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Functional Analysis MATH 36202 and MATH M6202∗ Functional Analysis MATH 36202 and MATH M6202∗ 1 Inner Product Spaces and Normed Spaces Inner Product Spaces Functional analysis involves studying vector spaces where we additionally have the notion of size of an element (the norm), such spaces are known as normed spaces. Sometimes we will have an additional notion of an inner product which can be informally thought of as a way of giving an angle beteen elements. Of particular interest will be infinite dimensional spaces. Remark. Throughout this unit we will be taking vector spaces over either R or C. We will use F to denote a field which is either R or C. n If we consider the vector space V = C then we can define the dot product for x; y 2 V by n X x · y = xkyk: k=1 This has the following properties for all x; y; z 2 V and λ 2 C. 1. (x + y) · z = x · z + y · z. 2. λ(x:y) = (λx):y 3. x · y = y · x 4. jxj = (x · x)1=2 (where j · j denote the Euclidean distance) 5. jx · yj ≤ jxjjyj. The notion of an inner product generalises this notion to general vector spaces. ∗This material is copyright of the University unless explicitly stated otherwise. It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only. 1 Definition 1.1. Let V be a vector space over F we say that h·; ·i : V ×V ! F is an inner product if the following properties are satisfied 1. For all x; y; z 2 V hx + y; zi = hx; zi + hy; zi. 2. For all x; y 2 V and λ 2 F we have that hλx, yi = λhx; yi. 3. For all x; y 2 V we have that hx; yi = hy; xi. 4. For all x 2 V we have that hx; xi ≥ 0 with equality if and only if x = 0. When we have a vector space V over F with an inner product h·; ·i : V ×V ! F we'll refer to V as an inner product space over F. n n Example. We can easily check that the dot product on either C or R satisfies these properties. However if we consider the vector spaces CN or RN we cannot generalise this definition because we would not always have convergence. So for either F = C or F = R let 1 2 1 X 2 ` = ffxngn=1 : xn 2 F and jxnj < 1g: n=1 2 2 We'll leave it as an exercise to check that ` is a supspace of FN. For x; y 2 ` we let 1 X hx; yi = xnyn: n=1 We first have to show that this is well defined. To see this we use the n 2 properties of the dot product on F . We have that for any x; y 2 ` and k 2 N that !1=2 !1=2 k k k X X 2 X 2 jxnynj ≤ jxnj jynj n=1 n=1 n=1 1 !1=2 1 !1=2 X 2 X 2 ≤ jxnj jynj : n=1 n=1 Thus this is bounded independently of k and so is absolutely convergent. Thus the definition makes sense. It is now routine to check that this is an inner product. We can now state some simple properties of the inner product Theorem 1.2. Let V be an inner product space over F. We have that 1. For x; y; z 2 V that hx; y + zi = hx; yi + hx; zi. 2. For x; y 2 V and λ 2 F that hx; λyi = λhx; yi 2 3. For any x 2 V we have that hx; 0i = h0; xi = 0. 4. If x; z 2 V satisfies that hx; yi = hz; yi for all y 2 V then x = z. Proof. Let x; y; z 2 V and λ 2 F. For part 1 we have hx; y + zi = hy + z; xi = hy; xi + hz; xi = hx; zi + hy; zi: For part 2 hx; λyi = hλy; xi = λhy; xi = λhy; xi = λhx; yi: For part 3 we have that hx; 0i = hx; 0 · xi = 0hx; xi = 0 = 0hx; xi = h0; xi: Finally hx; yi = hz; yi for all y 2 V implies that hx − z; yi = 0 for all y 2 V . Thus hx − z; x − zi = 0 and so x − z = 0 which means x = z. We can use the notion of an inner product to give a notion of size and distance on V . Definition 1.3. For v 2 V we define kvk = (hv; vi)1=2: Theorem 1.4 (Cauchy-Schwarz inequality). Let V be an inner product space over F. We have that for all x; y 2 V jhx; yij ≤ kxkkyk: Moreover, when x and y are linearly independent, the inequality is strict. Proof. We first suppose there exists λ 2 F such that x = λy. We then have that jhλy; yij = jλjhy; yi = (λλhy; yihy; yi)1=2 = (hλy; λyihy; yi)1=2 = kykkλyk: Now we suppose that x and y are linearly independent. So for all λ 2 F we have that hy + λx, y + λxi > 0. We can write 0 < hy + λx, y + λxi = kyk2 + hλx, yi + hy; λxi + jλj2kxk2 = kyk2 + 2Re(hλx, yi) + jλj2kxk2 = kyk2 + 2Re(λhx; yi) + jλj2kxk2 jhx;yij We now let λ = tu where t 2 R and u = hx;yi , note that juj = 1. Substi- tuting this in gives that 0 > kyk2 + 2tjhx; yij + t2kxk2: 3 Thus the quadratic kyk2 + 2tjhx; yij + t2kxk2 has no roots and so must have discriminant less than 0. Thus 4jhx; yij2 < 4kxk2kyk2 and the result follows. We now can deduce that k·k has the properties we would wish a notion of size to have. Theorem 1.5. Let V be an inner product space over F. For v 2 V let kvk = (hv; vi)1=2. We have that 1. For all v 2 V kvk > 0 unless v = 0. 2. For all v 2 V and λ 2 F we have kλvk = jλjkvk. 3. For all x; y; z 2 V we have that kx + yk ≤ kxk + kyk 4. If we define d(x; y) = kx − yk then this gives a metric on V . Proof. The first and fourth parts are obvious. For the second part write kλvk = (hλv; λvi)1=2 = (λλhv; vi)1=2 = λkvk: Finally for the third part we use Cauchy-Schwarz. We fix x; y 2 V and write kx + yk2 = hx + y; x + yi = hx; xi + hy; yi + 2Re(hx; yi) ≤ hx; xi + hy; yi + 2jhx; yij ≤ kxk2 + kyk2 + 2kxkkyk = (kxk + kyk)2: Example. Let V be the vector space of all continuous functions f : [0; 1] ! F. We can define Z 1 hf; gi = fgdx: 0 This gives an inner product space (see exercises). We have seen that an inner product gives a way of definining a size and a distance between elements. This means inner product spaces give examples of what are called normed spaces, however not all normed spaces come from inner products. 4 Normed spaces Definition 1.6. Let V be a vector space over F we say that k·k : V ! R is a norm if 1. kvk ≥ 0 for all v 2 V with equality if and only if v = 0. 2. For all v 2 V and λ 2 F we have that kλvk = jλjkvk. 3. For all x; y 2 V we have that kx + yk ≤ kxk + kyk: We then call V a normed space with respect to k·k. Example. Consider C([0; 1]) the space of real valued continuous functions (we could also take complex valued functions). We define kfk = sup fjf(x)jg: x2[0;1] We can easily see this gives a norm. Note that if we use the inner product Z 1 hf; gi = fgdx 0 then we get a different norm Z 1 2 1=2 kfk2 = ( jf(x)j ) : 0 Definition 1.7 (`p space). We now consider certain spaces of sequences on which there will be a natural definition of a norm. Let 1 ≤ p < 1 and define 1 p X p p ` = ffxngn2N : xn 2 F and jxnj < 1g ⊂ ` : n=1 For x 2 `p we define 1 !1=p X p kxkp = jxnj : n=1 Note that for p = 2 this agrees with the norm induced by the inner product. We can also define 1 ` = ffxngn2N : xn 2 F and sup jxnj < 1g n2N and jxk1 = supfkxnjg: n2N 5 Theorem 1.8. For all 1 ≤ p ≤ 1 we have that `p is a normed space. p Proof. It is an exercise to show that ` is a subspace of FN. It is clear that p for all 1 ≤ p ≤ 1 that for nonzero x 2 ` , kxk > 0 and that for all λ 2 F we have kλxk = jλjkxk. So we need to show the triangle inequality. For p = 1 and p = 1 it is easy to show. p So now let 1 < p < 1 and q = p−1 . We have Young's inequality (see exercise sheet) that for all A; B ≥ 0 that Ap Bq AB ≤ + : p q H¨older'sinequality We now show that for x; 2 `p and y 2 `q that 1 X jxkykj ≤ kxkpkykq: k=1 To do this fix x; y 2 `p and k 2 . We use A = jxkj and B = jykj in Young's N kxkp kyk inequality to get p q jxkjjykj 1 jxkj 1 jykj ≤ p + q : kxkpkykq p kxkp q kykq We can now sum over all k 2 N to get 1 1 X 1 1 jxkykj ≤ + kxkpkykq p q k=1 and finally 1 X jxkykj ≤ kxkpkykq: k=1 Minkowski's inequality We can now prove the triangle inequality for `p also known as Minkowski's inequality.
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