Ch4. Hilbert Space: An Introduction (Ex.10 ~ Ex.20)
10. Let \(S\) denote a subspace of a Hilbert space \(H\). Show that \(S^{\perp\perp}\) is the smallest closed subspace of \(H\) such that \(S\subset \mathcal{H}\).
proof ① \(S\subset S^{\perp\perp}\)
Let \(f\in S^\perp\) and \(g\in S\), then \((f,g)=0\).
\(S^{\perp\perp}\) can be defined as
follows: \(S^{\perp\perp}=\{g\in H:(g,f)=0\
\forall f\in S^\perp\}\).
When \(g\in S\), \((g,f)=0\) for all \(f\in S^\perp\). So, \(g\in S^{\perp\perp}\) and thus \(S\subset S^{\perp\perp}\).
Note that \((S^\perp)^\perp\) is a
closed subspace of \(H\).
② \(A\subset B\subset H\), then \(B^\perp\subset A^\perp\)
Let \(u\in B^\perp\), then \((u,v)=0\) for all \(v\in B\).
If \(a\in A\), \(a\in B\) as well so \((u,a)=0\) for all \(a\in A\). Thus, \(u\in A^\perp\).
③ \(F\) is closed subspace, then \(F^{\perp\perp}=F\)
It suffices to show \(F^{\perp\perp}\subset F\) (\(\because\) ①)
\((F^\perp)^\perp\) is a closed subspace, which is a Hilbert space, and \(F\subset F^{\perp\perp}\) is also a closed subspace of \(F^{\perp\perp}\). So \(F^{\perp\perp}=F\oplus F^\perp\) (prop 4.4.2): that is, \(x\in F^{\perp\perp}\) can be uniquely written as \(a+b\), where \(a\in F\) and \(b\in F^\perp\). Due to the definition of \(F^\perp\), \((a,b)=0\). Then, \[ \begin{aligned} 0&=(a+b,b)=(a,b)+\|b\|^2=\|b\|^2\ (\because a\perp b), \end{aligned} \] and \(b=0\). Thus, \(x=a\in F\) and \(F^{\perp\perp}\subset F\).
④ conclusion : \(S^{\perp\perp}\subset F\)
Suppose \(F\subset H\) is another closed subspace containing \(S: S\subset F\). Due to ②, \(F^\perp\subset S^\perp\) and \(S^{\perp\perp}\subset F^{\perp\perp}=F\) (\(\because\) ③). Since \(S^{\perp\perp}\subset F\) for all such closed subspaces containing \(S\), \(S^{\perp\perp}\) is the smallest closed subspace containing \(S\).
11. Let \(P\) be the orthogonal projection onto a closed subspace \(S\subset \mathcal{H}\).
(a)-1 Show that \(P^2=P\)
proof \(f\in\mathcal{H}\) can be uniquely written
as \(f_S+f_{S^\perp}\), where \(f_S\in S\) and \(f_{S^\perp}\in S^\perp\).
\(P(f)=f_S\) and \(P^2(f)=P(P(f))=P(f_S)=f_S\). Thus, \(P^2=P\).
(a)-2 Show that \(P^*=P\)
proof Let \(g=g_S+g_{S^\perp}\), where \(g\in\mathcal{H}\), \(g_S\in S\), and \(g_{S^\perp}\in S^\perp\).
\((Pf,g)=(f_S,g_S+g_{S^\perp})=(f_S,g_S)=(f_S+f_{S^\perp},g_S)=(f,Pg)\).
Thus, \(P^*=P\).
(b) Conversely, \(P\) is a bounded operator satisfying \(P^2=P\) and \(P^*=P\). Then, show that \(P\) is the orthogonal projection onto some closed subspace of \(\mathcal{H}\).
proof Let \(S=\{Pf:f\in\mathcal{H}\}\); for \(\forall g\in S\), there is \(f\in\mathcal{H}\) such that \(g=Pf\).
① \(P(v)=v\) for \(\forall v\in S\)
Choose \(f\in\mathcal{H}\) such that \(v=P(f)\): \(P(v)=P^2(f)=P(f)\) (\(\because P^2=P\)) \(=v\).
② \(Pw=0\) for \(\forall w\in S^\perp\)
For \(\forall v\in S\), \((v,Pw)=(Pv,w)\) (\(\because P^*=P\)) \(=^{\cdot ①} (v,w)=0\), so \(Pw\in S^\perp\).
Since \(Pw\in S\) by def of \(S\), \(Pw=0\).
④ \(S\) is a subspace of \(\mathcal{H}\) (\(f,g\in S,\ \alpha,\beta\in\mathbb{C}\))
There exist \(u,v\in\mathcal{H}\)
such that \(f=Pu\) and \(g=Pv\).
\(\alpha f+\beta g=\alpha Pu+\beta Pv=P(\alpha
u+\beta v)\) (\(\because P\) is
operator).
\(\alpha u+\beta v\in\mathcal{H}\)
(\(\because \mathcal{H}\) is a vector
space), so \(\alpha f+\beta g\in
S\).
Thus, \(S\) is a subspace of \(\mathcal{H}\).
⑤ \(S\) is closed
Suppose \(\{z_n\}\subset S\) such
that \(z_n\to x\in\mathcal{H}\). \(P\) is continuous, so \(Pz_n\to Px\).
This means \(z_n\to Px\), so \(x=Px\in S\) and thus \(S\) is closed.
\(\because ①\)
⑥ conclusion
Consider \(y\in\mathcal{H}\). Then
\(y=y_S+y_{S^\perp}\), where \(y_S\in S\) and \(y_{S^\perp}\in S^\perp\).
\(P\) is an operator, so \(Py=Py_S+Py_{S^\perp}=Py_S\) (\(\because ②\)) \(=y_S\) (\(\because ①\)).
This holds for all \(y\in\mathcal{H}\),
so \(P\) is an orthogonal projection
onto \(S\).
(c) Show that \(S\) is separable \((\|\cdot\|_{\mathcal{H}}\to \|\cdot\|)\)
proof \(\mathcal{H}\) is separable, so there is \(\{\phi_k\}\subset\mathcal{H}\) such that its finite linear combinations are dense in \(\mathcal{H}\); for \(f\in\mathcal{H}\) and \(\forall\varepsilon>0\), there is \(\psi_\varepsilon=\sum_{i=1}^N a_i\phi_i\) (\(a_i\in\mathbb{C}\)) such that \(\|\psi_\varepsilon-f\|<\varepsilon\). Consider the case \(f\in S\subset\mathcal{H}\).
\(P(\psi_\varepsilon)=P\!\left(\sum_{i=1}^N
a_i\phi_i\right)=\sum_{i=1}^N a_i\cdot P(\phi_i)\), finite linear
combination of \(\{P(\phi_i)\}\subset
S\).
Then, \(\|P(\psi_\varepsilon)-f\|=\|P(\psi_\varepsilon)-P(f)\|=\|P(\psi_\varepsilon-f)\|\le
\|\psi_\varepsilon-f\|<\varepsilon\).
This holds for all \(\varepsilon>0\) and all \(f\in S\), so \(\{P(\phi_i)\}_i\) is dense in \(S\) and \(S\) is separable.
12. Let \(E\) be a measurable subset of \(\mathbb{R}^d\), and \(S\) is a subspace of \(L^2(\mathbb{R}^d)\) whose functions vanish a.e. \(x\in E^c\). Show that the orthogonal projection \(P\) on \(S\) is \(P(f)=\chi_E f\) (\(f\in L^2(\mathbb{R}^d)\))
proof Let \(T(f(x))=\chi_E(x)f(x)\) (\(f\in L^2\), \(T:L^2\to L^2\), \(\|\cdot\|_{L^2}\to\|\cdot\|\)). Then \(\{Tf:f\in L^2\}=S\).
Note that \(Tf(x)\) vanishes for a.e.
\(x\in E^c\).
\(T^2(f)=T(Tf)=T(\chi_E f)=\chi_E(\chi_E f)=\chi_E f=T(f)\), so \(T^2=T\).
\((Tf,g)=\int \chi_E f\cdot \overline{g}=\int_E f\cdot \overline{g}=(f,\chi_E g)=(f,Tg)\), so \(T^*=T\).
\(\|T(f)\|^2=\|\chi_E f\|^2=\int_{\mathbb{R}^d}|\chi_E f|^2=\int_E |f|^2\le \int_{\mathbb{R}^d}|f|^2=\|f\|^2\).
Then, \(\|T(f)\|\le \|f\|\), so \(T\) is bounded \((\|T\|=1)\). So, \(T\) is an orthogonal projection onto a closed subspace \(S\) (\(\because\) ex 11)linear \((f,g\in\mathcal{H},\ \alpha,\beta\in\mathbb{C})\) \(T(\alpha f+\beta g)=\chi_E(x)(\alpha f(x)+\beta g(x))=\alpha\cdot \chi_E f+\beta\cdot \chi_E g=\alpha Tf+\beta Tg\)
13. Suppose \(P_1,P_2\) are orthogonal projections on \(S_1,S_2\subset \mathcal{H}\) respectively.
(1) Show that \(P_1P_2\) is an orthogonal projection if \(P_1\) and \(P_2\) commute
proof ① \((AB)^*=B^*A^*\), where \(A,B\) are linear operator on \(\mathcal{H}\) \((v,w\in\mathcal{H})\)
\((ABv,w)=(A(Bv),w)=(Bv,A^*w)=(v,B^*A^*w)\),
so \((AB)^*=B^*A^*\).
\(\because Bv\in\mathcal{H}\) \(\because v\in\mathcal{H}\)
②→ : when \(P_1P_2\) is orthogonal projection
\(P_2P_1=P_2^*P_1^*=(P_1P_2)^*=P_1P_2\), so they commute.
③← : When \(P_1,P_2\) commute
\((P_1P_2)^2=P_1P_2P_1P_2=P_1P_1P_2P_2=P_1P_2\)
\((P_1P_2)^*=P_2^*P_1^*=P_2P_1=P_1P_2\)
\(\|P_1P_2 f\|=\|P_1(P_2f)\|\le \|P_2f\|\le \|f\|\to P_1P_2\) is bounded
So, \(P_1P_2\) is an orthogonal projection
(2) Show that \(P_1P_2\) projects onto \(S_1\cap S_2\) if they commute
proof ① \(\mathrm{Im}(P_1P_2)\subset S_1\cap S_2\)
\(P_1P_2 f=P_1(P_2f)\in S_1\), and \(P_1P_2 f=P_2(P_1f)\in S_2\). \(\therefore P_1P_2 f\in S_1\cap S_2\)
② \(S_1\cap S_2\subset \mathrm{Im}(P_1P_2),\ \forall f\in\mathcal{H}\)
Let \(s\in S_1\cap S_2\). \(P_1(s)=s\) (\(\because s\in S_1\)) and \(P_2(s)=s\) (\(\because s\in S_2\)).
Then, \(P_1P_2(s)=P_1(s)=s\). So, \(s\in \mathrm{Im}(P_1P_2)\)
\(\therefore \mathrm{Im}(P_1P_2)=S_1\cap S_2\)
14. Suppose \(\mathcal{H}\) and \(\mathcal{H}'\) are two completions of pre-Hilbert space \(\mathcal{H}_0\). Show that there is a unitary mapping from \(\mathcal{H}\) to \(\mathcal{H}'\) that is identity on \(\mathcal{H}_0\).
proof Choose a Cauchy sequence \(\{f_n\}\subset \mathcal{H}_0\) such that \(\|f_n-f\|_{\mathcal{H}}\to 0\) for \(f\in\mathcal{H}\). Define \(U:\mathcal{H}\to\mathcal{H}'\) as follows: \(\|f_n-U(f)\|_{\mathcal{H}'}\to 0\) for \(f\in\mathcal{H}\).
① \(U\) is identity on \(\mathcal{H}_0\) : \(Uf=f\) for \(f\in\mathcal{H}_0\)
Consider the case \(f\in\mathcal{H}_0\). \[
\begin{aligned}
\|f-U(f)\|_{\mathcal{H}'}
&\le \|f-f_n\|_{\mathcal{H}'}+\|f_n-U(f)\|_{\mathcal{H}'} \\
&= \|f_n-f\|_{\mathcal{H}_0}+\|f_n-U(f)\|_{\mathcal{H}'} \qquad
(\because f,f_n\in\mathcal{H}_0)\\
&= \|f_n-f\|_{\mathcal{H}}+\|f_n-U(f)\|_{\mathcal{H}'} \qquad
(\because \mathcal{H}_0\subset \mathcal{H})
\end{aligned}
\] So, \(\|f-U(f)\|_{\mathcal{H}'}\le
\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}+\lim_{n\to\infty}\|f_n-U(f)\|_{\mathcal{H}'}=0\).
This means \(\|f-U(f)\|_{\mathcal{H}'}=0\), so \(U(f)=f\).
This holds for all \(f\in\mathcal{H}_0\), so \(U\) is identity on \(\mathcal{H}_0\).
② \(U\) is unitary
Consider Cauchy sequences \(\{f_n\},\{g_n\}\subset \mathcal{H}_0\) such that \((f,g\in\mathcal{H})\) \(\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}=0\) and \(\lim_{n\to\infty}\|g_n-g\|_{\mathcal{H}}=0\).
\(U\) is linear \((f,g\in\mathcal{H},\ \alpha,\beta\in\mathbb{C})\) \(\lim_{n\to\infty}\|f_n-U(f)\|_{\mathcal{H}'}=0\) and \(\lim_{n\to\infty}\|g_n-U(g)\|_{\mathcal{H}'}=0\) are true. Then, \(\lim_{n\to\infty}\|\alpha f_n-\alpha U(f)\|_{\mathcal{H}'}=0\) and \(\lim_{n\to\infty}\|\beta g_n-\beta U(g)\|_{\mathcal{H}'}=0\) by def of \(U\). So, \[ \begin{aligned} \|(\alpha f_n+\beta g_n)-(\alpha U(f)+\beta U(g))\|_{\mathcal{H}'} &\le \|\alpha f_n-\alpha U(f)\|_{\mathcal{H}'}+\|\beta g_n-\beta U(g)\|_{\mathcal{H}'},\\ \lim_{n\to\infty}\|(\alpha f_n+\beta g_n)-(\alpha U(f)+\beta U(g))\|_{\mathcal{H}'}&=0. \end{aligned} \] \[ \begin{aligned} \|(\alpha f_n+\beta g_n)-(\alpha f+\beta g)\|_{\mathcal{H}} &\le \|\alpha f_n-\alpha f\|_{\mathcal{H}}+\|\beta g_n-\beta g\|_{\mathcal{H}} \to 0,\\ \lim_{n\to\infty}\|(\alpha f_n+\beta g_n)-(\alpha f+\beta g)\|_{\mathcal{H}}&=0, \end{aligned} \] so \(\lim_{n\to\infty}\|\alpha f_n+\beta g_n-U(\alpha f+\beta g)\|_{\mathcal{H}'}=0\) (\(\because\) def of \(U\)).
Thus, \(U(\alpha f+\beta g)=\alpha U(f)+\beta U(g)\) : linear\(U\) is bijection
- \(U\) is injective (1-1) Let \(U(f)=U(g)\in\mathcal{H}'\) for \(f,g\in\mathcal{H}\). \[
\begin{aligned}
\|f-g\|_{\mathcal{H}}
&\le
\|f-f_n\|_{\mathcal{H}}+\|f_n-g_n\|_{\mathcal{H}}+\|g_n-g\|_{\mathcal{H}}\\
&=
\|f-f_n\|_{\mathcal{H}}+\|f_n-g_n\|_{\mathcal{H}}+\|g_n-g\|_{\mathcal{H}}
\qquad (\because f_n,g_n\in\mathcal{H}_0\subset \mathcal{H}')\\
&\le
\|f-f_n\|_{\mathcal{H}}+\|f_n-U(f)\|_{\mathcal{H}'}+\|U(g)-g_n\|_{\mathcal{H}'}+\|g_n-g\|_{\mathcal{H}}
\qquad (\because U(f)=U(g))
\end{aligned}
\] Then, \[
\|f-g\|_{\mathcal{H}}\le
\lim_{n\to\infty}\Big\{\|f-f_n\|_{\mathcal{H}}+\|f_n-U(f)\|_{\mathcal{H}'}+\|U(g)-g_n\|_{\mathcal{H}'}+\|g_n-g\|_{\mathcal{H}}\Big\}=0.
\] So, \(f=g\) and thus \(U\) is injective (1-1).
- \(U\) is surjective (onto) Consider
\(f'\in\mathcal{H}'\) and \(\{f_n\}\), a Cauchy sequence in \(\mathcal{H}_0\), such that \(\lim_{n\to\infty}\|f_n-f'\|_{\mathcal{H}'}=0\).
\(\exists f\in\mathcal{H}\) s.t. \(\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}=0\).
\(\lim_{n\to\infty}\|f_n-U(f)\|_{\mathcal{H}'}=0\) by def of \(U\), then \[ \|U(f)-f'\|_{\mathcal{H}'}\le \|U(f)-f_n\|_{\mathcal{H}'}+\|f_n-f'\|_{\mathcal{H}'},\ \text{so }\|U(f)-f'\|_{\mathcal{H}'}=0. \] Thus, there is \(f\in\mathcal{H}\) such that \(f'=U(f)\Rightarrow\) surjective (onto). - \(\therefore\) \(U\) is bijective
- \(\|U(f)\|_{\mathcal{H}'}=\|f\|_{\mathcal{H}}\)
\(\ \forall f\in\mathcal{H}\) Choose
the same Cauchy sequence \(\{f_n\}\) in
\(\mathcal{H}_0\). \[
\begin{aligned}
\|U(f)\|_{\mathcal{H}'}
&\le
\|U(f)-f_n\|_{\mathcal{H}'}+\|f_n\|_{\mathcal{H}'}=\|U(f)-f_n\|_{\mathcal{H}'}+\|f_n\|_{\mathcal{H}}
\qquad (\because f_n\in\mathcal{H}_0\subset \mathcal{H})\\
&\le
\|U(f)-f_n\|_{\mathcal{H}'}+\|f_n-f\|_{\mathcal{H}}+\|f\|_{\mathcal{H}}.
\end{aligned}
\] Since \(\|U(f)\|_{\mathcal{H}'}\le
\lim_{n\to\infty}\|U(f)-f_n\|+\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}+\|f\|_{\mathcal{H}}=\|f\|_{\mathcal{H}}\),
\(\|U(f)\|_{\mathcal{H}'}\le
\|f\|_{\mathcal{H}}\).
Similarly, \[ \|f\|_{\mathcal{H}}\le \|f-f_n\|_{\mathcal{H}}+\|f_n-U(f)\|_{\mathcal{H}'}+\|U(f)\|_{\mathcal{H}'},\ \text{so }\|f\|_{\mathcal{H}}\le \|U(f)\|_{\mathcal{H}'}. \] Thus, \(\|U(f)\|_{\mathcal{H}'}=\|f\|_{\mathcal{H}}\) for all \(f\in\mathcal{H}\).
15. Let \(T\) be a linear transformation from \(H_1\) to \(H_2\). If we suppose that \(H_1\) is finite dimension, show that \(T\) is bounded.
proof Let \(\{e_i\}_{i=1}^n\) be an orthonormal basis for \(H_1\). \(f\in H_1\) can be written as \(f=\sum_{i=1}^n a_i e_i\), where \(a_i\in\mathbb{C}\). Let \(M=\max_{1\le i\le n}\|T(e_i)\|_{H_2}\). \[ \begin{aligned} \|T(f)\|_{H_2} &=\left\|T\!\left(\sum_{i=1}^n a_i e_i\right)\right\|_{H_2} =\left\|\sum_{i=1}^n a_i T(e_i)\right\|_{H_2} \\ &\le \sum_{i=1}^n |a_i|\,\|T(e_i)\|_{H_2}\qquad (\because \triangle\ \text{ineq})\\ &\le M\sum_{i=1}^n |a_i| \\ &\le M\left(\sum_{i=1}^n |a_i|^2\cdot n\right)^{1/2}\qquad (\because \text{Cauchy-Schwartz})\\ &=M\sqrt{n}\,\|f\|_{H_1}\qquad (\text{thm }4.2.3\ \text{iv})\ \ (\text{let }M'=M\sqrt{n}) \end{aligned} \] Since \(\|T(f)\|_{H_2}\le M'\|f\|_{H_1}\) for all \(f\in H_1\), \(T\) is bounded.
18. Let \(H\) denote a Hilbert space, and \(\mathcal{L}(H)\) the vector space of all bounded linear operators on \(H\).
(a) Show that \(\|T_1+T_2\|\le \|T_1\|+\|T_2\|\) \(\ \forall T_1,T_2\in\mathcal{L}(H)\)
proof \[ \begin{aligned} \|T_1+T_2\| &=\sup\Big\{|((T_1+T_2)f,g)|: f,g\in H\ \&\ \|f\|\le 1,\ \|g\|\le 1\Big\}\qquad \text{omitted}\\ &=\sup_{\|f\|,\|g\|\le 1}\Big\{|(T_1f,g)+(T_2f,g)|\Big\}\\ &\le \sup_{\|f\|,\|g\|\le 1}\Big\{|(T_1f,g)|+|(T_2f,g)|\Big\}\qquad (\because \sup_n(a_n+b_n)\le \sup_n a_n+\sup_n b_n)\\ &\le \sup_{\|f\|,\|g\|\le 1}|(T_1f,g)|+\sup_{\|f\|,\|g\|\le 1}|(T_2f,g)|=\|T_1\|+\|T_2\|. \end{aligned} \]
(b) Show that \(d(T_1,T_2)=\|T_1-T_2\|\) is a metric on \(\mathcal{L}(H)\)
proof ① \(d(T_1,T_2)\ge 0\)
\(\|T_1-T_2\|=\sup_{f,g\in H}|((T_1-T_2)f,g)|\) : supremum of non-negative numbers. So, \(\|T_1-T_2\|\ge 0\).
② \(d(T_1,T_2)=0\) iff \(T_1=T_2\)
\(\to\) : \(\|(T_1-T_2)f\|_H\le \|T_1-T_2\|\|f\|_H\),
so \(\|(T_1-T_2)f\|_H=0\) for all \(f\in H\) and \((T_1-T_2)f=0\) \(\forall f\in H\). Then \(T_1-T_2\) is a zero operator: \(T_1-T_2=0\), so \(T_1=T_2\).
\(\leftarrow\) : \(\|T_1-T_1\|=\sup_{\|f\|,\|g\|\le 1}|\langle
0f,g\rangle|=\sup_{\|f\|,\|g\|\le 1}|\langle
0,g\rangle|=\sup_{\|f\|,\|g\|\le 1}0=0\).
③ \(d(T_1,T_2)=d(T_2,T_1)\)
\(\|-T\|=\sup|(-Tf,g)|=\sup|(T(-f),g)|=\|T\|\ \
(\because \|-f\|_H=\|f\|_H)\).
Then, \(d(T_1,T_2)=\|T_1-T_2\|=\|T_2-T_1\|=d(T_2,T_1)\).
④ \(d(T_1,T_3)\le d(T_1,T_2)+d(T_2,T_3)\)
\(\|T_1-T_3\|=\|T_1-T_2+T_2-T_3\|\le \|T_1-T_2\|+\|T_2-T_3\|\) (\(\because\) (a))
(c) Show that \(\mathcal{L}(H)\) is complete in the metric ‘d’
proof Let \(\{T_n\}\subset\mathcal{L}(H)\) be a Cauchy sequence, then for every \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(n,m\ge N_\varepsilon\) implies \(\|T_m-T_n\|<\varepsilon\). Choose \(f\in H\).
① \(\{T_nf\}\) is Cauchy in \(H\), define \(T\) as \(F=T(f)\).
\(\|(T_m-T_n)f\|_H\le
\|T_m-T_n\|\|f\|_H\), so \(n,m\ge
N_\varepsilon\) implies \(\|T_mf-T_nf\|_H<\|f\|_H\varepsilon\).
This holds for all \(\varepsilon>0\), so \(\{T_nf\}\) is Cauchy sequence in \(H\).
Then, there is \(F\in H\) such that
\(\lim_{n\to\infty}\|T_n(f)-F\|_H=0\)
(\(\because H\) is complete).
Let \(F=T(f)\) \((T:H\to H)\).
② \(T\) is bounded
\(\|Tf\|_H\le
\|Tf-T_nf\|_H+\|T_nf\|_H\) and \(\|T_nf\|_H\le
\|T_nf-Tf\|_H+\|Tf\|_H\).
This leads to \(\|Tf\|_H=\lim_{n\to\infty}\|T_nf\|_H\)
(\(\because\) ①).
\(\|T_n\|\le
\|T_n-T_m\|+\|T_m\|<\|T_m\|+\varepsilon\) for \(n,m\ge N_\varepsilon\), so \(\lim_{n\to\infty}\|T_n\|\le
\|T_{N_\varepsilon}\|+\varepsilon\) let \(B\).
Then, \(\|Tf\|_H=\lim_{n\to\infty}\|T_nf\|_H\le
\lim_{n\to\infty}\|T_n\|\|f\|_H< B\cdot \|f\|_H\).
Thus, \(\|T\|\le B\) and \(T\) is bounded.
③ \(T\) is linear \((\alpha,\beta\in\mathbb{C})\)
When \(g\in H\), \(\{T_ng\}\) is a Cauchy sequence in \(H\): \(\lim_{n\to\infty}\|T_ng-Tg\|_H=0\). Then,
\[
\begin{aligned}
\lim_{n\to\infty}\|\alpha\cdot T_nf+\beta\cdot T_ng-(\alpha\cdot
Tf+\beta\cdot Tg)\|_H
&\le
\lim_{n\to\infty}\Big\{|\alpha|\|T_nf-Tf\|_H+|\beta|\|T_ng-Tg\|_H\Big\}=0.
\end{aligned}
\] \(\{T_n(\alpha f+\beta g)\}\)
is a Cauchy sequence in \(H\), so \(\lim_{n\to\infty}\|T_n(\alpha f+\beta g)-T(\alpha
f+\beta g)\|_H=0\).
\(T_n\) is linear for all \(n\in\mathbb{N}\), so \(T_n(\alpha f+\beta g)=\alpha\cdot
T_n(f)+\beta\cdot T_n(g)\). Then, \[
\lim_{n\to\infty}\|T_n(\alpha f+\beta g)-T(\alpha f+\beta g)\|
=\lim_{n\to\infty}\|\alpha\cdot T_n(f)+\beta\cdot T_n(g)-T(\alpha
f+\beta g)\|=0.
\] Thus, \(T(\alpha f+\beta
g)=\alpha\cdot T(f)+\beta\cdot T(g)\), and \(T\) is linear.
④ \(\lim_{n\to\infty}\|T_n-T\|=0\)
\(\lim_{m\to\infty}\|T_mf-Tf\|_H=0\), so
there is \(M_\varepsilon\in\mathbb{N}\)
such that \(m\ge M_\varepsilon\)
implies \(\|T_mf-Tf\|_H<\varepsilon\|f\|_H\) for
the given \(\varepsilon>0\) (&
fixed \(f\)).
Then, \(m\ge M_\varepsilon\) leads to
\(\|T_m-T\|\le \varepsilon\).
This holds for all \(\varepsilon>0\), so \(\lim_{m\to\infty}\|T_m-T\|=0\).
⑤ conclusion
\(\therefore\) \(T\in\mathcal{L}(H)\) and \(\mathcal{L}(H)\) is complete.
19. \(T\) is a bounded linear operator on a Hilbert space. Show that \(\|TT^*\|=\|T^*T\|=\|T\|^2=\|T^*\|^2\).
proof It suffices to show \(\|T^*T\|=\|T\|^2\) (\(\because \|T\|=\|T^*\|\) is known)
① \(\|T^*T\|\le \|T\|^2\)
\[ \begin{aligned} \|T^*T\| &=\sup_{\|f\|_H,\|g\|_H\le 1}|(T^*Tf,g)|=\sup_{f,g}|(Tf,Tg)| \\ &\le \sup_{f,g}\|Tf\|\cdot\|Tg\|\ \ (\because \text{Cauchy-Schwartz ineq}) \\ &=\|T\|^2 \end{aligned} \]
② \(\|T\|^2\le \|T^*T\|\)
\[ \begin{aligned} \|T\|^2 &=\sup_{\|x\|_H\le 1}\|Tx\|^2=\sup_{\|x\|_H\le 1}(Tx,Tx) \\ &=\sup_{\|x\|_H\le 1}(x,T^*Tx) \\ &\le \sup_{\|x\|_H\le 1}\|x\|\cdot\|T^*Tx\| \qquad (\because \text{Cauchy}) \\ &=\sup_{\|x\|_H\le 1}\|T^*Tx\|\\ &\le \sup_{\|x\|_H\le 1}\|T^*T\|\cdot\|x\|\\ &=\|T^*T\| \end{aligned} \]
So, \(\|T\|^2\le \|T^*T\|\).
③ \(\therefore\ \|T^*T\|=\|T\|^2\)
④ other equalities
Since \(\|T\|=\|T^*\|\), \(\|T\|^2=\|T^*\|^2\). Also, switching \(T\) with \(T^*\) vise versa, \(\|TT^*\|=\|T^*\|^2\).
Thus, \(\|TT^*\|=\|T^*T\|=\|T\|^2=\|T^*\|^2\).
cf) \(\|T\|=\sup_{\|x\|_H\le 1}\|Tx\|_H=\sup_{\|x\|_H=1}\|Tx\|_H=\sup_{x\ne 0}\dfrac{\|Tx\|_H}{\|x\|_H}\)
proof Let \(I=\inf\{C:\|Tx\|_H\le C\|x\|_H\}\).
\(S_1=\sup\{\|Tx\|_H:\|x\|_H\le
1\}\)
\(S_2=\sup\{\|Tx\|_H:\|x\|_H=1\}\)
\(S_3=\sup\left\{\dfrac{\|Tx\|_H}{\|x\|_H}:x\ne
0\right\}\)
\(\|Tx\|_H\le S_3\|x\|_H\ \to\ I\le S_3\)
\(\forall x,\ \exists v\ne 0\ \text{s.t.}\
S_3<\dfrac{\|Tv\|}{\|v\|}+\varepsilon\)
\(\because \dfrac{\|Tv\|}{\|v\|}\le I\ \ \}\to
S_3\le I\)
20. Suppose \(H\) is an infinite-dimensional Hilbert space. Show that for any \(\{f_n\}\subset H\) with \(\|f_n\|=1\) \(\forall n\in\mathbb{N}\), there is \(f\in H\) and a subsequence \(\{f_{n_k}\}\) such that
\[ \lim_{k\to\infty}(f_{n_k},g)=(f,g)\qquad \forall g\in H. \]
proof Let \(\{e_i\}\) be an orthonormal basis for \(H\).
① \(\exists\) convergent \(\{(f_n,e_1)\}_n\)
Consider a sequence \(\{(f_n,e_1)\}_n\subset \mathbb{C}\).
\(|(f_n,e_1)|\le \|f_n\|\|e_1\|=1\), so
\(\{(f_n,e_1)\}_n\) is a bounded
sequence in \(\mathbb{C}\).
Then, \(\{(f_n,e_1)\}_n\) has a
convergent subsequence \(\{(f_{1n},e_1)\}_n\) such that \(\lim_{n\to\infty}(f_{1n},e_1)=l_1\in\mathbb{C}\)
(PMA 3.6) (\(\{f_{1n}\}\subset\{f_n\}\)).
② \(\exists\) convergent \(\{(f_{2n},e_2)\}_n\)
Consider a sequence \(\{(f_{1n},e_2)\}_n\subset
\mathbb{C}\).
\(|(f_{1n},e_2)|\le
\|f_{1n}\|\|e_2\|=1\), so \(\{(f_{1n},e_2)\}_n\) is a bounded sequence
in \(\mathbb{C}\).
Then \(\{(f_{1n},e_2)\}_n\) has a
convergent subsequence \(\{(f_{2n},e_2)\}_n\) such that \(\lim_{n\to\infty}(f_{2n},e_2)=l_2\in\mathbb{C}\)
(\(\{f_{2n}\}\subset\{f_{1n}\}\)).
Note that \(\lim_{n\to\infty}(f_{2n},e_1)=l_1\) (PMA
3.5-1).
③ \(S_m=\{f_{m,n}\}_n\) s.t. \(\lim_{n\to\infty}(f_{m,n},e_i)=l_i\) \((i=1,\cdots,m)\)
Continuing the same process, we can obtain sequences \(S_0,S_1,S_2,\cdots\), where \(S_0=\{f_n\}\), \(S_m=\{f_{m,n}\}_n\), \(S_{m+1}\) is a subsequence of \(S_m\) obtained by the previous process, and \(\lim_{n\to\infty}(f_{m,n},e_i)=l_i\) \((i=1,\cdots,m,\ m\in\mathbb{N})\).
\(S_0:\ f_1,\ f_2,\ f_3,\ \cdots\)
\(S_1:\ f_{1,1},\ f_{1,2},\ f_{1,3},\ \cdots\ \ \text{s.t.}\ \lim_{n\to\infty}(f_{1,n},e_i)=l_i,\ i=1\)
\(S_2:\ f_{2,1},\ f_{2,2},\ f_{2,3},\ \cdots\ \ \text{s.t.}\ \lim_{n\to\infty}(f_{2,n},e_i)=l_i,\ i=1,2\)
\(\qquad \vdots\)
\(S_k:\ f_{k,1},\ f_{k,2},\ \cdots,\ f_{k,k},\ \cdots\ \ \text{s.t.}\ \lim_{n\to\infty}(f_{k,n},e_i)=l_i,\ i=1,\cdots,k\)
④ \(S=\{f_{k,k}\},\ f=\sum_{i=1}^\infty l_ie_i\in H\)
Let \(S=\{f_{k,k}\}\) \((k\in\mathbb{N})\). Since \(S\subset S_i\) for \(i\in\mathbb{N}\), \(\lim_{k\to\infty}(f_{k,k},e_i)=l_i\).
Relable \(f_{k,k}\) as \(f_{n_k}\), and define \(f=\sum_{i=1}^\infty l_ie_i\in H\).
\[
\begin{aligned}
\|f\|^2
&=\sum_{i=1}^\infty |l_i|^2
=\lim_{N\to\infty}\sum_{i=1}^N |l_i|^2\\
&=\lim_{N\to\infty}\sum_{i=1}^N \lim_{k\to\infty}|(f_{n_k},e_i)|^2
\\
&=\lim_{N\to\infty}\lim_{k\to\infty}\sum_{i=1}^N |(f_{n_k},e_i)|^2
\\
&\le \lim_{k\to\infty}\|f_{n_k}\|^2=1.
\end{aligned}
\] \(\sum_{i=1}^\infty
|(f_{n_k},e_i)|^2=\|f_{n_k}\|^2=1\) and \(\lim_{k\to\infty}\|f_{n_k}\|^2=1\), so
\(\|f\|<\infty\) and \(f\in H\).
Then, \(\lim_{k\to\infty}(f_{n_k},e_i)=l_i=(f,e_i)\)
\(\forall i\in\mathbb{N}\).
⑤ conclusion
Let \(g=\sum_{i=1}^\infty b_ie_i\in H\) where \(b_i\in\mathbb{C}\) for all \(i\in\mathbb{N}\). Thus, \[ \begin{aligned} \lim_{k\to\infty}(f_{n_k},g) &=\lim_{k\to\infty}\left(f_{n_k},\sum_{i=1}^\infty b_ie_i\right) =\lim_{k\to\infty}\sum_{i=1}^\infty \overline{b_i}(f_{n_k},e_i) \\ &=\sum_{i=1}^\infty \overline{b_i}\lim_{k\to\infty}(f_{n_k},e_i)\qquad (\because \text{DCT},\ \sum_{i=1}^\infty |b_i|^2|(f_{n_k},e_i)|^2\le \sum_{i=1}^\infty |b_i|^2\cdot 1<\infty\ \because g\in H)\\ &=\sum_{i=1}^\infty \overline{b_i}(f,e_i)=\left(f,\sum_{i=1}^\infty b_ie_i\right)=(f,g). \end{aligned} \]