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.
Taking square roots, we have
kx + yk ≤ kxk + kyk.
() Introduction to Functional Analysis June 22, 2021 7 / 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.
Taking square roots, we have
kx + yk ≤ kxk + kyk.
() Introduction to Functional Analysis June 22, 2021 7 / 11 Theorem 4.5. Parallelogram Law Theorem 4.5. Parallelogram Law
Proposition 4.5. Parallelogram Law. 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 + kx − yk2 = 2(kxk2 + kyk2).
Proof. Lemma 4.2 gives kx + yk2 = kxk2 + kyk2 + 2Re(hx, yi) and, by replacing y with −y, gives
kx − yk2 = kxk2 + kyk2 + 2Re(hx, −yi) = kxk2 + kyk2 − 2Re(hx, yi).
Adding these two equations give the result.
() Introduction to Functional Analysis June 22, 2021 8 / 11 Theorem 4.5. Parallelogram Law Theorem 4.5. Parallelogram Law
Proposition 4.5. Parallelogram Law. 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 + kx − yk2 = 2(kxk2 + kyk2).
Proof. Lemma 4.2 gives kx + yk2 = kxk2 + kyk2 + 2Re(hx, yi) and, by replacing y with −y, gives
kx − yk2 = kxk2 + kyk2 + 2Re(hx, −yi) = kxk2 + kyk2 − 2Re(hx, yi).
Adding these two equations give the result.
() Introduction to Functional Analysis June 22, 2021 8 / 11 Proposition 4.7. Polarization Identity Proposition 4.7. Polarization Identity
Proposition 4.7. Polarization 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 1 hx, yi = kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2 . 4
Proof. As in the proof of the Parallelogram Law kx + yk2 = kxk2 + kyk2 + 2Re(hx, yi) and kx − yk2 = kxk2 + kyk2 − 2Re(hx, yi). Subtracting these give
4Re(hx, yi) = kx + yk2 − kx − yk2. (∗)
As with any complex number, hx, yi = Re(hx, yi) + iIm(hx, yi) and so
Im(hx, yi) = Re(−ihx, yi) = Re(hx, iyi). (∗∗)
() Introduction to Functional Analysis June 22, 2021 9 / 11 Proposition 4.7. Polarization Identity Proposition 4.7. Polarization Identity
Proposition 4.7. Polarization 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 1 hx, yi = kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2 . 4
Proof. As in the proof of the Parallelogram Law kx + yk2 = kxk2 + kyk2 + 2Re(hx, yi) and kx − yk2 = kxk2 + kyk2 − 2Re(hx, yi). Subtracting these give
4Re(hx, yi) = kx + yk2 − kx − yk2. (∗)
As with any complex number, hx, yi = Re(hx, yi) + iIm(hx, yi) and so
Im(hx, yi) = Re(−ihx, yi) = Re(hx, iyi). (∗∗)
() Introduction to Functional Analysis June 22, 2021 9 / 11 Proposition 4.7. Polarization Identity Proposition 4.7 (continued)
Proposition 4.7. Polarization 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 1 hx, yi = kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2 . 4
Proof (continued). Using (∗) for Re(hx, yi) and using (∗) with y replaced by iy gives 1 Re(hx, yi) = (kx + yk2 − kx − yk2) 4 and 1 Re(hx, iyi) = (kx + iyk2 − kx − iyk2), 4 and this substituted into (∗∗) gives the result.
() Introduction to Functional Analysis June 22, 2021 10 / 11 Theorem 4.9. Continuity of Inner Product Theorem 4.9. Continuity of Inner Product
Theorem 4.9. Continuity of Inner Product. In any semi-inner product space, if the sequences (xn) → x and (yn) → y, then (hxn, yni) → hx, yi.
Proof. We have
|hxn, yni − hx, yi| = |hxn, yni − hxn, yi + hxn, yi − hx, yi|
≤ |hxn, yni − hxn, yi| + |hxn, yi − hx, yi|
≤ |hxn, yn − yi| + |hxn − x, yi|
≤ kxnkkyn − yk + kxn − xkkyk by Cauchy-Schwartz.
Since (xn) is convergent, then it is bounded, since (yn) → y then kyn − yk → 0, since (xn) → x then kxn − xk → 0. Therefore
|hxn, yni − hx, yi| → 0 and hxn, yni → hx, yi.
() Introduction to Functional Analysis June 22, 2021 11 / 11 Theorem 4.9. Continuity of Inner Product Theorem 4.9. Continuity of Inner Product
Theorem 4.9. Continuity of Inner Product. In any semi-inner product space, if the sequences (xn) → x and (yn) → y, then (hxn, yni) → hx, yi.
Proof. We have
|hxn, yni − hx, yi| = |hxn, yni − hxn, yi + hxn, yi − hx, yi|
≤ |hxn, yni − hxn, yi| + |hxn, yi − hx, yi|
≤ |hxn, yn − yi| + |hxn − x, yi|
≤ kxnkkyn − yk + kxn − xkkyk by Cauchy-Schwartz.
Since (xn) is convergent, then it is bounded, since (yn) → y then kyn − yk → 0, since (xn) → x then kxn − xk → 0. Therefore
|hxn, yni − hx, yi| → 0 and hxn, yni → hx, yi.
() Introduction to Functional Analysis June 22, 2021 11 / 11