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(1): The magical ingredient that makes Hilbert spaces behave so much nicer than general Banach spaces is the parallelogram identity: Lemma: Let $H$ be a Hilbert space and $x,y\in H$. Then $$\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$$ Proof: Verify using $\|x\|^2=\langle x,x\rangle$ and straightforward calculation. $\square$ Theorem: Let $H$ be a ...

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Bessaga showed something stronger, but only for Hilbert spaces. Generalization to certain Banach spaces (i.e., those which are linearly injectable into some $c_0(\Gamma)$) was given by Dobrowolski: In particular, in 1966 C. Bessaga [1] proved that every infinite-dimensional Hilbert space $H$ is $C^\infty$ diffeomorphic to its unit sphere. The key to prove ...

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Consider $H=L^2(\Omega,\mu)$ where $\Omega\subseteq\mathbb C$ is compact, and multiplication operator $T:H\to H,\ (Tf)(z)=zf(z).$ Then $T$ is bounded and normal. By the spectral theorem $T=\int_{\mathbb C}\lambda dE(\lambda).$ In this case you can give an explicit formula for the spectral measure $E:$ $$(E(X)f)(z)=\chi_X(z)f(z),$$ where $\chi_X$ is the ...

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This is an old joke that's done the rounds on the internet a few times. Stephan Rauh gave it a pretty good treatment on his blog last year, in which he concludes that it is indeed nonsense: Funny thing is it took me a while to figure out that the sentence really is utter nonsense. Most people immediately dismiss it as a joke – but they’ll never know ...

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$\forall x, y$, we have following two equations: \begin{align} <T(x+iy), x+iy> = 0 \\ <T(x+y), x+y> = 0 \end{align} Since $<Tx, x> = <Ty, y> = 0$, these two are equivalent to \begin{align} <Ty, x> - <Tx, y> = 0 \\ <Ty, x> + <Tx, y> = 0 \end{align} Because $x$ and $y$ are chosen arbitrarily, we ...

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In Banach space $X$ a sequence $\{f_n\}$ converges weakly to $f$ if $$\varphi(f_n)\to\varphi(f),$$ for all $\varphi\in X^*$, where $X^*$ is the dual of $X$. In the case of Hilbert space $H$, every element of the dual space is realized by an element of $H$ (Riesz Representation Theorem). Thus $f_n\to f$ weakly if and only if $$\langle ... 3 The shortest proof I know runs as follows. If f\in A^2 and if you write f(z)=\sum_0^\infty a_n(f) z^n then, using polar coordinates and Parseval formula you find that$$\Vert f\Vert_{A^2}^2=\pi\,\sum_{n=0}^\infty \frac{\vert a_n(f)\vert^2}{n+1}\cdot$$Conversely, if (a_n)_{n\geq 0} is a sequence of complex numbers such that \sum_0^\infty ... 2 The ingredients you have are thus S(w_n) is bounded, which guarantees the existence of weakly convergent subsequences, and S is such that the only possible limit of a subsequence is S(w) (by pointwise a.e. convergence of a further subsequence). That means the full sequence S(w_n) converges weakly to S(w), because of the Theorem: Let (x_n) be ... 2 The polynomial algebra \mathbb{C}[T,T^*] is dense in A. A polynomial p(T,T^*) is sent to the corresponding polynomial function p(z,\overline{z}) on the spectrum of T. The general description then comes from continuity. For example, when ||T|| < 1, the element \frac{1}{1-T} = \sum_{k=0}^{\infty} T^k is sent to \frac{1}{1-z} = ... 2 Note that \langle x, e_n \rangle = x_n since (e_n) forms an orthonormal basis. Similarly \langle y,e_n \rangle = y_n Then what you asked simplifies to$$ \langle x,y \rangle = \sum_{n} \langle x , e_n \rangle \langle y , e_n \rangle = \sum_{n} x_n y_n, $$as you are familiar with seeing in the Euclidean case, and have likely seen as the inner product ... 2 In order to make the inner product well-defined, we talk about L^2(\Omega,\mathcal F,\mu), where (\Omega,\mathcal F,\mu) is the underlying probability space. But we then extend condition expectation to integrable random variables. We use a projection over the closed subspace L^2(\Omega,\mathcal N,\mu), that is, the vector subspace which consists of ... 2 Assuming the axiom of choice, yes. Let A' be the unique bounded extension. Pick a Hamel basis B for \mathcal{D} and extend it to a Hamel basis B' for \mathcal{H}. Now define \tilde{A} on B' \setminus B any way you like, as long as it's different from A'. The result will necessarily be an unbounded linear operator. 2 No. But it has the following universal property: H represents the contravariant functor which maps a Hilbert space K to the set of bounded families of bounded linear maps g_i : K \to H_i which are l^2-summable in the sense that \sum_{i \in I} ||g_i(x)||^2<\infty for all x \in K. Because then we can define g : K \to H by g(k)=(g_i(x)). 2 A L^2-Riemann Cauchy sequence of L^2-Riemann functions can have a nonintegrable limit (see the example of David Mitra). In general, The Riemann integral behaves much worse that the Lebesge integral. Example: pick a enumeration of rationals in [0,1], r_1,r_2,\cdots. The sequence$$f_n=\chi_{\{r_1,\cdots,r_n\}}$$of Riemann integrable functions has a ... 2 The answer is yes. This is particularly easy to prove using the "multiplication operator" version of the spectral theorem, for which a good reference is Section VIII.3 of Reed and Simon. Consider the special case that \mathcal{H} = L^2(X,\mu) for some finite measure space (X,\mu), and that P is multiplication by some real-valued measurable function ... 2 An element a of C^*-algebra is called an isometry if a^* a = 1. This coincides with the usual notion when applied to B(H). Maybe we can just generalize this terminology to the setting of dagger categories. So call a morphism f an isometry if f^{\dagger} f = 1. I don't know if this is standard. 2 Fix a basis v_1,\ldots,v_m \in V_k. Then the map \sum \alpha_jv_j\to\left(\sum |\alpha_j|^2\right)^{1/2} defines a norm on V_k, and this norm is induced by the inner product \langle \sum \alpha_jv_j,\sum \beta_jv_j\rangle = \sum \alpha_j\overline{\beta_j}. In a finite-dimensional space all norms are equivalent, so the identity map becomes a ... 2 I started writing this as comments, but then ran out of space to give a satisfactory reply. (I rant too much to be confined to 400 characters!) At any rate, the answer is: no. But some clarification is necessary. I'm old and forgetful, so I will note John Baez describes the basic algorithm to geometric quantization fairly well. Examples Worth Considering ... 2 Somehow, on the whole internet, it seems that the simplest proof of Cauchy- Schwarz has yet to be recorded. At least I couldn't find it after several minutes of searching... The most prominent is certainly the proof mentioned by Daniel Fischer in this comment above, but that always seemed quite contrived to me. Here is the best'' proof imho: let x,y ... 1 No matter what, the projection is still orthogonal, so it still takes the form$$\tilde h = P_m(h) = \sum_{j=1}^m a_jv_j$$i.e. it is in the linear span of the v_j's. If the v_j's are orthonormal, then the coefficients are "easy" to calculate:$$ a_j = \left<h,v_j\right> $$In general, the coefficients have to satisfy a system of equations. In ... 1 It is true if you assume that T is self-adjoint (i.e. symmetric), meaning that$$ (Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1} $$and assuming that$$ |(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2} $$Note that your inequality holds even for T=-2I, and thus we NEED to assume these two additional things: (1) and (2). So ... 1 Your proof seems perfect. (Well, a small thing: \ker(L) is closed since L is continuous linear..) If x\in K\subseteq H, then let y:=L(x)\,\in H', then by definition we have S(y)=x. The last part (that S is determined uniquely by the mentioned properties) is still to be proved. 1 This is a very interesting issue. In fact, this isomorphism is not as "natural" as one might have thought. As an exercise, one should see that, given a map of Hilbert spaces V\to W it is rarely the case that the square of maps involving W^*\to V^* and the "Riesz-Fisher" dualities ... commutes. This is fairly crazy, yes, given the standard curriculum. A ... 1 The trick is here that the norms (and scalar products) of L^2(U) and H^1(U) are not the same, which means that the closure of the orthogonal basis (or complete orthogonal set) will be different, depending on the process of completion. The general idea is that the inclusion of the derivative in the Sobolev norm will not only make sure that for a Cauchy ... 1 The statement in the comment is correct. The reasoning in the question is also correct on the technical level, but I can't really tell how adequate it is in the context where you'll put it. Generally, by the time people learn what a Hilbert space is, they already know how to prove that a linear subspace of \mathbb R^n is closed (because they know enough ... 1 Let M be the diameter of B. Each of the closed sets has diameter < M, and is therefore contained in a closed ball that also has diameter <M. If the covering by arbitrary sets was finite, this would give a covering by finitely many closed balls each of whose diameters was less than M. 1 The notation \mathbb{H} = \operatorname{lin} \{ e_1, ..., e_n \} means that any point v \in \mathbb{H} can be written as a linear combination of the e_k. So let v \in \mathbb{H} and write v = \sum_k x_k e_k. Then note that \langle e_i , v \rangle = \sum_k x_k \langle e_i , e_k \rangle = x_i, since the e_k are orthonormal (and so \langle e_i , ... 1 Let \Phi:B(H,K)\rightarrow S(H,K) be the morphism you defined. Given \phi\in B(H,K), you can verify that \Vert\phi\Vert=\sup\left\{|\langle\phi(x),y\rangle|:x\in H, y\in K, \Vert x\Vert\leq 1,\Vert y\Vert\leq 1\right\}=\Vert\langle\phi(\cdot),\cdot\rangle\Vert=\Vert\Phi(\phi)\Vert, so \Phi is an isometry. Also, it follows (almost directly) from Riesz ... 1 I'll assume that you meant for \{ x_{n}\} to be a sequence of unit vectors. Otherwise, you could let x_{n}=0 for all n, and there would exist such a sequence regardless of whether or not i\lambda \in\sigma(T). Note that there is nothing gained by introducing i into the discussion, especially because you made no assumption about \lambda being ... 1 By the Lebesgue differentiation theorem,$$(Df)^\prime(t)=\frac{d}{dt} \int_0^t f(s)ds=f(t)$$for a.e. t\in[0,1]. Therefore,$$\langle Df, Dg\rangle_{C^\prime}=\int_0^1 (Df)^\prime(t) (Dg)^\prime(t) dt = \int_0^1 f(t) g(t) dt = \langle f,g\rangle_{L^2[0,1]} so $D$ preserves the inner product. By the characterization of absolutely continuous functions by ...

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