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4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


2

The indices are different because the sums are independent from one another. If one intends to multiply the sums, then this distinction is a critical one. Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction. Then, we would have incorrectly $$\begin{align} ...


2

Example, $$ A = \frac{1}{i}\frac{d}{dx} $$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...


2

If $s \in S$, then $\mathcal{N}_s = \{ x \in H : \langle x,s \rangle = 0 \}$ is the null space of a continuous linear functional, which is the inverse image of $\{0\}$ under this continuous linear functional. Hence $\mathcal{N}_s$ is closed, as is the intersection $$ S^{\perp} = \bigcap_{s\in S}\mathcal{N}_s. $$


2

This is not an answer, and don't take it as such. From a theoretical point of view, you have $A=\sum_{k=1}^{n}\lambda_k P_k$ where the $P_k$ are orthogonal projections. If you start iterating in $A$, or perform various functions of $A$, the problem is that $f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k$. You can get down below the level of a projection. You can form ...


1

Let $T$ be the Fr├ęchet derivative of $f$ at $e\in E$. Then $$ \frac{|f_n(e+h)-f_n(e)-\langle Th,e_n\rangle|}{\|h\|} =\frac{|\langle f(e+h),e_n\rangle-\langle f(e),e_n\rangle-\langle Th,e_n\rangle}{\|h\|} =\frac{|\langle f(e+h)- f(e)- Th,e_n\rangle}{\|h\|} \leq\frac{\| f(e+h)- f(e)- Th\|}{\|h\|}\to0. $$ So the derivative of $f_n$ is $h\longmapsto \langle ...


1

If it is infinite dimensional, we can find an infinite orthonormal basis, $(e_n)_{n \in \Bbb{N}}$. Notice $||e_j|| = 1$ and $||e_i - e_j|| = \sqrt{2}$ for $i \neq j$. (This is a very standard construction.)


1

Yes, it is fine if you interpret $\langle D^2 f(x) , e_n \rangle$ as the bilinear form $$(y,z) \mapsto \langle D^2 f(x)[y,z], e_n \rangle.$$ As you already said, this follows simply from the chain rule and the linearity of $L_n$.


1

a) We have that $U_sU_{-s}=Id=U_{-s}U_s$ and $$<U_sf,g>=\int_{\mathbb{R}} U_sf(x)\overline{g(x)}\mathrm{d}x=\int_{\mathbb{R}} f(x-s)\overline{g(x)}\mathrm{d}x=\int_{\mathbb{R}} f(x)\overline{g(x+s)}\mathrm{d}x=<f,U_{-s}g>.$$ Hence $U_s^*=U_{-s}$. By the first identity $U_s$ is unitary. Try $V_s$ yourself. c) We have that $$(U_tV_sf-V_sU_tf)(x)= ...


1

In any infinite-dimensional Hilbert space $\mathcal H$, an orthonormal basis is not a basis in the algebraic sense. Suppose not: let $B$ be such an orthonormal basis. Let $b_1, b_2, \ldots$ be a sequence of distinct members of $B$ and $x = \sum_{j=1}^\infty b_j/j$. If $x = \sum_{b \in B} c_b b$ (where only finitely many $c_b \ne 0$), then $c_b = \langle b, ...


1

It's true that $S^\perp$ is closed. This is trivial from the definition. This does not imply that $S$ is closed, because in general $S\ne S^{\perp\perp}$. In fact $S=S^{\perp\perp}$ if and only if $S$ is a closed subspace of $H$; in general $S^{\perp\perp}$ is the closure of the span of $S$.


1

Yes, this is true. Indeed, let $t_n$ be a sequence in $S^{\perp}$ and let $t$ be its limit. We want to show that $t \in S^{\perp}$. By the continuity of the inner product, we have that, for all $s \in S$ $$ \langle t, s \rangle = \lim_{n \to \infty} \langle t_n, s \rangle=0 $$ so that $t \in S^{\perp}$. The equality $(S^{\perp})^{\perp}=S$ only holds if ...


1

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$


1

Orthornormality refers to the basis $e_i$. When a basis is orthonormal it means the inner product between any two elements of the basis $e_i,e_j$ is $\langle e_i, e_j \rangle = \delta_{ij}$ (see kronecker delta). More generally, two vectors $u,v$ are orthogonal if $\langle u, v \rangle = 0$. The normality part comes from elements of the basis having norm ...


1

That's because it is notationally bad and depending on how you read it, it may even give different results. For example, consider simple vectors in $\mathbb{R}^{3}$ given by $v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}$. Then if we use same index, \begin{equation} \begin{aligned} ...


1

Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write $T = UDU^*$, then $T^* = UD^* U^*$, giving that $T^*T^2 = UD^*D^2U^*$. However $T^3 = UD^3 U^*$. Since unitary conjugation does not change the operator norm, this boils down to considering $D^*D^2$ and $D^3$. Here $Dg(x) = f(x)g(x)$ for ...


1

Yes, but you have to verify this. Suppose $x=\sum f(v)v\in\operatorname{span}\beta$, where $v\in\beta$ and the sum is finite. Show that $\Vert Tx\Vert^2=\Vert x\Vert^2= \sum|f(v)|^2$ (use orthonormality of $\beta$). Thus, if $Tx=0$, we must have $x=0$, which means that $T$ is injective.



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