Operators on Hilbert Spaces - user web page - AIMS

Section 3.3. Sesquilinear functionals and Riesz representation Page 13
2. h(x, y1 + y2) = h(x, y1) + h(x, y2),
3. h(αx, y) = αh(x, y),
4. h(x, βy) = ¯ βh(x, y).
Remark 3.3.2 Notice that h is linear in the first argument and conjugate linear in the second
one. If X and Y are real then (4) becomes h(x, βy) = βh(x, y). Clearly, h is bilinear in both
arguments.
Definition 3.3.3 ([Swa97], Definition 8 (Bounded sesquilinear functional)) A sesquilinear functional
h is bounded if there exists K > 0 such that
The norm of of h is defined to be
|h(x, y)| ≤ Kxy, for every x ∈ X, y ∈ Y. (3.11)
|h(x, y)|
h = sup
x=0 xy
y=0
= sup
x=1
y=1
|h(x, y)|. (3.12)
It follows that |h(x, y)| ≤ hxy for all x ∈ X, y ∈ Y. The inner product is sesquilinear
and it is bounded.
Lemma 3.3.4 ([Kre78], 3·8-2 Lemma (Equality))
If 〈x1, y〉 = 〈x2, y〉 for all y in an inner product space X, then x1 = x2. In particular if 〈x1, w〉 = 0
for all w ∈ X, then x1 = 0.
Proof.
For all y,
〈x1 − x2, y〉 = 〈x1, y〉 − 〈x2, y〉 = 0.
If we choose y = x1 − x2, then x1 − x2 2 = 0. Hence we have x1 − x2 = 0 implying that
x1 = x2. In particular, if 〈x1, y〉 = 0 with y = x1 we get x1 2 = 0, giving x1 = 0.
Theorem 3.3.5 ([Kre78], 3·8-4 Theorem (Riesz representation))
Let H1, H2 be Hilbert spaces and let h : H1 × H2 → K be a bounded sesquilinear form. Then h
has a representation
h(x, y) = 〈Sx, y〉 for x ∈ H1, y ∈ H2
(3.13)
where S : H1 → H2 is a bounded linear operator and S is uniquely determined and has norm
S = h.