AFA Hilbert spaces

Section 3.2 Hilbert spaces

Definition 3.2.1. Hilbert space.

A Hilbert space is a vector space which a scalar product that is complete with respect to the induced metric space structure.

Subsection 3.2.1 Example: scalar products on \(C^\infty([0,1])\)

The \(L^2\) scalar product. Set

\begin{equation*} \langle f,g\rangle_{L^2} = \int_0^1 f(s)g(s)\;ds \end{equation*}

if \(\bK=\bR\) and

\begin{equation*} \langle f,g\rangle_{L^2} = \int_0^1 f(s)\overline{g(s)}\;ds \end{equation*}

if \(\bK=\bC\text{.}\)

One can easily verify that \(\langle \cdot,\cdot\rangle_{L^2}\) is a scalar product on \(C^\infty([0,1])\text{.}\) The corresponding norm is the \(L^2\) norm and so the completion of this space is \(L^2([0,1])\text{.}\) Hence, \(L^2([0,1])\) is a Hilbert space.

The \(H^k\) scalar product. Set

\begin{equation*} \langle f,g\rangle_{H^k} = \langle f,g\rangle_{L^2} + \langle f',g'\rangle_{L^2} + \dots + \langle f^{(k)},g^{(k)}\rangle_{L^2}. \end{equation*}

The corresponding norm is the \(H^k\) norm and so the completion of this space is \(H^k([0,1])\) (see Subsection 3.1.2). Hence, \(H^k([0,1])\) is a Hilbert space.

Subsection 3.2.2 Some fundamental results on Hilbert spaces

The existence of a scalar product has several fundamental consequences.

Definition 3.2.2. Orthogonality.

If \(u,v\in H\) are such that \(\langle u,v \rangle = 0\text{,}\) we say that \(u\) and \(v\) are orthogonal and we write sometimes \(u\perp v\text{.}\) We say that two sets \(A,B\subset H\) are othogonal when all vectors of \(A\) are orthogonal to all vectors of \(B\text{.}\) The orthogonal complement of \(M\) in \(H\) is the set

\begin{equation*} M^\perp :=\{u\in H: \langle u,v\rangle =0, \quad\mbox{for all}\quad v\in M\}. \end{equation*}

Proposition 3.2.3. Orthogonal complement.

Let \(M\subset H\text{.}\) Its orthogonal complement \(M\perp\) is a closed linear subspace of \(H\text{.}\)

Proof.

We prove the theorem in \(\bR\) and leave to the reader the complex case. Let \(u,v\in M^\perp\text{.}\) Then, by the linearity of the scalar product, \(\lambda u + \mu v\in M^\perp\) for every \(u,v\in M^\perp\) and \(\lambda,\mu\in\bR\text{.}\) Hence, \(M^\perp\) is a linear subspace of \(H\text{.}\) Since the scalar product is continuous and \(M^\perp\) is defined by a close condition, \(M^\perp\) is closed.

Figure 3.2.4. Sketch of the shortest distance from \(x\) to a point \(u\in M\text{.}\)

Proposition 3.2.5. Proximality.

Let \(M\) be a closed linear subspace of \(H\text{.}\) Then:

\(M\) is proximal, i.e., for each \(v\in H\) there is a unique closest vector \(u\in M\) such that
\begin{equation*} \|v-u\| = \min_{w\in M}\|v-w\|. \end{equation*}
The \(u\) above is the unique element of \(M\) such \((v-u)\perp M\text{.}\)

Proof.

Let \(u_n\) be a sequence in \(M\) with \(\|v-u_n\|\to\inf_{w\in M}\|v-w\|\text{.}\) This sequence is Cauchy: indeed, by the parallelogram identity,

\begin{align*} \|u_n-u_m\|^2 & = 2\|v-u_n\|^2+2\|v-u_m\|^2-4\|v-\frac{u_n+u_m}{2}\|^2\leq\\ & \leq 2(\|v-u_n\|^2-d^2) + 2(\|v-u_m\|^2-d^2) \end{align*}

and, by construction, by taking \(m\) and \(n\) large enough, we can make the two summands above as small as we please.

Since \(M\) is closed and \(H\) is complete, there is a \(u\in M\) with \(u_n\to u\text{.}\) Hence, by continuity, \(\|v-u\| = \lim \|v-u_n\| = \inf_{w\in M}\|v-w\|\text{.}\)

Now we want an explicit parametrization of this minimum. For that we take a \(y\in M\) and consider

\begin{equation*} D(t) := \|v-(u-ty)\|^2 = \|v-u+ty\|^2 = \|v-u\|^2 + 2t(v-u,y) + t^2 \|y\|^2. \end{equation*}

We know that \(D(t)\) is minimal for \(t=0\text{,}\) hence

\begin{equation*} D′(t) = 2\langle v-u,y\rangle =0, \qquad \mbox{for all }\quad\quad y\in M. \end{equation*}

This means \(v-u\in M^\perp\text{.}\)

To show uniqueness of \(u\text{,}\) we assume that there in another minimizer \(u'\text{.}\) Then

\begin{equation*} 0 = \langle v-u,u-u'\rangle = \langle v-u',u-u'\rangle. \end{equation*}

By substracting the two, we get that \(\|u-u'\|=0\text{,}\) i.e. \(u=u'\text{.}\)

Proposition 3.2.6. Orthogonal decomposition.

Let \(M\subset H\) be a closed linear subspace. Then \(M\oplus M^\perp\text{.}\) In particular, every closed linear subset of a Hilbert space is complemented.

Proof.

Let \(v\in H\text{.}\) By Proposition 3.2.5, there is a unique \(u\in M\) such that \(v-u\in M^\perp\text{.}\) Let \(w=v-u\text{.}\) Then \(v=u+w\) with \(u\in M\) and \(w\in M^\perp\text{.}\) Let \(v=u'+w'\) with \(u'\in M\) and \(w'\in M^\perp\text{.}\) Then \(u-u'=v-v'\in M\cap M^\perp\text{.}\) Hence \(u=u'\) and \(v=v'\text{.}\)

The above proposition allows us to define orthogonal projections of \(x\) onto \(M\) as

\begin{equation*} P_M :H\to M, \quad P_M x = P_M(u+v) = u, \end{equation*}

where \(x=u+v\) is the unique decomposition of Proposition \ref{p:decompose}. The projection has the property

\begin{equation*} P_M^2=P_M, \qquad \mbox{and} \qquad \|P_M\|_{op} \leq 1, \end{equation*}

where we will introduce the operator norm in the next chapter.

We can take this orthogonal decomposition to the extreme, by decomposing with respect to one-dimensional subspaces. Doing this we quite naturally arrive at a Schauder basis.

Proposition 3.2.7. Schauder basis.

Let \(H\) be a Hilbert space. Then the following conditions are equivalent:

\(H\) has a Schauder basis;
\(H\) has a orthonormal Schauder basis;
\(H\) is separable.

Proof.

\(1\Longrightarrow2.\) This is just due to the Gram-Schmidt orthonormalization algorithm.

\(2\Longrightarrow3.\) Let \(e_i\) be a orthonormal basis of \(H\text{.}\) By hypothesis, \(v=\sum_{n=0}^\infty v_n e_n\text{,}\) with \(v_n=\langle v,e_n\rangle\text{,}\) for each \(v\in H\text{.}\) Since \(\bQ\) is dense in \(\bR\text{,}\) we can approximate \(v\) arbitrarily closely by \(v=\sum_{i=0}^\infty q_n e_n\text{,}\) where \(q_n\in\bQ\text{.}\) Since the set of all series with rational coefficients is countable, it follows that \(H\) is separable.

\(3\Longrightarrow1.\) Let \(s_n, n\geq1\text{,}\) be a countable dense set in \(H\text{.}\) Denote by \(e_n, n\geq1\text{,}\) an orthornormal basis extracted from the \(s_n\) via the Gram-Schmidt orthonormalization and set \(M=\overline{\span_{n\geq1}\{e_n\}\text{.}\) We claim that \(M^\perp=\{0\}\text{.}\) Indeed, if \(v\in M^\perp, v\neq0\text{,}\) by density there must be a \(s_k\) such that \(\|v-s_k\|\neq0\) -- if \(v\) were orthogonal to all \(s_k\) then, by continuity, it would be orthogonal to the whole \(H\) and therefore would be 0, against the assumption. At some step if the Gram-Schmidt algorithm, out of \(s_k\) was defined \(e_{k}\) as a finite linear combination of \(e_k\) with the \(e_i\) with \(i<k\text{,}\) so that \(\langle v,e_k\rangle=\langle v,s_k\rangle\neq0\text{,}\) which is incompatible with \(v\in M^\perp\text{.}\)

We claim that \(\lim_{k\to\infty}\sum_{n=0}^k \langle v,e_n\rangle e_n=v\text{.}\) Indeed,

\begin{equation*} \|v-\sum_{n=0}^k \langle v,e_n\rangle e_n\|^2 = \langle v-\sum_{n=0}^k \langle v,e_n\rangle e_n, v-\sum_{n=0}^k \langle v,e_n\rangle e_n\rangle \end{equation*}

Hence, \(\{e_n\}\) is a Schauder basis for \(H\text{.}\)

Proposition 3.2.8. Unit ball.

The unit sphere in an infinite dimensional Hilbert space is not compact.

Proof.

Let \(\{e_i\}\) be an orthonormal basis of the Hilbert space. Then each \(e_i\in \overline B_1(0)\) for all \(i\in \NN\text{.}\) But

\begin{equation*} \|e_n - e_m\|^2 = (e_n-e_m, e_n-e_m) = \|e_n\|^2 + \|e_m\|^2 = 2 \end{equation*}

for all \(n \neq m\text{.}\) Hence, \(\{e_j\}\) is a bounded sequence, but not a Cauchy sequence, and it does not converge in \(\overline{B_1(0)}\text{.}\)