Hilbert spaces, 1:
I’ve been reading Pedersen’s Analysis Now for a model-theoretic functional analysis reading course this term. We’re starting the section on Hilbert spaces (and the theory of operators on Hilbert spaces) and it’s my turn to present, so I’ve written some notes that I’ll also post here.
Definition. A sesquilinear form on a vector space \(X\) is a map \[( \cdot | \cdot) : X \times X \to \mathbb{F}\] which is linear in the first variable and conjugate linear in the second. To be precise, it’s additive in either variable, but compatible/\(\mathbb{F}\)-equivariant up to conjugation in the second variable.
Definition. To each sesquilinear form \((\cdot | \cdot)\) we define the adjoint form \(( \cdot | \cdot )^*\) by \[(x | y)^* \overset{\operatorname{df}}{=} \overline{(y | x)},\] for all \(x\) and \(y\) in \(X\). We say that the form is self-adjoint if it is its own adjoint. (When \(\mathbb{F} = \mathbb{R},\) the term symmetric is used.)
Claim. When \(\mathbb{F}\) is \(\mathbb{C},\) a calculation (or simply observing that \(S^1\) acts via isometries) shows that \[4(x | y) = \sum_{k = 0}^3 i^k(x + i^k y | x + i^k y).\]
Proof. Expanding the sum gives \[(x | x) + (y | x) + (x | y) + (y | y)\]\[+ i(x | x) + i(x|iy) + i(iy|x) + i(iy|iy)\]\[- (x|x) - (x|-y) - (-y|x) - (-y | -y)\]\[ -i (x | x) - i(x | -iy) - i(-iy | x) - i(-iy|-iy)\] \[= (x|x)(1 + i - 1 - i) + (y | x)(1 - 1 + 1 - 1)\]\[ + (x|y)(1 + 1 + 1 + 1) + (y|y)(1 -i^3 - 1 + i^3)\]\[= 4(x|y).\] \(\square\)
We say a sesquilinear form \(( \cdot | \cdot)\) is positive if \((x|x) \geq 0\) for every \(x \in X\); thus, for \(\mathbb{F} = \mathbb{C}\) a positive form is automatically self-adjoint, since the above claim shows (after a nearly-identical calculation) that a sesquilinear form is self-adjoint if and only if \((x | x) \in \mathbb{R}\) for every \(x \in X\).
On a real space, a positive form might not be symmetric. For example, take a nontrivial quadratic form.
Definition. An inner product on \(X\) is a positive, self-adjoint sesquilinear form such that \((x | x) = 0\) implies \(x = 0\) for all \(x \in X\), i.e. the form is positive-definite in addition to being positive.
Definition. From similar computations from the claim above, we obtain the following polarization identities: in \(\mathbb{C},\) \[4(x|y) = \sum_{k=0}^3 i^k || x + i^k y||^2,\] and in \(\mathbb{R}\), \[4(x|y) = || x + y||^2 - ||x - y||^2.\]
Lemma. (Cauchy-Schwarz inequality) Let \(\alpha\) be a scalar in \(\mathbb{F}\). The formula \[|\alpha|^2 ||x||^2 + 2 \Re \alpha(x | y) + ||y||^2 = ||\alpha x + y||^2 \geq 0\] for \(x\) and \(y\) in \(X\) immediately leads to the Cauchy-Schwarz inequality: \[|(x|y)| \leq ||x|| ||y||.\]
Inserting this into the above formula, it follows that the norm function \(|| \cdot ||\) is subadditive, and therefore a seminorm on \(X\). In the case when \(( \cdot | \cdot)\) is an inner product, the homogeneous function \(||x|| \overset{\operatorname{df}}{=} (x | x)^{½}\) is a norm, and equality holds in the Cauchy-Schwarz inequality if and only if \(x\) and \(y\) are scalar multiples of each other.
Lemma. Elementary computations show that the norm arising from an inner product space satisfies the parallelogram law, namely \[||x+y||^2 + ||x - y||^2 = 2(||x||^2 + ||y||^2).\]
Conversely, one way verify that if a norm satisfies the parallelogram law, then the polarization identities from before induce inner products on \(X\) (which one depending, of course, on the ground field.)
Definition. A Hilbert space is a vector space \(\mathfrak{H}\) with an inner product \(( \cdot | \cdot )\), such that \(\mathfrak{H}\) is a Banach space with respect to the associated norm.
Examples. - The Euclidean spaces are Hilbert spaces.
- The spaces of compactly-supported continuous functions on Euclidean spaces are a Hilbert space with inner product given by \((f | g ) \overset{\operatorname{df}}{=} \int f \overline{g} dx;\) the associated norm is the \(2\)-norm, and after completing we get the Hilbert space \(L^2 \mathbb{R}\).
(More generally, you could replace the Lebesgue integral above with a Radon integral on a locally compact Hausdorff space and the same construction gives you a Hilbert space \(L^2 X.\))
Algebraic direct sums iare formed as in \(\mathbf{Ban}\), with the new inner product being defined pointwise.
Lemma. If \(\mathbf{C}\) is a closed nonempty convex subset of a Hilbert space \(\mathfrak{H}\), there is for each \(y\) in \(\mathfrak{H}\) a unique \(x\) in \(\mathfrak{C}\) that minimizes the distance from \(y\) to \(\mathfrak{C}\).
Proof. By translation variance, we can assume \(y = 0\). Let \(\alpha\) be the infimum of the norms of elements in \(\mathfrak{C}\). Choose a sequence \((x_n)\) from \(\mathfrak{C}\) witnessing this.
For any \(y\) and \(z\) in \(\mathfrak{C}\), the parallelogram law (and convexity, whence \(½(y+z) \in \mathfrak{C}\)) gives \[2 \left( ||y||^2 + ||z||^2\right) = ||y+z||^2 \geq 4 \alpha^2 + ||y - z||^2,\] Hence, when \(y = x_n\) and \(z = x_m\), \((x_n)\) is a Cauchy sequence, so by completeness of a Hilbertspace there is a limit \(x\) with \(||x|| = \alpha\).
Furthermore, since the topology is Hausdorff, the limit is unique. \(\square\)
Theorem. For a closed subspace \(X\) of a Hilbert space \(\mathfrak{H}\), let \(X^{\perp}\) be the subspace of all elements orthogonal to all of \(X\). Then each vector \(y \in \mathfrak{H}\) has a unique decomposition \(y = x + x^{\perp}\), and \(x\) and \(x^{\perp}\) are the nearest points in \(X\) and \(X^{\perp}\) to \(y\); furthermore \(\mathfrak{H} = X \oplus X^{\perp}\) and \(\left(X^{\perp}\right)^{\perp} = X.\)
Proof. More or less automatic verification.
Given \(y\) in \(\mathfrak{H}\) we take \(x\) to be the nearest point in \(X\) to \(y\) gotten by the previous lemma; put \(x^{\perp}\) to be \(y - x\). For every \(z \in X\) and \(\epsilon > 0\),
\[||x^{\perp}||^2 = ||y - x||^2 \leq ||y - (x + \epsilon z)||^2\]
\[= ||x^{\perp} - \epsilon z||^2 = ||x^{\perp}||^2 - 2 \epsilon \Re(x^{\perp}|z) + \epsilon^2||z||^2.\]
It follows that \(2 \Re(x^{\perp}|z) \leq \epsilon ||z||^2\) for every \(\epsilon > 0\), whence \(\Re(x^{\perp}|z) \leq 0\) for every \(z\). Since \(X\) is a linear subspace, this implies that \((x^{\perp}|z)\) is zero, so that \(x^{\perp}\) deserves its name and is in \(X^{\perp}\). \(\square\)
Corollary. For every subset \(X \subset \mathfrak{H}\), the smallest closed subspace of \(\mathfrak{H}\) containing \(X\) is \(X\)-double perp. In particular, if \(X\) is a subspace of \(\mathfrak{H}\), then \(\overline{X}\) is \(X\)-double perp.
Proposition. The maps \(\Phi \overset{\operatorname{df}}{=} ( \cdot | x )\) are conjugate linear isometries of \(\mathfrak{H}\) onto \(\mathfrak{H}^*\).
Proof. Tautologically conjugate-linear; \(\Phi\) is maximized by CZ-inequality on \(x/||x||\) on the unit ball, so the norm of \(\Phi\) is precisely \(||x||\). \(\square\)
Remark. Take \(\varphi\) in \(\mathfrak{H}^* \backslash \{0\}\), and put \(X = \ker \varphi\). Since \(X\) is a proper closed subspace of \(\mathfrak{H}\), there is a vector \(x \in X^{\perp}\) such that \(\varphi(x) = 1.\) For every \(y \in \mathfrak{H}\), we see that \(y - \varphi(y) x \in X\), hence \[(y | x) = (y - \varphi(y) x + \varphi(y) x | x) = \varphi(y) ||x||^2.\] It follows that \(\varphi = \Phi(||x||^{-2} x).\)
Definition. We define the weak topology on a Hilbert space \(\mathfrak{H}\) as the initial topology corresponding to the family of functionals \(x \mapsto (x | y),\) so essentially the weak topology it induces…on itself. Since it is isometric with its dual, the weak topology on \(\mathfrak{H}\) is the weak-star topology on \(\mathfrak{H}^*\) pulled back via the isometry, and the unit ball in \(\mathfrak{H}\) is weakly compact.
We’ll see that every operator in \(\mathbf{B}(\mathfrak{H})\) is continuous as a function \(\mathfrak{H} \to \mathfrak{H}\) when both copies have the weak topology; we say that \(T\) is weak-weak continuous.
Conversely, if \(T\) is a weak-weak cts operator, we can invoke the closed graph theorem to see that \(T \in \mathbf{B}(\mathfrak{H})\). Indeed, if \(x_n \to x\) and \(T x_n \to y\) in the norm topology, then \(x_n \to x\) and \(T x_n \to y\) weakly since the norm topology refines the weak topology; by assumption \(T x_n \to Tx\) weakly, whence \(Tx = y,\) as was required.
Similarly, every norm-weak continuous operator on \(\mathfrak{H}\) is bounded. We’ll see later operator that is weak-norm continuous must have finite rank, which is neat; we’ll later also modify this to give a characterization of the compact operators (which are a good candidate for treating infinitesimal elements.)
