1 Inner Products and Norms on Real Vector Spaces

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1 Math 373: Principles Techniques of Applied Mathematics Spring 29 The 2 Inner Product 1 Inner Products Norms on Real Vector Spaces Recall that an inner product on a real vector space V is a function from V V to R such that for any v, w, x V scalars c R, we have v, w w, v cv, w c v, w v, w + x v, w + v, w v, v > for v Given a vector space V with an inner product, it is possible to define notions such as lengths an angles. The length or norm of an element v V is given by v v, v. It is not difficult to prove that inner products lengths satisfy the Cauchy-Schwarz inequality: v, w v w From this it follows that it makes sense to define the angle between v w to be the angle θ such that v, w θ v w, ce the right h side is between 1 1. We then say that two vectors v w are orthogonal if v, w. It also follows from the definition of an inner product, together with the Cauchy-Schwarz inequality, that for all v, w V scalars c R, we have cv c v v > for v v + w v + w These are actually the defining properties of a norm. 2 Orthogonal Bases Suppose that we have a basis {v 1, v 2,..., v n } for a vector space V consisting of vectors that are mutually orthogonal. That is, suppose that v j, v k for every j k. Now suppose we are given a vector w that we wish to write as a linear combination of the v k s. First write w c 1 v 1 + c 2 v c n v n

2 for some unknown coefficients c 1 through c n. Now fix k take the inner product of both sides with v k to get w, v k c 1 v 1, v k + c 2 v 2, v k + + c n v n, v k Now because of the orthogonality of the basis vectors, every one of the inner products on the right h side is zero, except for one, namely v k, v k. Thus the equation reduces to thus We therefore have the general formula w, v k c k v k, v k c k w, v k v k, v k. w w, v 1 v 1, v 1 v 1 + w, v 2 v 2, v 2 v w, v n v n, v n v n. Thus if we have a basis consisting of orthogonal vectors, then it is very simple to express a given vector in terms of the basis vectors. Simply take inner products to find the coefficients. 3 The 2 Inner Product Now consider the vector space V of real-valued continuous functions on an interval [a, b]. The 2 inner product on V is defined by f, g b a f(xg(x dx It is left as an exercise to verify that this satisfies the properties of an inner product. The 2 norm of a function is then defined by f ( b 1/2 f, f [f(x] dx 2. The thing that makes Fourier series work so well is that the basis functions are all orthogonal with respect to the 2 inner product. Example 1. Consider the Fourier e series φ(x A n n1 of a function φ on the interval [, ]. Denoting by we proved in class that s m, s n s n, s n s n (x a 2 dx 2, dx

3 From this we can derive formulas for the coefficients A n as we did for the c k above: A n φ, s n s n, s n 2 φ(x dx Example 2. Similar calculations apply to the Fourier ine series φ(x 1 2 A + A n of a function φ on the interval [, ]. Here the basis functions are c n (x, together with the constant function 1 (which can be thought of as ( πx from which we get c m, c n c n, c n 1, dx A n φ, c n c n, c n 2 n1 dx 2 φ, 1 A 2 1, 1 2 dx φ(x n 1, 2,... dx φ(x dx.. we have This explains the reason for the factor 1 in front of A 2. It just ensures that the formulas for all the coefficients have the same form. Example 3. Finally, consider the full Fourier series φ(x 1 2 A + A n n1 + B n of a function φ on the interval [, ]. The reason for ug the entire interval [, ] instead of [, ] is that the functions s n c n are not all mutually orthogonal on [, ], but they are orthogonal on [, ]. That is, c m, c n s m, s n c m, s n dx dx dx for all m, n

4 Since we then have c n, c n s n, s n 1, 1 2, A n φ, c n c n, c n 1 B n φ, s n s n, s n 1 A 1 φ(x dx φ(x dx φ(x dx. 4 Hermitian Inner Products et V be a complex vector space (the field of scalars is C. Then a Hermitian inner product is function V V to C such that v, w w, v cv, w c v, w v, w + x v, w + v, w v, v > for v The notions of orthogonality length are defined in the same way as we did ug inner products on real vector spaces. Also, given an orthogonal basis v 1,..., v n for V we can derive the coefficients c k in the expansion in the same way as before: w c 1 v c n v n c k w, v k v k, v k The 2 inner product for complex-valued functions on an interval [a, b] is given by f, g f(xg(x dx It is not hard to prove that it satisfies the four conditions above.

5 Example 4. given by The functions The complex form of the full Fourier series of a function φ on [, ] is φ(x n C n e inπx/ e n (x e inπx/ are orthogonal with respect to the 2 inner product on [, ]. To see this, we can use the fact that e iπ 1 to compute e m, e n e imπx/ e inπx/ dx e imπx/ e inπx/ dx e i(m nπx/ dx iπ(m n ei(m nπx/ ( e i(m nπ e i(m nπ iπ(m n ( ( 1 m n ( 1 n m iπ(m n Therefore e n, e n e inπx/ e inπx/ dx 1 dx 2 C n φ, e n e n, e n 1 φ(xe inπx/ dx. 2