# Tag Info

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. $$2 This is not an answer, and don't take it as such. From a theoretical point of view, you have A=\sum_{k=1}^{n}\lambda_k P_k where the P_k are orthogonal projections. If you start iterating in A, or perform various functions of A, the problem is that f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k. You can get down below the level of a projection. You can form ... 1 Let T be the Fréchet derivative of f at e\in E. Then$$ \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$v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}. Then if we use same index, \begin{aligned} ... 1 Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) WriteT = UDU^*$, 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.

Only top voted, non community-wiki answers of a minimum length are eligible