From Euclidean to Hilbert Spaces. Edoardo Provenzi

      Use this result to deduce that Φ is definite over E × E.

      2) Prove that Φ is an inner product over E, which we shall note 〈 , 〉.

      3) For n ∈ ℕ, let T_n be the n-th Chebyshev polynomial, that is, the only polynomial such that ∀θ ∈ ℝ, T_n(cos θ) = cos(nθ). Applying the substitution t = cos θ, show that (T_n)_n∈ℕ is an orthogonal family in E. Hint: use the trigonometric formula [1.13]:

[1.13]

      4) Prove that for all n ∈ ℕ, (T₀, . . . , T_n) is an orthogonal basis of ℝ_n[X], the vector space of polynomials in ℝ[X] of degree less than or equal to n. Deduce that (T_n)_n∈ℕ is an orthogonal basis in the algebraic sense: every element in E is a finite linear combination of elements in the basis of E.

5) Calculate the norm of T_n for all n and deduce an orthonormal basis (in the algebraic sense) of E using this result.

       Solution to Exercise 1.3

      1) We write . Since P and Q are polynomials, the function is continuous in a neighborhood V₁(1) and thus, according to the Weierstrass theorem, it is bounded in this neighborhood, that is, Ǝ C₁ > 0 such that . Similarly, the function is continuous in a neighborhood V₂(−1), thus Ǝ C₂ > 0 such that . This gives us:

      and:

      This implies that the integral defining Φ is definite; f(t) is continuous over (−1, 1) and therefore can be integrated. The result which we have just proved shows that f(t) is integrable in a right neighborhood of –1 and a left neighborhood of 1, as the integral of its absolute value is incremented by an integrable function in both cases.

      2) The bilinearity of Φ is obtained from the linearity of the integral using direct calculation. Its symmetry is a consequence of that of the dot product between functions. The only property which is not immediately evident is definite positiveness. Let us start by proving positiveness:

      and⁹:

      but the only polynomial with an infinite number of roots is the null polynomial 0(t) ≡ 0, so P = 0. Φ is therefore an inner product on E.

3) For all n, m ∈ ℕ:

      So, for all n ≠ m, we have:

      that is, Chebyshev polynomials form an orthogonal family of polynomials in relation to the inner product defined above.

      4) The family (T₀, T₁, . . . , T_n) is an orthogonal (and thus free) of n+1 elements of ℝ[X], which is of dimension n + 1, meaning that it is an orthogonal basis of ℝ_n[X]. To show that (T_n)_n∈ℕ is a basis in the algebraic sense of E, consider a polynomial P ∈ E of an arbitrary degree d ∈ ℕ, i.e. P ∈ ℝ_d[X], and note that (T₀, T₁, . . . , T_d) is an orthogonal (free) family of generators of ℝ_d[X], that is, a basis in the algebraic sense of the term.

      5) The norm of T_n is calculated using the following equality:

      which was demonstrated in point 3. Taking n = m, we have:

hence . Finally, the family:

      is an orthonormal basis of the vector space of first-order polynomials with real coefficients E□

1.9. Summary

      In this chapter, we have examined the properties of real and complex inner products, highlighting their differences. We noted that the symmetrical and bilinear properties of the real inner product must be replaced by conjugate

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