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