Sum (discrete) inequality

flip2004

Let $f(x)=\displaystyle Ax^3+Bx^2+CX+D$ with real coefficients. Then
$\sum\limits_{j=1}^{n}f(j) \le n^2(8n^2-7)\cdot\max\limits_{x\in [0,1]}|f(x)|\; \; \; ,\; \; \; \; n=1,2,...\; .$
Prove or disprove that this inequality cannot be improved.

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Alex. Lupas

harazi

No, I think the inequality cannot be improved. For this take the polynomial $f(x)=32x^3-48x^2+18x-1$ ( guess how I found it). And after many many computations I think we get equality in the inequality. The problem is really difficult and nice, but I solved it immediately. Do you remember that problem I asked you, flip 2004? And I still do not know your answer. The main thing is here to consider the polynomial $g(x)=f(\frac{1-x}{2})$ and to suppose without loss of generaliy that the maximal value of the absolute value of g on [-1,1] is 1. Then ( the main step) is to prove using Lagrange technique that for all x having modulus at least 1 we have that the absolute value of g(x) is at most the absolute value of $4x^3-3x$ and after that to use this inequality and sevral computations to finish the proof.

**Posted:** Thu Sep 16, 2004 11:41 am

flip2004

Hi Harazi, nice remarks. Indeed, let us try to extend my proposed question:
Denote by $T_n(z)$ the Chebychev polynomial of degree $n$ , i.e. $T_n(z)=\cos{(n\cdot \arccos{z})}$ when $z\in [-1,1] .$ Let $Y$ be a linear space of real functions defined on $I$ with $I \subset [1,\infty)$ . Suppose that the linear space of all real polynomials of degree $\le n$ is a subspace of $Y .$

Generalization of proposed question: If $A: Y\to {\mathbb R}$ is a linear positive functional , then for any polynomial $h$ of degree $n$ we have

$A[h]\le A[T_n^*] \cdot ||h||$
where $\; \displaystyle ||h||=\max\limits_{x\in [0,1]}|h(x)|\; \;$ and $\; \; \displaystyle T_n^*(x)= T_n(2x-1)\; .$

In the proposed question $Y =\left\{ f:[1,n]\to {\mathbb R}$ and $A[f]=\sum\limits_{k=1}^{n}f(k) ,$ $T_3(x)=\displaystyle 4x^3-3x$ as observed by Harazi.

Try to consider a linear functional $A$ having images
$A[f]=\frac{\Gamma(p+q)}{\Gamma(p)\Gamma(q)}\int\limits_0^1 t^{p-1}(1-t)^{q-1} f\left(1+at \right )\; dt \; \; \; ,\; \; \; \; a>0 , p>0,q>0 .$

**Posted:** Sat Sep 18, 2004 11:53 pm