Let \(B(H)\) denote the Banach space of all bounded linear operators on a Hilbert space \(H \neq \{0\}\), with the operator norm \[\|T\|=\|T\|_{H \to H}=\sup \{\|Tx\|: x \in H \text{ and } \|x\| \leq 1\}.\]
Proposition (A). If \(T \in B(H)\) and if \(\langle Tx,x\rangle=0\) for every \(x \in H\) then \(T=0\). In particular, if \(S \in B(H)\) and \(T \in B(H)\) and \[\langle Sx,x\rangle=\langle Tx,x\rangle\] for all \(x \in H\), then \(S=T\).
Remark. This proposition fails if the scalar field is \(\mathbb{R}\).
Proof. Since \(\langle T(x+y),x+y \rangle=0\) we see that \[\langle Tx,y\rangle+\langle Ty,x\rangle=0 \quad \text{ for all } \quad x,y \in H.\]
If \(y\) is replaced by \(iy\), then \[-i\langle Tx,y\rangle+i\langle Ty,x\rangle=0 \quad \text{ for all }\quad x,y \in H.\] Multiplying by \(i\) we have \[\langle Tx,y\rangle -\langle Ty,x\rangle=0.\] Thus \[2 \langle Tx,y\rangle=0,\] so taking \(y=Tx\) we obtain \(\|Tx\|^2=0\) hence \(Tx=0\).$$\tag*{$\blacksquare$}$$
Let \((H, \|\cdot\|)\) be a Hilbert space. If \(f:H \times H \to \mathbb{C}\) is a antilinear form, i.e.
\(f(\alpha x +\beta y,z)=\alpha f(x,z)+\beta f(y,z)\),
\(f(x,\alpha y+\beta z)=\overline{\alpha}f(x,y)+\overline{\beta}f(x,z)\),
\(f\) is bounded in the sense \[M=\sup\{|f(x,y)|:\|x\|=\|y\|=1\}<\infty.\]
Then there is a unique \(S \in B(H)\) such that \[{\color{red}f(x,y)=\langle x,Sy\rangle \quad \text{ and } \quad \|S\|=M}.\]
Proof. Since \[|f(x,y)| \leq M\|x\|\|y\| \quad \text{ for all } \quad x, y\in H,\] the mapping \[{\color{blue}x \mapsto f(x,y) \quad \text{ for each fixed } \quad y\in H,}\] is a bounded linear functional on \(H\) with norm at most \(M\|y\|\).
From the Riesz representation theorem it follows that for each \(y \in H\) we find a unique element \(Sy \in H\) such that \[{\color{blue}f(x,y)=\langle x,Sy\rangle},\] also \[\|Sy\|\leq M\|y\| \quad \text{ for all } \quad y\in H.\]
It is easy to see that \(S: H \to H\) is linear \[\begin{aligned} \langle x,S(\alpha y) \rangle=f(x,\alpha y)=\overline{\alpha}f(x,y) =\overline{\alpha}\langle x,Sy\rangle=\langle x,\alpha S(y )\rangle \end{aligned}\] so \(S(\alpha y)=\alpha S(y)\) by Proposition (A), and \[\begin{aligned} \langle x,S(y_1+y_2)\rangle=f(x,y_1+y_2)=&f(x,y_1)+f(x,y_2)\\ &=\langle x,Sy_1 \rangle+\langle x,Sy_2\rangle=\langle x,Sy_1+Sy_2\rangle \end{aligned}\] so \(S(y_1+y_2)=Sy_1+Sy_2\), again by Proposition (A). Thus \(S \in B(H)\) and we have \(\|S\| \leq M\). Moreover, we also have \[\begin{aligned} |f(x,y)|=|\langle x,Sy\rangle| \leq \|x\|\|Sy\| \leq \|x\|\|S\|\|y\|, \end{aligned}\] which implies that \(M\le \|S\|\), so we are done.$$\tag*{$\blacksquare$}$$
Fact. Let \((H, \|\cdot\|)\) be a Hilbert space. If \(T \in B(H)\) then \(\langle Tx,y \rangle\) is linear in \(x\) and antilinear in \(y\) and bounded \[\Lambda(x,y)=\langle Tx,y\rangle, \quad \text{ for all } \quad \|\Lambda\|=\|T\|.\]
Proof. Indeed, \(|\langle Tx,y\rangle| \leq \|T\|\|x\|\|y\|\) so \(\|\Lambda\| \leq \|T\|\) and also \[|\langle Tx,Tx\rangle|=\|Tx\|^2\] hence \(\sup_{\|x\|=1}\|Tx\| \leq \|\Lambda\|,\) and \(\|T\| \leq \|\Lambda\|\).
By the previous theorem there is a unique \({\color{purple}T^* \in B(H)}\) such that \[\langle Tx,y\rangle=\langle x,T^*y\rangle \quad \text{ and }\quad \|T\|=\|T^*\|.\] The operator \(T^*\) is called the adjoint operator to \(T\).
Proposition. Let \((H, \|\cdot\|)\) be a Hilbert space. If \(T,S \in B(H)\), then \[\begin{gathered} (T+S)^*=S^*+T^*,\\ (\alpha T)^*=\overline{\alpha}T^*,\\ (ST)^*=T^*S^*,\\ T^{**}=T. \end{gathered}\] Moreover, we have \[{\color{red}\|T\|_{H\to H}^2=\|T^*T\|_{H\to H}.}\]
Remark.
The last identity is very important in the so-called \(T^*T\) methods of bounding linear operators in Hilbert spaces.
Proof. We only prove the second part, as the first part easily follows from the definition of adjoint operators and Proposition (A).
We have
\[\begin{aligned} \|Tx\|^2=\langle Tx,Tx\rangle=\langle T^*Tx,x\rangle \leq \|T^*T\|_{H\to H}\|x\|^2, \end{aligned}\] so \[\|T\|_{H\to H}^2 \leq \|T^*T\|_{H\to H}.\]
On the other hand, \[\begin{aligned} \|T^*T\|_{H\to H} \leq \|T^*\|_{H\to H}\|T\|_{H\to H}=\|T\|_{H\to H}^2, \end{aligned}\] so \[{\color{red}\|T\|_{H\to H}^2=\|T^*T\|_{H\to H}.}\] This completes the proof.$$\tag*{$\blacksquare$}$$
If \(T \in B(H)\) then the kernel and range of \(T\) is defined, respectively, by \[\begin{gathered} {\color{purple}N(T)=\ker(T)=\{x \in H: Tx=0\},}\\ {\color{blue}R(T)={\rm range}(T)=\{y \in H: y=Tx \text{ for some } x \in H\}.} \end{gathered}\]
Let \((H, \|\cdot\|)\) be a Hilbert space. If \(T \in B(H)\) then the \[N(T^*)=R(T)^{\perp} \quad \text{ and } \quad N(T)=R(T^*)^{\perp}.\]
Proof. Note that \[\begin{aligned} {\color{purple}T^*y=0} \iff {\color{blue}\langle x,T^*y\rangle=0 \quad \text{ for all } \quad x \in H}\iff {\color{purple}y \in R(T)^{\perp}} \end{aligned}\] Thus \(N(T^*)=R(T)^{\perp}\) and \(N(T^{**})=R(T^*)^{\perp}\), since \(T^{**}=T\). $$\tag*{$\blacksquare$}$$
A linear map \(U:H_1\to H_2\) between two Hilbert spaces \((H_1, \langle\cdot, \cdot\rangle_{H_1})\) and \((H_2, \langle\cdot, \cdot\rangle_{H_2})\) is called unitary or orthogonal if it is invertible and if \[\langle Ux, Uy\rangle_{H_2}=\langle x, y\rangle_{H_1} \quad \text{ for all } \quad x, y\in H_1.\] Thus, a unitary operator is one-to-one and onto, and preserves the inner product.
Let \((H, \|\cdot\|)\) be a Hilbert space. An operator \(T \in B(H)\) is said to be:
normal if \(T^*T=TT^*\),
self-adjoint if \(T=T^*\),
a projection if \(P^2=P\).
Let \((H, \|\cdot\|)\) be a Hilbert space. If \(U \in B(H)\) the following are equivalent:
\(U\) is unitary,
\(U^*U=UU^*=I\), where \(I:H\to H\) is the identity operator on \(H\),
\(R(U)=H\) and \(\|Ux\|=\|x\|\) for all \(x \in H\).
Proof (c) \(\Longrightarrow\) (b). If (c) holds, then \[\begin{aligned} \langle U^*Ux,x\rangle=\langle Ux,Ux\rangle=\|Ux\|^2=\|x\|^2=\langle x,x\rangle,\quad \text{ for all } \quad x\in H. \end{aligned}\] Hence \(U^*U=I\). But (c) also implies that \(U\) is a linear isometry of \(H\) onto \(H\) so that \(U\) is invertible in \(B(H)\) and \(U^{-1}=U^*\), since \(U^*U=I\).
Proof (b) \(\Longrightarrow\) (a). If \(U^*U=UU^*=I\), then \(U\) is invertible and we have \[\langle Ux,Uy\rangle=\langle x,U^*Uy\rangle=\langle x,y\rangle \quad \text{ for all } \quad x, y\in H.\]
Proof (a) \(\Longrightarrow\) (c). This implication is obvious. $$\tag*{$\blacksquare$}$$
Let \((u_\alpha)_{\alpha \in A}\) be an orthonormal basis for \((H, \|\cdot\|)\), then the correspondence \[{\color{red}x \mapsto \hat{x} \qquad \text{ defined by } \qquad \hat{x}(\alpha)=\langle x,u_\alpha\rangle_H}\] is a unitary map from \(H\) to \(\ell^2(A)\). Therefore, by this identification the \(\ell^2(A)\) spaces are (sometimes) called universal Hilbert spaces.
Proof. We define \(U:H \to \ell^2(H)\) by \[U(x)=\hat{x} \quad \text{ for }\quad x\in H.\] This map is clearly linear and it is an isometry from \(H\) to \(\ell^2(A)\), since by the Parseval identity we have \[\|Ux\|_{\ell^2(A)}^2=\sum_{\alpha \in A}|\hat{x}(\alpha)|^2=\|x\|_H^2.\] Thus \(U\) is an isometry, and \(\langle Ux, Uy\rangle_{\ell^2(A)}=\sum_{\alpha\in A}\hat{x}(\alpha)\overline{\hat{y}(\alpha)}=\langle x, y\rangle_{H}\).
We now prove that \(R(U)=\ell^2(A)\). Indeed, if \(f \in \ell^2(A)\), then \[\sum_{\alpha \in A}|f(\alpha)|^2<\infty.\] The partial sums of the series \[\sum_{\alpha \in A}f(\alpha)u_\alpha\] are Cauchy. Hence \[x_f=\sum_{\alpha \in A}f(\alpha)u_\alpha\in H,\] and \[U(x_f)(\alpha)=\hat{x}_f(\alpha)=\langle x_f,u_\alpha\rangle_H=f(\alpha) \quad \text{ for all } \quad \alpha\in A.\]
So \(U\) is the isometry and onto, hence invertible and unitary as claimed.$$\tag*{$\blacksquare$}$$
Let \((H, \|\cdot\|)\) be a Hilbert space and let \(P \in B(H)\) be a projection \(P^2=P\). Each of the following four properties of \(P\) implies the other three:
\(P\) is self-adjoint, \(P=P^*\),
\(P\) is normal,
\(R(P)=N(P)^\perp\),
\(\langle Px,x\rangle=\|Px\|^2\) for all \(x \in H\).
Property (c) is usually used to say that \(P\) is an orthogonal projection.
Proof \({\color{blue}(a) \implies (b)}\). Since \(P=P^*\), then obviously \(PP^*=P^*P\). $$\tag*{$\blacksquare$}$$
Proof \({\color{blue}(b) \implies (c)}\). If \(P\) is normal then \[\|Px\|^2=\langle Px,Px\rangle=\langle P^*Px,x\rangle=\langle PP^*x,x\rangle=\|P^*x\|^2.\] so \(N(P)=N(P^*)\) hence \(N(P)=N(P^*)=R(P)^{\perp}\). Since \(P\) is a projection \(R(P)=N(I-P)\) so that \(R(P)\) is closed so \[\qquad\qquad N(P)^{\perp}=R(P)^{\perp \perp }=R(P). \qquad\qquad\tag*{$\blacksquare$}\] Proof \({\color{blue}(c) \implies (d)}\). For every \(x \in H\) has the form \(x=y+z\) with \(y \perp z\), and \(P(y)=0\) and \(P(z)=z\) and \(\langle Px,x\rangle=\langle z,z\rangle\) so \[\qquad\qquad\langle Px,x \rangle=\langle z,z\rangle=\|Pz\|^2=\|Px\|^2.\qquad\qquad\tag*{$\blacksquare$}\] Proof \({\color{blue}(d) \implies (a)}\). Note that \[\|Px\|^2=\langle Px,x\rangle=\langle x,P^*x\rangle=\langle P^*x,x\rangle.\] Since \(\|Px\|\) is real thus \(P=P^*\).$$\tag*{$\blacksquare$}$$
Let \((H, \|\cdot\|)\) be a Hilbert space. Let \(U:H\to H\) be a unitary operator, then for any \(f \in H\), we have \[{\color{purple}A_Nf=\frac{1}{N}\sum_{n=0}^{N-1}U^nf \ _{\overrightarrow{N\to \infty}} \ P_If,}\] where \(P_I\) is the projection onto \(I=\{x \in H: Ux=x\}\).
Proof. We begin with establishing Riesz decomposition \[H=I\oplus \overline{R(U-I)}^{H}.\] We claim that \(I=R(U-I)^\perp\). Since \(U\) is unitary, thus normal, by Theorem (C), observe that \[I=N(U-I)=N((U-I)^*)=R(U-I)^{\perp}.\]
Using Riesz’s decomposition, any \(f \in H\) can be written as \(f=P_Ih+h\), where \(h \in \overline{R(U-I)}^H\). Note that \(A_NP_If=P_If\). We claim that \[A_Nh \ _{\overrightarrow{N\to \infty}} \ 0.\] For \(h \in R(U-I)\) so that \(h=Ug-g\), by telescoping, we have \[{\color{blue}\|A_Nh\|=\bigg\|\frac{1}{N}\sum_{n=0}^{N-1}U^n(Ug-g)\bigg\|=\frac{1}{N}\|U^Ng-g\| \ _{\overrightarrow{N\to \infty}} \ 0.}\] Let \((g_n)_{n \in \mathbb{N}} \subseteq H\) be such that \(h_m \ _{\overrightarrow{m\to \infty}} \ h\), where \(h_m=Ug_m-g_m\). Thus \[\|A_Nh\| \leq \|A_N(h-h_m)\|+{\color{blue}\|A_Nh_m\|}\le \|h-h_m\|+{\color{blue}\|A_Nh_m\|}.\] Fix \(\varepsilon>0\) then there is \(N_{\varepsilon}\in \mathbb N\) such that \(\|h-h_m\|<\frac{\varepsilon}{2}\) and \({\color{blue}\|A_Nh_m\|<\frac{\varepsilon}{2}}\) whenever \(m, N\ge N_{\varepsilon}\), then \(\|A_Nh\|<{\varepsilon}\) for all \(N\ge N_{\varepsilon}\) as desired.$$\tag*{$\blacksquare$}$$
Let \((u_n)_{n \in \mathbb{N}}\) be a bounded sequence in a Hilbert space \((H, \|\cdot\|)\). Define a sequence \((s_n)_{n \in \mathbb{N}}\) of real numbers by \[s_m = \limsup_{N \rightarrow \infty} \Big| \frac{1}{N} \sum_{n=1}^N \langle u_{n+m}, u_n \rangle \Big|.\] If \[\lim_{H \rightarrow \infty} \frac{1}{H} \sum_{h=0}^{H-1} s_h = 0,\] then \[\lim_{N \rightarrow \infty} \Big\|\frac{1}{N} \sum_{n=1}^N u_n \Big\| = 0.\]
Proof. Assume that \(M=\sup_{n \in \mathbb{N}} \|u_n\| < \infty\). Let \(0 < H \leq N\), then \[\begin{gathered} \frac{1}{N} \sum_{n=1}^N u_n =\frac{1}{N} \frac{1}{H} \sum_{h=1}^H\sum_{n\in \mathbb Z}u_{n+h}\mathbf{1}_{{[1,N]}}(n+h)=\\ \frac{1}{H} \sum_{h=1}^H \bigg[ \frac{1}{N} \sum_{n=1}^h u_{n+h} - \frac{1}{N} \sum_{n = N+1}^{N+h} u_{n+h} + \frac{1}{N} \sum_{n \in \mathbb{Z}} u_{n+h} \mathbf{1}_{{[h+1,N+h]}}(n+h) \bigg]. \end{gathered}\] By the triangle inequality \[\begin{gathered} \bigg\| \frac{1}{N} \sum_{n=1}^N u_n \bigg\| \leq \frac{1}{NH} \bigg\| \sum_{h=1}^H \sum_{n \in [N]} u_{n+h} \bigg\| + \frac{2MH}{N}, \end{gathered}\] since \[\begin{aligned} \bigg\|\frac{1}{H} \frac{1}{N}\sum_{h=1}^H \sum_{n=1}^h u_{n+h} \bigg\| +\bigg\|\frac{1}{H} \frac{1}{N} \sum_{h=1}^H\sum_{n = N+1}^{N+h} u_{n+h} \bigg\|\le \frac{2MH}{N}. \end{aligned}\]
Let \([N]=[1, N]\) for \(N\in \mathbb N\). By the Cauchy–Schwarz inequality, we obtain \[\begin{aligned} \bigg\| \frac{1}{N} \sum_{n \in [N]} \frac{1}{H} &\sum_{h \in [H]} u_{n+h} \bigg\|^2 \leq \frac{1}{N} \sum_{n \in [N]} \bigg\|\frac{1}{H} \sum_{h \in [H]} u_{n+h}\bigg\|^2\\ &= \bigg|\frac{1}{N} \sum_{n \in [N]} \frac{1}{H^2} \sum_{h_1,h_2 \in [H]} \langle u_{n+h_1}, u_{n+h_2} \rangle \bigg|\\ &\leq \frac{2}{H^2} \sum_{\substack{h_1,h_2 \in [H]\\h_2\ge h_1}} \bigg| \frac{1}{N}\sum_{n \in [N]+h_1} \langle u_{n}, u_{n+h_2-h_1} \rangle\bigg|\\ &\leq \frac{4M^2 H}{N} + \frac{2}{H^2} \sum_{\substack{h_1,h_2 \in [H]\\h_2\ge h_1}} \bigg| \frac{1}{N}\sum_{n \in [N]} \langle u_{n}, u_{n+h_2-h_1} \rangle \bigg|\\ &\leq \frac{4M^2 H}{N} + \frac{2}{H^2} \sum_{h_1\in [H]}\sum_{h\ge0}\mathbf{1}_{{[H]}}(h+h_1) \bigg| \frac{1}{N}\sum_{n \in [N]} \langle u_{n}, u_{n+h} \rangle \bigg|. \end{aligned}\]
Thus we proved that \[\begin{aligned} \bigg\| \frac{1}{N} \sum_{n \in [N]} u_n \bigg\|\le \bigg(\frac{2}{H} \sum_{h=0}^{H-1} \bigg| \frac{1}{N}\sum_{n \in [N]} \langle u_{n}, u_{n+h} \rangle \bigg|\bigg)^{1/2} +\frac{4MH^{1/2}}{N^{1/2}}. \end{aligned}\] Now, given \(\varepsilon > 0\), we find \(H_\varepsilon \in \mathbb{N}\) so that, for \(H \geq H_\varepsilon\), we have \[\frac{1}{H} \sum_{h=0}^{H-1} s_h < \varepsilon.\] Then \[\limsup_{N \rightarrow \infty} \bigg\|\frac{1}{N} \sum_{n \in [N]} u_n \bigg\| \leq \bigg(\frac{2}{H} \sum_{h=0}^{H-1} s_h\bigg)^{1/2} \leq (2\varepsilon)^{1/2}.\] Letting \(\varepsilon\to 0\) we conclude \[\qquad \qquad \lim_{N \rightarrow \infty} \bigg\| \frac{1}{N} \sum_{n \in [N]} u_n \bigg\| = 0.\qquad \qquad \tag*{$\blacksquare$}\]
Remark. In fact, the proof gives a much stronger quantitative statement. Namely, for any \(0 < H \leq N\), one has \[\begin{aligned} \bigg\| \frac{1}{N} \sum_{n \in [N]} u_n \bigg\|\le \bigg(\frac{2}{H} \sum_{h=0}^{H-1} \bigg| \frac{1}{N}\sum_{n \in [N]} \langle u_{n}, u_{n+h} \rangle \bigg|\bigg)^{1/2} +\frac{4H^{1/2}}{N^{1/2}} \sup_{n \in \mathbb{N}} \|u_n\|. \end{aligned}\]