5.4 Hilbert Spaces
Definition 5.19. Two Hilbert spaces \(H_1\) and \(H_2\) are isomorphic if there is a linear transformation \(T:H_1\rightarrow H_2\) such that for all \(x,y\in H_1\) , \((Tx,Ty)_{H_2}=(x,y)H_1\)
Such a transformation is called unitary.
We now elaborate these ideas and give some examples of Hilbert spaces.
Example 5.20. It was shown in example 5.2 that \(\mathbb {C}^n\) with \(\displaystyle {(x,y)=\sum ^n_{i=1}x_i\overline {y_i}}\hspace {0.2cm}\) is an inner product space. Show that this in fact a Hilbert space.
Solution. Only completeness is left to check. Let \(\{x^{(k)}\}\) be Cauchy in \(\mathbb {C}^n\), where \(x^{(k)}=\big (x^{(k)}_1,\dots ,x^{(k)}_n\big )\). For each fixed coordinate \(i\), \[\big |x^{(k)}_i-x^{(l)}_i\big |\leq \left (\sum _{j=1}^{n}\big |x^{(k)}_j-x^{(l)}_j\big |^2\right )^{\frac 12} =||x^{(k)}-x^{(l)}|| ,\] so \(\{x^{(k)}_i\}_k\) is Cauchy in \(\mathbb {C}\), which is complete; say \(x^{(k)}_i\rightarrow x_i\). Put \(x=(x_1,\dots ,x_n)\). Then \[||x^{(k)}-x||^2=\sum _{i=1}^{n}\big |x^{(k)}_i-x_i\big |^2\rightarrow 0\] because it is a sum of finitely many terms each tending to \(0\). Hence \(x^{(k)}\rightarrow x\) in \(\mathbb {C}^n\) and the space is complete.
Finiteness of the sum is what makes this easy; the same argument for \(l^2\) needs more care, as the next example shows.
Example 5.21. It was shown in example 5.3 that \(l^2\) with \(\displaystyle {(x,y)=\sum ^{\infty }_{n=1}x_n\overline {y_n}}\hspace {0.2cm}\) is an inner product space. Show that this is in fact a Hilbert space.
Solution. Let \(\{x^{(k)}\}\) be Cauchy in \(l^2\) and let \(\varepsilon >0\). Choose \(N\) with \(||x^{(k)}-x^{(l)}||<\varepsilon \) for \(k,l\geq N\).
A candidate limit
For each coordinate \(i\), \(\big |x^{(k)}_i-x^{(l)}_i\big |\leq ||x^{(k)}-x^{(l)}||\), so \(\{x^{(k)}_i\}_k\) is Cauchy in \(\mathbb {C}\) and converges to some \(x_i\). Put \(x=(x_1,x_2,\dots )\).
The limit lies in \(l^2\) and is approached in norm
For \(k,l\geq N\) and every \(m\), \[\sum _{i=1}^{m}\big |x^{(k)}_i-x^{(l)}_i\big |^2\leq ||x^{(k)}-x^{(l)}||^2 <\varepsilon ^2 .\] Holding \(k\) and \(m\) fixed and letting \(l\rightarrow \infty \) — a limit of finitely many terms, so it may be taken inside the sum — gives \[\sum _{i=1}^{m}\big |x^{(k)}_i-x_i\big |^2\leq \varepsilon ^2 .\] This holds for every \(m\), so letting \(m\rightarrow \infty \), \[||x^{(k)}-x||^2=\sum _{i=1}^{\infty }\big |x^{(k)}_i-x_i\big |^2 \leq \varepsilon ^2\qquad \text {for all }k\geq N .\] In particular \(x^{(k)}-x\in l^2\), and since \(l^2\) is a vector space \(x=x^{(N)}-\big (x^{(N)}-x\big )\in l^2\). The displayed bound says exactly that \(x^{(k)}\rightarrow x\) in \(l^2\).
The order of the limits is the whole point: passing to a finite partial sum first is what makes the coordinatewise limit legitimate, and only afterwards is \(m\) allowed to grow.
These two examples are very important in the analysis of Hilbert spaces. It will be shown later that any Hilbert that has a countable dense set and is finite dimensional is isomorphic to \(\mathbb {C}^n\), and if not finite dimensional, then isomorphic to \(l^2\).
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