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Representations of functionals

Lemma 1: let \((X, \langle \cdot, \cdot \rangle)\) be an inner product space, if

\[ \forall z \in X: \langle x, z \rangle = \langle y, z \rangle \implies x = y, \]

and if

\[ \forall z \in X: \langle x, z \rangle = 0 \implies x = 0. \]
Proof:

Will be added later.

Lemma 1 will be used in the following theorem.

Theorem 1: for every bounded linear functional \(f\) on a Hilbert space \((X, \langle \cdot, \cdot \rangle)\), there exists a \(z \in X\) such that

\[ f(x) = \langle x, z \rangle, \]

for all \(x \in x\), with \(z\) uniquely dependent on \(f\) and \(\|z\| = \|f\|\).

Proof:

Will be added later.

Sequilinear form

Definition 1: let \(X\) and \(Y\) be vector spaces over the field \(F\). A sesquilinear form \(h\) on \(X \times Y\) is an operator \(h: X \times Y \to F\) satisfying the following conditions

  1. \(\forall x_{1,2} \in X, y \in Y: h(x_1 + x_2, y) = h(x_1, y) + h(x_2, y)\).
  2. \(\forall x \in X, y_{1,2} \in Y: h(x, y_1 + y_2) = h(x_1, y_1) + h(x_2, y_2)\).
  3. \(\forall x \in X, y \in Y, \alpha \in F: h(\alpha x, y) = \alpha h(x,y)\).
  4. \(\forall x \in X, y \in Y, \beta \in F: h(x, \beta y) = \overline \beta h(x,y)\).

Hence, \(h\) is linear in the first argument and conjugate linear in the second argument. Bilinearity of \(h\) is only true for a real field \(F\).

Definition 2: let \(X\) and \(Y\) be normed spaces over the field \(F\) and let \(h: X \times Y \to F\) be a sesquilinear form, then \(h\) is a bounded sesquilinear form if

\[ \exists c \in F: |h(x,y)| \leq c \|x\| \|y\|, \]

for all \((x,y) \in X \times Y\) and the norm of \(h\) is given by

\[ \|h\| = \sup_{\substack{x \in X \backslash \{0\} \\ y \in Y \backslash \{0\}}} \frac{|h(x,y)|}{\|x\| \|y\|} = \sup_{\|x\|=\|y\|=1} |h(x,y)|. \]

For example, the inner product is sesquilinear and bounded.

Theorem 2: let \((X, \langle \cdot, \cdot \rangle_X)\) and \((Y, \langle \cdot, \cdot \rangle_Y)\) be Hilbert spaces over the field \(F\) and let \(h: X \times Y \to F\) be a bounded sesquilinear form. Then there exists a bounded linear operators \(T: X \to Y\) and \(S: Y \to X\), such that

\[ h(x,y) = \langle Tx, y \rangle_Y = \langle x, Sy \rangle_X, \]

for all \((x,y) \in X \times Y\), with \(T\) and \(S\) uniquely determined by \(h\) with norms \(\|T\| = \|S\| = \|h\|\).

Proof:

Will be added later.