Bessel's inequality

Let H be a Hilbert space and let e_1, e_2, \dots be an orthonormal sequence in H. Then for any vector x in H, Bessel's inequality states: :\sum_{k=1}^{\infty}\left\vert\left\langle x,e_k\right\rangle \right\vert^2 \le \left\Vert x\right\Vert^2 where ⟨·,·⟩ denotes the inner product in the Hilbert space H, and \|\cdot\| denotes the norm induced by the inner product. The terms \langle x,e_k\rangle are the Fourier coefficients of x with respect to the sequence (e_k). The inequality implies that the series of the squared magnitudes of these coefficients converges. This allows for the definition of the vector x', which is the projection of x onto the subspace spanned by the orthonormal sequence: :x' = \sum_{k=1}^{\infty}\left\langle x,e_k\right\rangle e_k Bessel's inequality guarantees that this series converges. If the sequence (e_k) is a complete orthonormal basis, then x' = x, and the inequality becomes an equality known as Parseval's identity. Proof The inequality follows from the non-negativity of the norm of a vector. For any natural number n, let :x_n = \sum_{k=1}^n \langle x, e_k \rangle e_k This vector x_n is the projection of x onto the subspace spanned by the first n basis vectors. The vector x-x_n is orthogonal to this subspace, and thus orthogonal to x_n itself. By the Pythagorean theorem for inner product spaces, we have \|x\|^2 = \|x_n\|^2 + \|x-x_n\|^2. The proof proceeds by computing \|x - x_n\|^2: :\begin{align} 0 \leq \left\| x - \sum_{k=1}^n \langle x, e_k \rangle e_k\right\|^2 &= \left\langle x - \sum_{k=1}^n \langle x, e_k \rangle e_k, \; x - \sum_{j=1}^n \langle x, e_j \rangle e_j \right\rangle \\ &= \|x\|^2 - \sum_{k=1}^n \overline{\langle x, e_k \rangle} \langle x, e_k \rangle - \sum_{j=1}^n \langle x, e_j \rangle \langle e_j, x \rangle + \sum_{k=1}^n \sum_{j=1}^n \overline{\langle x, e_k \rangle} \langle x, e_j \rangle \langle e_k, e_j \rangle \\ &= \|x\|^2 - \sum_{k=1}^n |\langle x, e_k \rangle |^2 - \sum_{j=1}^n |\langle x, e_j \rangle |^2 + \sum_{k=1}^n | \langle x, e_k \rangle |^2 \\ &= \|x\|^2 - \sum_{k=1}^n | \langle x, e_k \rangle |^2 \end{align} This holds for any n \ge 1. Since the partial sums are non-negative and bounded above by \|x\|^2, the series \sum_{k=1}^{\infty}|\langle x,e_k\rangle|^2 converges and its sum is less than or equal to \|x\|^2. ==Fourier series==

In the theory of Fourier series, in the particular case of the Fourier orthonormal system, we get if f \colon \mathbb{R} \to \mathbb{C} has period T, : \sum_{k \in \mathbb{Z}} \left\vert \int_0^T e^{-2 \pi i k t/T} f (t) \,\mathrm{d}t\right\vert^2 \le T \int_0^T \vert f (t)\vert^2 \,\mathrm{d}t. In the particular case where f \colon \mathbb{R} \to \mathbb{R} , one has then : \left\vert \int_0^T f (t) \,\mathrm{d}t\right\vert^2 + 2 \sum_{n = 1}^\infty \left\vert \int_0^T \cos (2 \pi k t/T) f (t) \,\mathrm{d}t\right\vert^2 + 2 \sum_{n = 1}^\infty \left\vert \int_0^T \sin (2 \pi k t/T) f (t) \,\mathrm{d}t\right\vert^2 \le T \int_0^T \vert f (t)\vert^2 \,\mathrm{d}t. ==Non countable case==

More generally, if H is a pre-Hilbert space and (e_\alpha)_{\alpha \in A} is an orthonormal system, then for every x \in H : \sum_{\alpha \in A} | \langle x, e_\alpha \rangle |^2 \le \lVert x \rVert^2 This is proved by noting that if F \subseteq A is finite, then : \sum_{\alpha \in F} | \langle x, e_\alpha \rangle |^2 \le \lVert x \rVert^2 and then by definition of infinite sum : \sum_{\alpha \in A} | \langle x, e_\alpha \rangle |^2 = \sup \Bigl\{\sum_{\alpha \in F} | \langle x, e_\alpha \rangle |^2 : F \subseteq A \text{ is finite}\Bigr\} \le \lVert x \rVert^2. ==See also==