Problem 11056. Proposed by John H. Lindsey II, Cambridge, MA. If P is a polynomial of degree at most n, and if I is any finite interval then supx∈I |P (x)| is at most 2n2 times the average on I of x 7→ |P (x)|. Solution, by Omran Kouba Higher Institute for Applied Sciences And Technology, Damascus, Syria. We obtain a better estimate than the one in the statement of the problem. That is, if P is a polynomial of degree at most n, and if I is any finite interval then supx∈I |P (x)| is at most (n + 1)2 times the average on I of x 7→ |P (x)|. ) 1 dn ( 2 (X − 1)n . In fact, we will We will use Legendre polynomials, defined, for n ∈ N, by Pn (X) = n n 2 n! dX make use of the following two properties which can be found in [1] for example. { ∫ 1 0 if n ̸= m 2 1. Orthogonality : for (n, m) ∈ N , we have Pn (x)Pm (x) dx = 2 if n = m −1 2n+1 2. for n ∈ N, we have sup |Pn (x)| = Pn (1) = 1. −1≤x≤1

First, asume that I = [−1, 1], and let P be a polynomial of degree at most n. Using orthogonality we find that ∫ 1 n ∑ 2k + 1 P (X) = λk (P )Pk (X), with λk (P ) = P (t)Pk (t) dt 2 −1 k=0

Then, using 2., we have ∀ x ∈ [−1, 1],

|P (x)| ≤ ≤

n ∑ 2k + 1 k=0 n ∑ k=0

2

|λk (P )| |Pk (x)| ≤

n ∑ 2k + 1

2

k=0

|λk (P )| sup |Pk (t)| −1≤t≤1

2k + 1 |λk (P )| 2

and using 2. again, we obtain (∫ |λk (P )| ≤

1

−1

)

∫

|P (t)| dt · sup |Pk (t)| = −1≤t≤1

finally,

( ∀ x ∈ [−1, 1],

|P (x)| ≤

n ∑ 2k + 1

2

k=0

that is sup |P (x)| ≤ (n + 1)2 ·

−1≤x≤1

)∫

1 2

−1

|P (t)| dt

1 −1

∫

1

|P (t)| dt

1

−1

|P (t)| dt

In general, if I = [a, b], of degree n, we just apply the preceeding result to the ( and P is a polynomial ) a+b b−a polynomial Q(X) = P + X and we find 2 2 sup |P (x)| ≤ (n + 1)2 · x∈I

1 |I|

∫ |P (t)| dt I

which is the desired result. Reference: [1] G.B. Arfken, H.J. Weber, Mathematical Methodes for Physicists, 4th edition, Chapter 12, Academic Press, 1995. My solution was published in AMM Oct 2005 page 752. 1