Introduction to Functional Analysis Chapter 4. Hilbert Spaces 4.2. Semi-Inner Products—Proofs of Theorems June 22, 2021 () Introduction to Functional Analysis June 22, 2021 1 / 11 Table of contents 1 Lemma 4.2. Basic Identity 2 Theorem 4.3. Cauchy-Schwartz Inequality 3 Theorem 4.4. Triangle Inequality 4 Theorem 4.5. Parallelogram Law 5 Proposition 4.7. Polarization Identity 6 Theorem 4.9. Continuity of Inner Product () Introduction to Functional Analysis June 22, 2021 2 / 11 Lemma 4.2. Basic Identity Lemma 4.2. Basic Identity Lemma 4.2. Basic Identity. Let X be a semi-inner product space. Then the semi-norm induced by the semi-inner product satisfies: for all x, y ∈ X , we have kx + yk2 = kxk2 + kyk2 + 2Rehx, yi. Proof. We have kx + yk2 = hx + y, x + yi = hx, x + yi + hy, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + hx, yi + hx, yi + kyk2 = kxk2 + 2Re(hx, yi) + kyk2. () Introduction to Functional Analysis June 22, 2021 3 / 11 Lemma 4.2. Basic Identity Lemma 4.2. Basic Identity Lemma 4.2. Basic Identity. Let X be a semi-inner product space. Then the semi-norm induced by the semi-inner product satisfies: for all x, y ∈ X , we have kx + yk2 = kxk2 + kyk2 + 2Rehx, yi. Proof. We have kx + yk2 = hx + y, x + yi = hx, x + yi + hy, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + hx, yi + hx, yi + kyk2 = kxk2 + 2Re(hx, yi) + kyk2. () Introduction to Functional Analysis June 22, 2021 3 / 11 Suppose kyk 6= 0. If hx, yi = 0, the result trivially follows. If hx, yi= 6 0 then x and y can be scaled so that kyk = 1 and hx, yi = 1 (the second claim possibly requiring the use of a complex factor). Then by Lemma 4.2, with y replaced by −y, we have 0 ≤ kx−yk2 = kxk2+k−yk2+2Re(hx, −yi) = kxk2+1+2(−1) = kxk2−1. So kxk ≥ 1 and the result holds in the case that kyk= 6 0 (or symmetrically, in the case that kxk= 6 0). Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof. Notice that if we scale x or y by a nonzero amount, the Cauchy-Schwartz Inequality remains unchanged (the scaling factor is introduced on both sides and can be divided out). () Introduction to Functional Analysis June 22, 2021 4 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof. Notice that if we scale x or y by a nonzero amount, the Cauchy-Schwartz Inequality remains unchanged (the scaling factor is introduced on both sides and can be divided out). Suppose kyk= 6 0. If hx, yi = 0, the result trivially follows. If hx, yi 6= 0 then x and y can be scaled so that kyk = 1 and hx, yi = 1 (the second claim possibly requiring the use of a complex factor). Then by Lemma 4.2, with y replaced by −y, we have 0 ≤ kx−yk2 = kxk2+k−yk2+2Re(hx, −yi) = kxk2+1+2(−1) = kxk2−1. So kxk ≥ 1 and the result holds in the case that kyk 6= 0 (or symmetrically, in the case that kxk 6= 0). () Introduction to Functional Analysis June 22, 2021 4 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof. Notice that if we scale x or y by a nonzero amount, the Cauchy-Schwartz Inequality remains unchanged (the scaling factor is introduced on both sides and can be divided out). Suppose kyk= 6 0. If hx, yi = 0, the result trivially follows. If hx, yi 6= 0 then x and y can be scaled so that kyk = 1 and hx, yi = 1 (the second claim possibly requiring the use of a complex factor). Then by Lemma 4.2, with y replaced by −y, we have 0 ≤ kx−yk2 = kxk2+k−yk2+2Re(hx, −yi) = kxk2+1+2(−1) = kxk2−1. So kxk ≥ 1 and the result holds in the case that kyk 6= 0 (or symmetrically, in the case that kxk 6= 0). () Introduction to Functional Analysis June 22, 2021 4 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3 (continued 1) Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof (continued). If kxk = kyk = 0 but hx, yi= 6 0, then we can scale x and y such that hx, yi = 1. Then by Lemma 4.2, 0 ≤ kx − yk2 = kxk2 + k − yk2 + 2Re(hx, −yi) = 0 + 0 − 2, a contradiction (so it must be that hx, yi = 0) and the inequality follows. If y = αx then |hx, yi| = |hx, αxi| = |αhx, xi| = |α|kxk2 = kxk|α|kxk = kxkkyk. () Introduction to Functional Analysis June 22, 2021 5 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3 (continued 1) Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof (continued). If kxk = kyk = 0 but hx, yi= 6 0, then we can scale x and y such that hx, yi = 1. Then by Lemma 4.2, 0 ≤ kx − yk2 = kxk2 + k − yk2 + 2Re(hx, −yi) = 0 + 0 − 2, a contradiction (so it must be that hx, yi = 0) and the inequality follows. If y = αx then |hx, yi| = |hx, αxi| = |αhx, xi| = |α|kxk2 = kxk|α|kxk = kxkkyk. () Introduction to Functional Analysis June 22, 2021 5 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3 (continued 2) Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof (continued). Conversely, suppose that equality holds. Notice that h0, xi = hx − x, xi = hx, xi − hx, xi = 0, so if either kxk = 0 or kyk = 0, then x and y are linearly dependent since x = 0 or y = 0 (we are assuming h·, ·i is an inner product and k · k is a norm in the exploration of equality). If neither x nor y is 0, then again we can scale y so that kyk = 1 and then scale x so that hx, yi = 1. Then equality in the Cauchy-Schwartz Inequality implies that kxk = 1. Then, as above, with −y replacing y in Lemma 4.2, 0 ≤ kx − yk2 = kxk2 + kyk2 + 2Re(hx, −yi) = 1 + 1 + 2(−1) = 0, and so kx − yk = 0 and x = y () Introduction to Functional Analysis June 22, 2021 6 / 11 Theorem 4.3. Cauchy-Schwartz Inequality Theorem 4.3 (continued 2) Theorem 4.3. Cauchy-Schwartz Inequality. Given a semi-inner product on X , for all x, y ∈ X we have |hx, yi| ≤ kxkkyk. If h·, ·i is an inner product, then equality holds if and only if x is a scalar multiple of y (that is, x and y are linearly dependent). Proof (continued). Conversely, suppose that equality holds. Notice that h0, xi = hx − x, xi = hx, xi − hx, xi = 0, so if either kxk = 0 or kyk = 0, then x and y are linearly dependent since x = 0 or y = 0 (we are assuming h·, ·i is an inner product and k · k is a norm in the exploration of equality). If neither x nor y is 0, then again we can scale y so that kyk = 1 and then scale x so that hx, yi = 1. Then equality in the Cauchy-Schwartz Inequality implies that kxk = 1. Then, as above, with −y replacing y in Lemma 4.2, 0 ≤ kx − yk2 = kxk2 + kyk2 + 2Re(hx, −yi) = 1 + 1 + 2(−1) = 0, and so kx − yk = 0 and x = y () Introduction to Functional Analysis June 22, 2021 6 / 11 Theorem 4.4. Triangle Inequality Theorem 4.4. Triangle Inequality Theorem 4.4. Triangle Inequality. Let X be a semi-inner product space. Then the semi-norm induced by the semi-inner product satisfies: for all x, y ∈ X , we have kx + yk ≤ kxk + kyk. Proof. We simply have kx + yk2 = kxk2 + kyk2 + 2Re(hx, yi) by Lemma 4.2 ≤ kxk2 + kyk2 + 2|hx, yi| ≤ kxk2 + kyk2 + 2kxkkyk by Cauchy-Schwartz = (kxk + kyk)2.