P_y''\right\rVert_{C[-1,1]} \leq \sum_v\lambda_v\left\lVert P_v''\right\rVert_{C[-1,1]} \leq 80.\] Thus every vertex with \(\lambda_v>0\) must be a Chebyshev vertex; any other vertex would make the last inequality strict. Hence \(y\) lies on the segment joining \((1,-1,1,-1,1)\) to \((-1,1,-1,1,-1)\). In other words, \(P=\alpha T_4\) with \(|\alpha|\leq1\), and equality in the norm forces \(|\alpha|=1\). ◻
Set \[q_*(x):=T_4''(x)=96x^2-16.\] For a measurable function \(q\) on \([-1,1]\), define the level-set function \[\Lambda_q(t):=\operatorname{meas}\{x\in[-1,1]: |q(x)|>t\}.\] Then \[\label{eq:tail-level} \int_{-1}^{1}(|q(x)|-t)_+\,dx =\int_t^\infty \Lambda_q(s)\,ds .\tag{8}\] We write \(\Lambda_*:=\Lambda_{q_*}\). The polynomial \(q_*\) vanishes at \(|x|=1/\sqrt6\), takes the value \(-16\) at \(x=0\), and takes the value \(80\) at \(x=\pm1\). Hence \[\label{eq:Lambda-star} \Lambda_*(t)= \begin{cases} \displaystyle 2-2\sqrt{\frac{16+t}{96}} +2\sqrt{\frac{16-t}{96}}, &0\leq t\leq16,\\[1.1em] \displaystyle 2-2\sqrt{\frac{16+t}{96}}, &16\leq t\leq80,\\[1.1em] 0, &t\geq80. \end{cases}\tag{9}\] Equivalently, the integrated Chebyshev tail is \[\label{eq:R-star} \int_{-1}^{1}(|q_*(x)|-t)_+\,dx= \begin{cases} \displaystyle 32-2t+ \frac{(16+t)^{3/2}+(16-t)^{3/2}}{3\sqrt6}, &0\leq t\leq16,\\[1.1em] \displaystyle 32-2t+ \frac{(16+t)^{3/2}}{3\sqrt6}, &16\leq t\leq80,\\[1.1em] 0, &t\geq80. \end{cases}\tag{10}\]
Lemma 4. Let \[q(x)=a(x-h)^2-v, \qquad a>0, \qquad -1<h\leq0, \qquad v>0,\] and set \[L:=q(-1),\qquad R:=q(1).\] Assume \[0<v\leq L\leq R.\] Then \[\Lambda_q(t)= \begin{cases} \displaystyle 2-2\sqrt{\frac{v+t}{a}}+2\sqrt{\frac{v-t}{a}}, &0\leq t\leq v,\\[1.1em] \displaystyle 2-2\sqrt{\frac{v+t}{a}}, &v\leq t\leq L,\\[1.1em] \displaystyle 1-h-\sqrt{\frac{v+t}{a}}, &L\leq t\leq R,\\[1.1em] 0,&t\geq R. \end{cases}\]
Proof. The minimum of \(q\) is \(-v\), attained at \(x=h\). If \(0\leq t\leq v\), then the set where \(q<-t\) is \[h-\sqrt{\frac{v-t}{a}}<x<h+\sqrt{\frac{v-t}{a}},\] and the set where \(q>t\) is the union of the two outer intervals cut off by \[x=h\pm\sqrt{\frac{v+t}{a}}.\] The assumptions \(0<v\leq L\leq R\) ensure that the relevant roots lie in \([-1,1]\) for this range of \(t\). Adding the lengths gives the first formula.
If \(v\leq t\leq L\), the negative interval has disappeared, and both roots of \(q=t\) still lie in \([-1,1]\). This gives the second formula. If \(L\leq t\leq R\), the left root of \(q=t\) has moved past \(-1\), while the right root remains in \([-1,1]\) until \(t=R\). Thus only the right outer interval remains, which gives the third formula. For \(t\geq R\) the set is empty. ◻
The remaining point is the following finite comparison.
Lemma 5. For each non-Chebyshev vertex derivative \(q\) in 7 , \[\label{eq:level-dom} \Lambda_q(t)\leq \Lambda_*(t),\qquad t\geq0.\qquad{(5)}\] Consequently, \[\label{eq:tail-dom-vertices} \int_{-1}^{1}(|q(x)|-t)_+\,dx \leq \int_{-1}^{1}(|T_4''(x)|-t)_+\,dx, \qquad t\geq0.\qquad{(6)}\]
Proof. The zero vertex is immediate, so we consider the remaining cases.
Let \[q(x)=12\sqrt2\,x+4, \qquad m:=12\sqrt2.\] Then \[\Lambda_q(t)= \begin{cases} \displaystyle 2-\frac{2t}{m}, &0\leq t\leq m-4,\\[0.9em] \displaystyle \frac{m+4-t}{m}, &m-4\leq t\leq m+4,\\[0.9em] 0,&t\geq m+4. \end{cases}\] For \(0\leq t\leq m-4\), we need to prove \[\frac{t}{m} \geq \sqrt{\frac{16+t}{96}}- \sqrt{\frac{16-t}{96}}.\] For \(t>0\) the difference of square roots on the right is \[\frac{2t}{\sqrt{96}\bigl(\sqrt{16+t}+\sqrt{16-t}\bigr)}.\] It is therefore enough that \[\sqrt{96}\bigl(\sqrt{16+t}+\sqrt{16-t}\bigr) \geq 2m,\] which holds because the left-hand side is at least \(16\sqrt6\), while \(2m=24\sqrt2\). The endpoint \(t=0\) follows by continuity.
For \(m-4\leq t\leq m+4\), we have \(\Lambda_q(t)\leq 8/m=\sqrt2/3\). Since \(\Lambda_*\) is decreasing, it remains only to check \[\frac{\sqrt2}{3}\leq \Lambda_*(m+4) =2-2\sqrt{\frac{20+12\sqrt2}{96}}.\] Both sides are nonnegative; after squaring, this is equivalent to \(61>33\sqrt2\). Hence ?? holds for the linear vertex.
Consider the six rows \[Q_1,\quad Q_2, \quad Q_3, \quad Q_4, \quad Q_5, \quad Q_7.\] Write \[q(x)=a(x-h)^2-v, \qquad a>0, \qquad v>0, \qquad h\leq0,\] and set \[L:=q(-1),\qquad R:=q(1),\qquad L\leq R.\] For these six rows, Table 1 gives the inequalities \[\label{eq:standard-conditions} \begin{gather} 0<v\leq16,\qquad a\leq96,\qquad -1<h\leq0,\\ a\leq3(v+16),\qquad 0<v\leq L\leq R<80. \end{gather}\tag{11}\] Thus Lemma 4 applies to each of the six rows. We compare the level sets on the intervals \([0,v]\), \([v,L]\), \([L,R]\), and \([R,\infty)\).
If \(0\leq t\leq v\), then Lemma 4 gives \[\Lambda_q(t)=2-2\sqrt{\frac{v+t}{a}}+2\sqrt{\frac{v-t}{a}}.\] We have to show \[\sqrt{\frac{v+t}{a}}- \sqrt{\frac{v-t}{a}} \geq \sqrt{\frac{16+t}{96}}- \sqrt{\frac{16-t}{96}}.\] The function \(s\mapsto\sqrt{s+t}-\sqrt{s-t}\) is decreasing for \(s\geq t\). Since \(v\leq16\) and \(a\leq96\), the required inequality follows.
If \(v\leq t\leq L\), then Lemma 4 gives \[\Lambda_q(t)=2-2\sqrt{\frac{v+t}{a}}.\] First assume \(t\leq16\). It remains to prove \[\sqrt{\frac{v+t}{a}} \geq \sqrt{\frac{16+t}{96}}- \sqrt{\frac{16-t}{96}}.\] Using \[\left( \sqrt{\frac{16+t}{96}}- \sqrt{\frac{16-t}{96}} \right)^2 \leq \frac{t^2}{768},\] it is enough to show \[\frac{v+t}{a}\geq \frac{t^2}{768}.\] The function \(t^2/(v+t)\) is increasing for \(t\geq0\), and \(t\leq16\), so \[\frac{t^2}{v+t}\leq \frac{256}{v+16}.\] Thus the claim follows from \(a\leq3(v+16)\).
If \(16\leq t\leq L\), then \[\Lambda_*(t)=2-2\sqrt{\frac{16+t}{96}},\] and we only have to prove \[\frac{v+t}{a}\geq \frac{16+t}{96}.\] Equivalently, \[96(v+t)\geq a(16+t).\] Because \(a\leq96\), the difference between the left and right sides is increasing in \(t\). It is therefore enough to check \(t=16\), where the condition becomes \(a\leq3(v+16)\).
It remains to consider \(L\leq t\leq R\). In this interval, Lemma 4 gives \[\Lambda_q(t)=1-h-\sqrt{\frac{v+t}{a}}.\] For the two rows with \(L<16\), namely \[q(x)=24x^2+12x-2, \qquad q(x)=24x^2+12(\sqrt2-1)x-6,\] we have \(\Lambda_q(t)\leq -2h\). Since \(\Lambda_*\) is decreasing, it remains to verify \(-2h\leq\Lambda_*(R)\).
For \(q(x)=24x^2+12x-2\), this is \[\frac{1}{2}\leq 2-2\sqrt{\frac{50}{96}}.\] Both sides are nonnegative, and squaring reduces this to \(25\leq27\). For \(q(x)=24x^2+12(\sqrt2-1)x-6\), the desired inequality is \[\frac{\sqrt2-1}{2} \leq 2-2\sqrt{\frac{22+12\sqrt2}{96}}.\] Both sides are nonnegative, and after squaring this becomes \(72\sqrt2\leq140\).
For the two rows with \(16<L<R\), namely \[q(x)=48x^2+12\sqrt2\,x-8, \qquad q(x)=72x^2+12x-14,\] put \(E(t):=\Lambda_*(t)-\Lambda_q(t)\). On \([L,R]\), one has \(t\geq16\), and \(E'(t)<0\) provided \[a(v+t)>24(16+t).\] This condition need only be checked at \(t=L\). For the two rows, the left side minus the right side equals, respectively, \[1032-288\sqrt2>0, \qquad 2868>0.\] Hence \(E\) is decreasing on \([L,R]\), and so \[E(t)\geq E(R)=\Lambda_*(R)>0,\] because \(R<80\) and \(\Lambda_q(R)=0\). The remaining two standard rows, \(48x^2-12\) and \(48x^2-4\), have \(L=R\), so there is no endpoint interval to check. This proves the comparison for the six standard quadratic vertices.
It remains to handle the row \[q(x)=24x^2+12(\sqrt2+1)x-6.\] Write \[q(x)=24(x-h)^2-v, \qquad h=-\frac{1+\sqrt2}{4}, \qquad v=\frac{21}{2}+3\sqrt2.\] Then \[q(-1)=6-12\sqrt2<0, \qquad q(1)=30+12\sqrt2<80.\] Set \[s_0:=12\sqrt2-6.\] We have \[0<s_0<v<16<30+12\sqrt2<80.\]
For \(0\leq t\leq s_0\), \[\Lambda_q(t)=2+ \sqrt{\frac{v-t}{24}}- \sqrt{\frac{v+t}{24}}.\] Since \(v<16\) and \(s\mapsto\sqrt{s+t}-\sqrt{s-t}\) is decreasing in \(s\), we have \[\sqrt{v+t}-\sqrt{v-t} \geq \sqrt{16+t}-\sqrt{16-t}.\] Dividing by \(\sqrt{24}\) gives \[\sqrt{\frac{v+t}{24}}- \sqrt{\frac{v-t}{24}} \geq 2\left( \sqrt{\frac{16+t}{96}}- \sqrt{\frac{16-t}{96}} \right),\] and hence \(\Lambda_q(t)\leq\Lambda_*(t)\) on \([0,s_0]\).
For \(s_0\leq t\leq v\), the negative part is fully inside \([-1,1]\), and \[\Lambda_q(t)=1-h- \sqrt{\frac{v+t}{24}}+ 2\sqrt{\frac{v-t}{24}}.\] Let \(E(t):=\Lambda_*(t)-\Lambda_q(t)\). Then \[E'(t)=\frac{1}{\sqrt{96}} \left( -\frac{1}{\sqrt{16+t}} -\frac{1}{\sqrt{16-t}} +\frac{1}{\sqrt{v+t}} +\frac{2}{\sqrt{v-t}} \right)>0,\] because \(v<16\). Since the comparison has already been proved at \(t=s_0\), it holds on \([s_0,v]\).
For \(v\leq t\leq16\), \[\Lambda_q(t)=1-h-\sqrt{\frac{v+t}{24}}.\] Again set \(E(t):=\Lambda_*(t)-\Lambda_q(t)\). Then \[E'(t)=\frac{1}{\sqrt{96}} \left( -\frac{1}{\sqrt{16+t}} -\frac{1}{\sqrt{16-t}} +\frac{1}{\sqrt{v+t}} \right)<0,\] since \(t\geq v\) implies \(v+t>16-t\). Thus \(E\) is decreasing on \([v,16]\), so it remains to check \(E(16)>0\). At \(t=16\), \[E(16)= 2-\frac{2}{\sqrt3} -\frac{5+\sqrt2}{4} + \sqrt{\frac{53+6\sqrt2}{48}}.\] Using \[\sqrt{\frac{53+6\sqrt2}{48}}>\frac{11}{10}, \qquad \sqrt2<\frac{3}{2}, \qquad \frac{2}{\sqrt3}<\frac{7}{6},\] we get \[E(16)> 2-\frac{7}{6}-\frac{5}{4}-\frac{3}{8}+\frac{11}{10} =\frac{37}{120}>0.\] Hence the comparison holds on \([v,16]\).
For \(16\leq t\leq30+12\sqrt2\), \[\Lambda_q(t)=1-h-\sqrt{\frac{v+t}{24}}, \qquad \Lambda_*(t)=2-2\sqrt{\frac{16+t}{96}}.\] Thus \[\frac{d}{dt}(\Lambda_*(t)-\Lambda_q(t)) = \frac{1}{\sqrt{96}} \left( \frac{1}{\sqrt{v+t}}- \frac{1}{\sqrt{16+t}} \right)>0,\] because \(v<16\). Since the difference is positive at \(t=16\), it remains positive up to \(30+12\sqrt2\). For larger \(t\), \(\Lambda_q(t)=0\), so the comparison is trivial. This completes the proof of ?? .
Finally, ?? follows from 8 . ◻
Lemma 6. If \(P\in\pi_4^\circ\) and \(P\neq\pm T_4\), then \[\label{eq:L1-strict} \int_{-1}^{1}|P''(x)|\,dx < \int_{-1}^{1}|T_4''(x)|\,dx .\qquad{(7)}\]
Proof. Let \(q\) be a non-Chebyshev vertex derivative. Lemma 5 gives \(\Lambda_q\leq\Lambda_*\), and Table 1 gives \(\left\lVert q\right\rVert_{C[-1,1]}<80\). Since \(\Lambda_*(s)>0\) for \(0\leq s<80\), formula 8 with \(t=0\) gives \[\int_{-1}^{1}|q(x)|\,dx < \int_{-1}^{1}|T_4''(x)|\,dx\] for every non-Chebyshev vertex. The two Chebyshev vertices give equality.
The functional \(F_0(y)=\int_{-1}^{1}|P_y''(x)|\,dx\) is convex. If equality held in ?? , then in a convex decomposition of the nodal vector \(y=(P(x_0),\ldots,P(x_4))\) no non-Chebyshev vertex could appear with positive weight; otherwise the convexity estimate would be strict. Thus \(y\) would lie on the segment between \((1,-1,1,-1,1)\) and its negative. Hence \(P=\alpha T_4\) with \(|\alpha|\leq1\), and equality in ?? would force \(|\alpha|=1\). This contradicts \(P\neq\pm T_4\), and the lemma follows. ◻
Fix \(t\geq0\). Lemma 5, together with the row \(q=T_4''\), shows that every vertex \(v\in\{-1,1\}^5\) satisfies \[F_t(v) \leq \int_{-1}^{1}\bigl(|T_4''(x)|-t\bigr)_+\,dx .\] By Lemma 2, the same estimate holds for all \(y\in[-1,1]^5\). Now let \(P\in\pi_4^\circ\) and take \(y_j=P(x_j)\). Then \(y\in[-1,1]^5\) and \(P=P_y\), so \[\int_{-1}^{1}\bigl(|P''(x)|-t\bigr)_+\,dx \leq \int_{-1}^{1}\bigl(|T_4''(x)|-t\bigr)_+\,dx.\] This proves ?? .
By Lemma 3, both \(|P''|\) and \(|T_4''|\) take their values in \([0,80]\). We may therefore apply Lemma 1 with \[f(x)=|P''(x)|, \qquad g(x)=|T_4''(x)|.\] This gives ?? for every nondecreasing convex \(\varphi\).
Finally suppose that \(P\neq\pm T_4\). Lemma 6 gives \[\int_{-1}^{1}|P''(x)|\,dx < \int_{-1}^{1}|T_4''(x)|\,dx .\] The strict part of Lemma 1 then gives strict inequality in ?? for every strictly increasing convex \(\varphi\). This completes the proof.
Remark 2. The equality cases for the whole tail family are also exactly \(P=\pm T_4\). Indeed, if equality held in ?? for all \(t\), then the highest tail levels would force \(\left\lVert P''\right\rVert_{C[-1,1]}=80\). Lemma 3 would then give \(P=\pm T_4\).
This appendix gives the finite enumeration used in the reduction to 7 . Reflection of the interval sends a nodal vector \[(y_0,y_1,y_2,y_3,y_4) \quad\text{to}\quad (y_4,y_3,y_2,y_1,y_0),\] and multiplication of the polynomial by \(-1\) sends \(y\) to \(-y\). Both operations preserve the level-set and integrated-tail quantities, since they replace \(q(x)\) by \(q(-x)\) or by \(-q(x)\). The representatives below, together with these symmetries, account for all \(32\) vertices; the orbit sizes in the third column sum to \(32\).
| Type | representative \(y\) | orbit size | \(P_y''(x)\) |
|---|---|---|---|
| \(0\) | \((-1,-1,-1,-1,-1)\) | \(2\) | \(0\) |
| \(L\) | \((1,-1,-1,1,1)\) | \(4\) | \(12\sqrt2\,x+4\) |
| \(Q_1\) | \((1,-1,-1,-1,-1)\) | \(4\) | \(24x^2+12x-2\) |
| \(Q_2\) | \((1,-1,1,1,1)\) | \(4\) | \(48x^2+12\sqrt2\,x-8\) |
| \(Q_3\) | \((-1,-1,1,1,1)\) | \(4\) | \(24x^2+12(\sqrt2-1)x-6\) |
| \(Q_4\) | \((-1,-1,1,-1,-1)\) | \(2\) | \(48x^2-12\) |
| \(Q_5\) | \((1,-1,1,-1,-1)\) | \(4\) | \(72x^2+12x-14\) |
| \(Q_6\) | \((1,-1,1,1,-1)\) | \(4\) | \(24x^2+12(\sqrt2+1)x-6\) |
| \(Q_7\) | \((1,-1,-1,-1,1)\) | \(2\) | \(48x^2-4\) |
| \(Q_*\) | \((1,-1,1,-1,1)\) | \(2\) | \(96x^2-16\) |
Each row is obtained by substituting the representative vector into 4 –6 . Thus Table 2 gives a direct check of the reduction from the \(32\) vertex derivatives to the list 7 .