Problems to Solve

$I^{*} = I, 0^{*} = 0$;

$(\alpha A+\beta B)^{*} = \bar{\alpha } A^{*}+\bar{\beta } B^{*}, \alpha , \beta \in \mathbb {C}$;

$(A B)^{*} = B^{*} A^{*}$;

if $A = A^{*}, B^{*} = B$, then $(A B)^{*} = A B \Leftrightarrow A B = B A$;

if $A^{-1} \in \mathcal{L}(H)$ exists, then $\left(A^{*}\right)^{-1} \in \mathcal{L}(H)$ also exists, and $\left(A^{*}\right)^{-1} = \left(A^{-1}\right)^{*}$.

if $\mathbb {K} = \mathbb {C}$, then

(polarization identity);

if $\mathbb {K} = \mathbb {R}$, then the form $b$ is symmetric (i.e., $b(x, y) = b(y, x), x, y \in H$) if and only if $4 b(x, y) = b\left[x+y\right]-b\left[x-y\right], x, y \in H$.

the bilinear form $b$ is bounded;

there exists an operator $A \in \mathcal{L}(H)$ such that $b(x, y) = (A x, y), x, y \in H$; in this case $\left\| b\right\| = \left\| A\right\|$;

there exists an operator $C \in \mathcal{L}(H)$ such that $b(x, y) = (x, C y), x, y \in H$; in this case $\left\| b\right\| = \left\| C\right\|$.

$b(x, y) = \frac{1}{N} \sum_{k = 1}^{N} b\left[x+e^{\frac{2 \pi i k}{N}} y\right] e^{\frac{2 \pi i k}{N}}, x, y \in H, N \in \mathbb {N}, N \geq 3$;

$b(x, y) = \frac{1}{2 \pi } \int_{0}^{2 \pi } b\left[x+e^{i \varphi } y\right] e^{i \varphi } d \varphi , x, y \in H$.

operator $A \in \mathcal{L}(H)$ is self-adjoint;

the bilinear form $b_{A}$, induced by the operator $A$, is Hermitian;

the quadratic form $b_{A}\left[x\right] = b_{A}(x, x), x \in H$, attains only real values.

Let $H$ be a complex Hilbert space, $A \in \mathcal{L}(H)$. Prove that $A = 0$ if and only if $(A x, x) = 0$ for all $x \in H$.

Provide an example of a real Hilbert space $H$ and a nonzero operator $A \in \mathcal{L}(H)$ such that $(A x, x) = 0$ for all $x \in H$.

if $A^{*} = A$, then $A^{2} \geq 0$;

the condition $A^{*} = A$ in item (1) is essential;

if $A \leq B, B \leq A$, then $A = B$;

if $A \leq B, B \leq C$, then $A \leq C$;

if $A \leq C, B \leq D$, then $A+B \leq C+D$;

if $A \leq B, \lambda \geq 0$, then $\lambda A \leq \lambda B$;

if $A \leq B$, then $-B \leq -A$;

if $A^{*} = A, B^{*} = B$ and $A B = B A$, then $A^{2}+B^{2} \geq 2 A B$;

if $A^{*} = A$, then there exist $m, M \in \mathbb {R}$ such that $m I \leq A \leq M I$;

if $A^{*} = A, B^{*} = B$, then it is not necessarily that $A \leq B$ or $B \leq A$;

if $A^{*} = A, A \geq 0$ and $A^{-1} \in \mathcal{L}(H)$, then $A^{-1} \geq 0$;

if $A$ and $B$ are unitary operators, then $A B$ is also a unitary operator;

if $A^{*} = A, B^{*} = B, A \geq B$, then the operators $C^{*} A C, C^{*} B C$ are selfadjoint and $C^{*} A C \geq C^{*} B C$;

if $A^{2} = A$, then $R(A) = \operatorname {Ker}(I-A)$, in particular, the set $R(A)$ is a subspace of $H$;

if $\mathbb {K} = \mathbb {C}, \alpha , \beta \in \mathbb {C},\left|\alpha \right| = \left|\beta \right| = 1$, then the operator $\alpha A+\beta A^{*}$ is normal;

if $A$ is an orthoprojector, then $A \geq 0$;

if $A$ is an orthoprojector, then $I-A$ is also an orthoprojector;

if $\mathbb {K} = \mathbb {C}, A = A^{*}, B = B^{*}$, then the operator $C = i(A B-B A)$ is selfadjoint.

$A$ is isometric $\Leftrightarrow \left\| A x\right\| = \left\| x\right\| , x \in H$;

if $H$ is a finite-dimensional space, then $A$ is unitary if and only if $A$ is isometric;

$A$ is isometric $\Leftrightarrow A^{*} A = I$;

$A$ is unitary $\Leftrightarrow A^{*} A = I$ and $A A^{*} = I$;

if $H$ is an infinite-dimensional space, there exists $A$ which is isometric but not unitary.

$A$ is self-adjoint $\Leftrightarrow \mathcal{A}^{*} = \mathcal{A}$;

$A$ is unitary $\Leftrightarrow A$ is isometric $\Leftrightarrow \mathcal{A}^{*} \mathcal{A} = \mathcal{J}$;

$A$ is nonnegative $\Leftrightarrow \mathcal{A}^{*} = \mathcal{A}$ and all roots of the equation $\operatorname {det}\left(\mathcal{A}-\lambda \mathcal{J}\right) = 0$ are nonnegative;

if operator $A$ in Hilbert space $\mathbb {R}^{m}$ is represented by matrix $\mathcal{A}$ and $\mathcal{A}^{*} = \mathcal{A}^{T}$ is the transposed matrix, then statements (1)-(3) are true, but the condition $(A x, x) \geq 0, x \in \mathbb {R}^{m}$, does not imply that operator $A$ is self-adjoint, because it is not necessarily $\mathcal{A}^{*} = \mathcal{A}$.

$A x = \left(\alpha_{1} x_{1}, \alpha_{2} x_{2}, \ldots \right)$;

$A x = \left(x_{2}, x_{3}, \ldots \right)$;

$A x = \left(x_{1}, \ldots , x_{j}, 0,0, \ldots \right)$;

$A x = (\underbrace{0, \ldots , 0}_{j-1}, x_{1}, 0,0, \ldots )$;

$A x = (\underbrace{x_{1}, \ldots , x_{1}}_{j}, 0,0, \ldots )$;

$A x = \left(x_{2}, x_{1}, x_{4}, x_{3}, x_{6}, x_{5}, \ldots \right)$;

$A x = \left(\alpha_{j} x_{j}, \alpha_{j+1} x_{j+1}, \ldots \right)$;

$A x = \left(0,0, \alpha_{1} x_{1}, \alpha_{2} x_{2}, \ldots \right)$;

$A x = \left(2 x_{1}+3 x_{2}, 3 x_{1}+5 x_{2}, x_{3}, 0,0, \ldots \right)$;

$A x = \left(x_{1}, 4 x_{2}+7 x_{3}, 7 x_{2}+2 x_{3}, 0, x_{5}, 0,0, \ldots \right)$;

$A x = \left(3 x_{1}-2 x_{2}, x_{2}, x_{3}, x_{4}, \ldots \right)$, where $j \in \mathbb {N}$ is fixed, and $\left\{ \alpha_{n}: n \geq 1\right\} \subset \mathbb {C}$ is a fixed bounded sequence.

$H = L_{2}(\mathbb {R}),(A x)(t) = \chi_{\left[-\alpha , \alpha \right]}(t) x(t)$, where $\alpha > 0$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = t x(t)$;

$H = L_{2}(\mathbb {R}),(A x)(t) = a(t) x(t+\tau )$, where $a \in L_{\infty }(\mathbb {R})$ is a given function, $\tau \in \mathbb {R} ;$

$H = L_{2}(\mathbb {R}),(A x)(t) = \frac{1}{2}(x(t)+x(-t))$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = \int_{0}^{1} e^{t+s} x(s) d s$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = t^{3} \int_{0}^{t^{2}} x(s) d s$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = \int_{0}^{1} t^{2} s^{3} x(s) d s$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = \int_{t^{5}}^{t} t s x(s) d s$;

$H = L_{2}(\left[-1,1\right]),(A x)(t) = \int_{t^{2}}^{1}\left(t-3 s^{2}\right) x(s) d s$;

$H = L_{2}(\left[0,1\right]),(A x)(t) = x\left(t^{\alpha }\right)$, where $0 < \alpha \leq 1$;

$A x = (x, y) z$, where $y, z \in H$ are fixed elements.

$H = L_{2}(\mathbb {R}),(A x)(t) = \sqrt{7} t^{3} x\left(t^{7}-1\right)$;

$H = L_{2}([0, \infty )),(A x)(t) = 2 t \sqrt{t} x\left(t^{4}\right)$.

$A$ is self-adjoint;

$A$ is normal;

$A$ is unitary;

$A$ is isometric;

$A$ is an orthogonal projector;

$\operatorname {Ker} A = \left\{ 0\right\}$.

$A = A^{*} \Leftrightarrow$ the function $p$ is almost everywhere real;

$A$ is unitary $\Leftrightarrow \left|p\right| = 1$ almost everywhere;

$A \geq 0 \Leftrightarrow p \geq 0$ almost everywhere;

$A$ is an orthogonal projector $\Leftrightarrow p^{2} = p$ almost everywhere, i.e. $p = \chi_{C}(\bmod \mu )$, where $C \in \mathcal{F}$ is a measurable set.

Find $A_{s}^{*}$ and $A_{s}^{-1}$ and prove that $A_{s}$ is a unitary operator.

Let $A = \frac{1}{2}\left(A_{s_{0}}+A_{s_{1}}\right), s_{0}, s_{1} \in \mathbb {R}$. Prove that $A^{*} = A$ if and only if $s_{0}+s_{1} = 0$.

Find the orthogonal projector onto the linear span of the set $\left\{ e_{1}, \ldots , e_{k}\right\} , k \in \mathbb {N}$.

Prove that the operator $A x = \sum_{n = 1}^{\infty } x_{\sigma (n)} e_{n}, x = \sum_{n = 1}^{\infty } x_{n} e_{n} \in H$, where $\sigma : \mathbb {N} \rightarrow \mathbb {N}$ is a fixed bijection, is a unitary operator.

$A$ is normal, but not self-adjoint;

$A^{2} = A$, but $A$ is not an orthogonal projector;

$A$ is a unitary operator, but not an orthogonal projector;

$A^{*} = A$, but does not satisfy either of the inequalities $A \geq 0, A \leq 0$;

$A^{2} = I$, but does not satisfy either of the inequalities $A \geq 0, A \leq 0$;

$A^{*} = A$, but the range $R(A)$ is not closed in $H$;

$A^{*} = A$, but the set $\left\{ (A x, x) \mid x \in H,\left\| x\right\| = 1\right\}$ is not closed in $\mathbb {R}$;

$A \geq 0, B \geq 0$, but $A B$ is not positive;

$A$ is not normal;

$A \neq 0$, but $A^{2} = 0$;

$A = A^{*}, B = B^{*}, A^{2} = B^{2}$, but $A B \neq B A$.

$\left|(A x, y)\right|^{2} \leq (A x, x)(A y, y), x, y \in H$;

$\left\| A x\right\|^{2} \leq \left\| A\right\| (A x, x), x \in H$.

$A^{*} = A$;

$A \geq 0$?

$A^{n} x \neq 0$ for all $n \in \mathbb {N}$;

the sequence $\left\{ \alpha_{n} = \frac{\left\| A^{n+1} x\right\| }{\left\| A^{n} x\right\| }: n \geq 1\right\}$ converges.

$\left\| A\right\| = \sup_{x, y \in S(0,1)}\left|(A x, y)\right|$;

if $A^{*} = A$, then $\left\| A\right\| = \sup_{x \in S(0,1)}\left|(A x, x)\right|$;

if $A^{*} = A$, then $A = 0$ if and only if $(A x, x)$ = 0 for all $x \in H$.

the operators $\mathrm{Re}\left(A\right) = \frac{1}{2}\left(A+A^{*}\right), \mathrm{Im}\left(A\right) = \frac{1}{2 i}\left(A-A^{*}\right)$ are self-adjoint;

the operator $A$ can be uniquely represented in the form $A = B+i C$, where $B$ and $C$ are self-adjoint operators;

the operator $A$ is normal $\Leftrightarrow \mathrm{Re}\left(A\right) \cdot \mathrm{Im}\left(A\right) = \mathrm{Im}\left(A\right) \cdot \mathrm{Re}\left(A\right)$;

the operator $A$ is unitary if and only if the following conditions are fulfilled:

(a) $A$ is normal;

(b) $(\mathrm{Re}\left(A\right))^{2}+(\mathrm{Im}\left(A\right))^{2} = I$.

$\operatorname {Ker} A = \operatorname {Ker} A^{*} = (R(A))^{\perp }$;

$\left\| A^{2}\right\| = \left\| A\right\|^{2}$;

if $A^{2} = 0$, then $A = 0$;

if $A^{-1} \in \mathcal{L}(H)$ exists, then $A^{-1}$ is a normal operator.

$P$ is an orthoprojector;

$P = P^{*}$;

$P P^{*} = P^{*} P$;

$\operatorname {Ker} P = (R(P))^{\perp }$;

$(P x, x) = \left\| P x\right\|^{2}, x \in H$.

$(P x, x) \geq 0, x \in H$.

$P_{1}+P_{2}$ is an orthogonal projector $\Leftrightarrow P_{1} P_{2} = 0 \Leftrightarrow H_{1} \perp H_{2}$; in this case $P_{1}+P_{2}$ is an orthogonal projector onto $H_{1} \oplus H_{2}$;

$P_{1} P_{2}$ is an orthogonal projector $\Leftrightarrow P_{1} P_{2} = P_{2} P_{1} \Leftrightarrow$ $\Leftrightarrow \left(H_{1} \ominus H_{0}\right) \perp \left(H_{2} \ominus H_{0}\right)$; in this case $P_{1} P_{2}$ is an orthogonal projector onto $H_{0}$;

$P_{1}-P_{2}$ is an orthogonal projector $\Leftrightarrow P_{1} \geq P_{2} \Leftrightarrow H_{1} \supset H_{2}$; in this case $P_{1}-P_{2}$ is an orthogonal projector onto $H_{1} \ominus H_{0}$;

$P_{1} P_{2} = P_{2} \Leftrightarrow P_{2} P_{1} = P_{2} \Leftrightarrow P_{1} \geq P_{2} \Leftrightarrow H_{1} \supset H_{2}$.

For an operator $A \in \mathcal{L}(H)$ express in terms of algebraic relations between $A$ and $P_{1}$ the following statements:

$H_{1}$ is invariant with respect to $A$ (i.e., $A\left(H_{1}\right) \subset H_{1}$);

both $H_{1}$ and $H_{1}^{\perp }$ are invariant with respect to $A$.

$A_{n} \rightrightarrows A \Leftrightarrow A_{n}^{*} \rightrightarrows A^{*}$;

$A_{n} \xrightarrow {w} A \Leftrightarrow A_{n}^{*} \xrightarrow {w} A^{*}$;

$A_{n} \xrightarrow {s} A \Leftrightarrow A_{n}^{*} \xrightarrow {s} A^{*}$.

if $A_{n} = A_{n}^{*}, n \geq 1, A_{n} \xrightarrow {w} A, n \rightarrow \infty$, then $A = A^{*}$;

if $A_{n} \geq 0, n \geq 1, A_{n} \xrightarrow {w} A, n \rightarrow \infty$, then $A \geq 0$;

if $A_{n} \xrightarrow {w} A$ and $\left\| A_{n} x\right\| \rightarrow \left\| A x\right\|$ for all $x \in H$, then $A_{n} \xrightarrow {s} A$; in particular, if $A, A_{n}, n \geq 1$, are isometric, then $A_{n} \xrightarrow {w} A \Leftrightarrow A_{n} \xrightarrow {s} A$;

if $A_{n}, n \geq 1$, and $A$ are normal, $A_{n} \xrightarrow {s} A$, then $A_{n}^{*} \xrightarrow {s} A^{*}$;

if $A_{n}$ are isometric for each $n \geq 1$, and $A_{n} \xrightarrow {w} A$, then $A$ is not necessarily isometric;

if $A_{n}$ are unitary for each $n \geq 1$, and $A_{n} \xrightarrow {w} A$, then $A$ is not necessarily unitary.

if $P_{n+1} \geq P_{n}, n \geq 1$, then $P_{n} \xrightarrow {s} P$, where $P$ is an orthogonal projector onto $\operatorname {span}\left(\cup_{n = 1}^{\infty } H_{n}\right) ;$

if $P_{n+1} \leq P_{n}, n \geq 1$, then $P_{n} \xrightarrow {s} P$, where $P$ is an orthogonal projector onto $\bigcap_{n = 1}^{\infty } H_{n} ;$

if $H_{k} \perp H_{j}, k \neq j, k, j \geq 1$, then the series $\sum_{n = 1}^{\infty } P_{n}$ converges strongly to the orthogonal projector onto $\overline{\operatorname {span}\left(\bigcup_{n = 1}^{\infty } H_{n}\right)}$;

if $P_{n} \xrightarrow {s} P$, then $P$ is an orthogonal projector;

if the sequence $\left\{ P_{n}: n \geq 1\right\}$ is monotone and $P_{n} \rightrightarrows P, n \rightarrow \infty$, then there exists $n_{0} \in \mathbb {N}$ such that $P_{n} = P, n \geq n_{0}$. Is the monotonicity essential in this statement?

if $P_{n} \xrightarrow {w} P$, then $P$ is not necessarily an orthoprojector;

if $P_{n} \xrightarrow {w} P$ and $P$ is an orthoprojector, then $P_{n} \xrightarrow {s} P$.

if $A \geq 0$, then $A^{-1} \in \mathcal{L}(H)$ exists if and only if there exists $\lambda > 0$ such that $A \geq \lambda I ;$

there exists $\left(I+A A^{*}\right)^{-1} \in \mathcal{L}(H)$;

if $A^{*} = A, m = \inf_{x \in S(0,1)}(A x, x), M = \sup_{x \in S(0,1)}(A x, x)$, then the operator $A-\lambda I$ is continuously invertible for all $\lambda \in \mathbb {R} \backslash [m, M]$ and is not continuously invertible for $\lambda \in \left\{ m, M\right\}$;

if $A^{*} = A, \lambda = \alpha +\beta i \in \mathbb {C}$, then