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We have $$\require{cancel}P^2(v)=P(P(v))=P\left(\frac{\langle v,w\rangle }{||w||^2}w\right)=\frac{\langle v,w\rangle }{||w||^2}P\left(w\right)=\frac{\langle v,w\rangle }{||w||^2}\cancelto{=1}{\frac{\langle w,w\rangle }{||w||^2}}w=P(v)$$ Moreover, we have by the Cauchy-Schwarz inequality $$||P(v)||=\left|\frac{\langle v,w\rangle ... 3 Suppose that f=\{ f_{0},f_{1},f_{2},\cdots\} \in l^{2} is orthogonal to every such f_{p}. Let$$ F(z)=\sum_{n=0}^{\infty}f_{n}z^{n},\;\;\; |z| < 1. $$F converges absolutely in |z| < 1 because the Cauchy-Schwartz inequality implies$$ \left[\sum_{n=0}^{\infty}|f_{n}||z|^{n}\right]^{2} \le ...

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One can prove strong convergence following this article: The product of projection operators I. Halperin http://acta.fyx.hu/acta/showCustomerArticle.action?id=7164&dataObjectType=article The original result is due to Neumann. Halperin's note generalizes this to $m$ projections: There it is proven that $(P_1P_2\dots P_m)^n x \to z$ with ...

2

As $A$ and $B$ commute, you have $\langle (A^2-B^2)x,x\rangle=0$ for all $x$, so $A^2=B^2$. Now you can use that both $A,B$ are positive, so $A$ is the unique positive square root of $A^2$, and likewise with $B$. So $A=B$. I don't immediately see how to prove that $A=B$ in the kernel of $A+B$ other than doing what I did above, which shows $A=B$ everywhere. ...

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I think I have answered this for you before, but I'll explain it again in more detail. The proof hinges on the following lemma, whose proof is given at the end. Perhaps you'll accept this answer. Lemma Let $X$ be a complex Hilbert space and let $T\in\mathcal{L}(X)$ satisfy $(Tx,x) \ge 0$ for all $x \in X$. Then $(Tx,x)=0$ iff $Tx=0$. That is, ...

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This is correct with the right interpretation of the symbol $\oplus$. On the left it should denote the disjoint union, on the right the Hilbert space sum. You should also be more explicit about the measures you are using (perhaps the spaces are Riemannian manifolds with their natural measures).

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For selfadjoint operators we know that $$\inf\sigma(T)=\inf\{\langle Tx,x\rangle:x\in S_H\}\\ \sup\sigma(T)=\sup\{\langle Tx,x\rangle:x\in S_H\}$$ It is remains to apply result of this answer.

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I don't see the point of your last computation: you have already obtained what you are looking for before the "And if we take..." You got your $e_i$ and $f_i$ in the wrong spots, but that's just a labeling issue: you should have started with $\{f_i\}$ as the orthonormal basis. On a deeper side, why would $\{T^*e_i\}$ and orthonormal basis? In general, it ...

1

Let $\{ e_{j}\}_{j=1}^{N}$ be an orthonormal basis of $\mathcal{R}(T^{\star}T)$--which is finite-dimensional and, hence, closed--and extend this to a full orthonormal basis of $X$ by adding an orthonormal basis $\{ e_{j} \}_{j=N+1}^{\infty}$ of $\mathcal{R}(T^{\star}T)^{\perp}=\mathcal{N}(T^{\star}T)=\mathcal{N}(T)$. Then $Te_{j}=0$ for $j > N$, and $$... 1 The idea is to divide the inequality$$ |\zeta|^2 \|z\|^2 + 2\Re \langle y ,\zeta z\rangle \ge0 $$by \zeta for suitable values of \zeta. First, take \zeta\in \mathbb R, \zeta> 0, then divide by \zeta, which gives$$ |\zeta| \|z\|^2 + 2\Re \langle y , z\rangle \ge0, $$then \zeta\searrow 0. Do the same for a negative \zeta. At the end, ... 1 It is like you are saying. If you write P_V for the orthogonal projection onto V, then$$ \min\{\|x-x_0\|:\ x\in V\}=\|P_Vx_0-x_0\|=\|(I-P_V)x_0\|=\|P_{V^\perp}x_0\|=\max\{|\langle P_{V^\perp}x_0,y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\} =\max\{|\langle x_0,y\rangle|:\ ...

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Let $H$ be a complex Hilbert space and let $T \in B(H)$ be an isometry. We claim that $T^\ast T = I$. To this end let $h \in H$ and note that by the Hilbert space Riesz representation theorem a linear functional in $H^\ast$ corresponds to an element $h \in H$ ($h \mapsto \langle \cdot, h \rangle$). Also note that if $\varphi (x) = 0$ for all $\varphi \in ... 1 In fact if$T$is non-expansive then the proposed inequality follow from the Cauchy-Schwartz inequality, since with notation of the proposer we have $$\langle T(x)-T(y),x-y\rangle= \langle A(x-y),x-y\rangle\leq\Vert A(x-y)\Vert\,\vert x-y\Vert\leq\Vert x-y\Vert^2.$$ Now, the converse is not correct in general without supplementary assumptions, for ... 1 There are certainly instances where this is false. As Jochen said, if this is an orthonormal basis, then$span(\beta-\beta_1) \subset \{\beta_1\}^\perp$and so is not dense. There are bases where it is possible to remove one or any finite number of terms and keep density, these are called over complete bases. For instance in a RKHS, the span of the kernel ... 1 I think the author is being a little sloppy in this paragraph, which raised your question: Lower semicontinuity was earlier defined by the openness of upper level sets. The weak topology being non-metrizable, one should not casually insert "in other words" followed by the definition of sequential weak lower semicontinuity. For example, Ciarlet ... 1 A Hilbert space is an inner product space that is complete with respect to the norm. Completeness is what differentiates the two. Not every metric space can be defined by an inner product, for instance the space of continuous functions on$[0,1]$with the supremum norm as its metric doesn't come from an inner product. In particular, a Banach space (a ... 1 This question has been addressed ad infinitum. For proof that$C^0$is complete with the supremum norm (hence is a Banach space) see here: How to show that$C=C[0,1]$is a Banach space For proof that the supremum norm arises from no inner product (and hence is not a Hilbert space) see here:$C[0,1]$is not Hilbert space 1 The notation$|T|$means the unique positive square root of$T^{\star}T$. Notice that $$\||T|x\|^{2}=(|T|x,|T|x)=(|T|^{2}x,x)=(T^{\star}Tx,x)=(Tx,Tx)=\|Tx\|^{2}.$$ So$\mathcal{N}(T)=\mathcal{N}(|T|)$. Let$Y=\mathcal{R}(|T|)^{c}$. Then$|T| : Y\rightarrow Y$is invertible because$\mathcal{N}(|T|)\cap ...

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I misinterpreted your question. I thought you wanted to see the standard construction of the $V$ where $V^{\star}V=P$ is the orthogonal projection onto $\mathcal{R}(|T|)^{c}$. This $V$ is unique in all cases, even where $\mathcal{N}(T)\ne \{0\}$, whereas a general partial isometry $V$ is not uniquely determined if $T$ has a non-trivial null space. $V$ is ...

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