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5

Expanding gives us $f(x,y) = \displaystyle\sum_{n = 0}^{\infty}2^{-n}\left(n^4-2 n^3 x+n^2 x^2-2 n^2 y+2 n x y+y^2\right)$ We can show that $\displaystyle\sum_{n = 0}^{\infty}2^{-n} = 2$, $\displaystyle\sum_{n = 0}^{\infty}n2^{-n} = 2$, $\displaystyle\sum_{n = 0}^{\infty}n^22^{-n} = 6$, $\displaystyle\sum_{n = 0}^{\infty}n^32^{-n} = 26$, ...


5

Consider $\newcommand{\inner}[2]{\langle #1 \mspace{-3mu}\mid\mspace{-3mu} #2\rangle}$ $$\inner{u}{v} = \int_0^\infty u'(t)v'(t)p(t)\,dt.\tag{1}$$ $\inner{\cdot}{\cdot}$ is a bilinear form, and $\inner{u}{u} = \lVert u\rVert^2$ is zero only for $u \equiv 0$. Hence $\inner{\cdot}{\cdot}$ is an inner product inducing the norm. Every space whose norm is ...


4

Orthogonal projections are Hermitian positive semi-definite (with eigenvalues $0$ and/or $1$). The product of two positive semi-definite matrices (operators) is diagonalisable (see, e.g., Corollary 7.6.2 here -- unfortunately, it is not part of the preview). Therefore, the product of two orthogonal projections is diagonalisable. The proof is based on the ...


4

Yes, this follows from Wiener's Tauberian theorem, the $L^2$ case. Since the answer is short, I'll keep the old version below the cut. It was based on misreading of the question (RBF allow horizontal scaling). Yes, this is true. Key terms: radial basis function, universal approximation property. A classical reference (with a proof of more general ...


3

Being identified with is a loose term, it does not have a precise definition. It is used when we have a canonical isomorphism, i.e., an isomorphism that is naturally defined in terms of the objects themselves, without any choices made on our part. In our case, there is one canonical map from $X$ to $X^{**}$, namely $J$. If this map is an isomorphism, we can ...


3

Jimmy's response shows how to sum the series to find the polynomial $f(x,y)$, and suggests using standard derivative methods to minimize the function. Here's a hint for a faster way of finding the minimum of $f(x,y)$ without calculus: $$\begin{align} f(x,y)&=150-52x+6x^2-12y+4xy+2y^2\\ &=\frac23(3x+y-13)^2+\frac43(y^2+4y+28)\\ ...


3

No, let $V$ be separable and $e_n$ form an orthonormal basis in $V$, set $Ae_n=\frac1n e_n$. Then $\langle Au,u \rangle>0$ since every non-zero vector has at least one non-zero coefficient in its expansion, but there is no $\alpha$ since otherwise $\|Ae_n\|\geq\alpha$ for all $n$. The first condition is sometimes called "strictly positive definite" and ...


3

Clearly this is not answer you requested, but may be this will be helpful for someone else. We may regard $c_0$ as a closed subspace of $\mathcal{B}(H)$. Indeed, take any orthnormal basis of vectors $(e_i)_{i\in I}\subset H$ and for each $s\in c_0(I)$ consider bounded linear operator $T_s:H\to H$ well defined by $T(e_i)=s_ie_i$. Consider map ...


2

There is a more general fact: A C*-algebra $A$ is reflexive if and only if it is finite-dimensional. In the infinite-dimensional case, there is a normal element $x\in A$ with infinite spectrum (actually, one can find a positive element with infinite spectrum). Therefore, by the spectral theorem, $C^*(x)$ is an infinite-dimensional commutative C*-algebra, ...


2

If the $e_n$ are orthonormal, and $\lambda_{n_k}$ is a sequence with $|\lambda_{n_k}| \ge \delta >0 $, then $A e_{n_k}$ is a sequence such that the distance between any two elements is $\ge \sqrt{2} \delta$, hence cannot have a convergent subsequence.


2

Hilbert spaces are vector spaces, i.e. modules over a field, with an inner product. So the "smallest" must have smallest possible dimension for certain, if not a proper subspace exists: i.e. this "smallest" space is just a field. But then, the cardinalities of fields can go all the way down to $2$--if you allow finite fields--whence the answer would be ...


2

The point is, when the authors of your book write "hence $X$ can be identified with $X^{**}$" they mean that $J$ is onto (and hence an isomorphism, as $J$ is isometric by Hahn-Banach), but want to say that it is more or less straightforward to see it. Let us show this explicitly for Hilbert spaces. Let $H$ be a Hilbert space with inner product ...


2

For the first problem, use that any $L^2$-convergent sequence has a subsequence that converges almost everywhere. Alternatively, you can use that any finite-dimensional subspace of any normed vector space is closed. This essentially follows (in your case), because $$\Bbb{R} \rightarrow \Bbb{R}_+, x \mapsto \Vert x \cdot \chi_{[0,1]} \Vert_{L^2}$$ gives a ...


2

No. $A$ and $B$ span a $2$ dimensional subspace in $H$, the whole triangle will be in it. Restriction of inner product on $H$ to a $2$-dimensional subspace is very much Euclidean, and so are all triangles in it, with angle sum $180^\circ$. If one could get non-Euclidean geometry by restricting to a plane in a Hilbert space one could get it just as well by ...


2

@Vladmir asks if $T$ is bounded or closed, since this does not hold in general for linear operators. If the domain of your operator $T$ is all of $H$ and it is closed, then it is also bounded. For bounded linear operators, it is certainly true. That is because these operators are continuous and $$Tx = T\left(\lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n ...


2

Suggestion: start with a symmetric operator $C$ that is not essentially self adjoint, and $A$ with ${\mathcal D}(A) \subset {\mathcal D}(C)$ that is self adjoint. Take $B = C - A$.


2

Note that ${\rm Ker}\,U={\rm Ker}\, U^*U$. Thus $({\rm Ker}\,U)^\bot={\rm Im}\,(U^*U)$ because $U^*U$ is an orthogonal projection. So, for $x\in({\rm Ker}\,U)^\bot$ we have $U^*Ux=x$ that is $\langle x,U^*Ux\rangle=\langle x,x\rangle$ or equivalently $\Vert Ux\Vert=\Vert x\Vert$.


2

Assuming $\phi_k$ are complete, i.e. span a dense subspace in $H$, by the Parseval identity $\sum_{k=1}^{\infty}|\langle \psi_n-\psi,\phi_k\rangle_{H}|^2=\|\psi_n-\psi\|^2$. So the limit is $0$ if and only if $\psi_n$ converges to $\psi$ strongly. For weak convergence this is false. Take $\psi_n=\phi_n$, it converges weakly to $0$ like any orthonormal basis, ...


2

The polar decomposition of $T \in M$ is given by $T=U|T|$, where U is a partial isometry, $|T|=(T^{*}T)^{\frac{1}{2}}$ and $\text{ker }T= \text{ker }|T|=\text{ker }U$. Clearly, $|T| \in M$ since it is an SOT closed subalgebra of $B(H)$. To show that $U \in M$, it suffices to show $U \in M''$ by the double commutant theorem. Let $S\in M'$. $TSx=STx=SU|T|x$ ...


2

No, it is not true. In the edit to this question on mathoverflow there is a simple and explicit counterexample. Consider the pre-hilbert space $H = \ell^2_{f}$ consisting of sequences of finite support in $\ell^2$. Let $x \in \ell^2$ be the sequence $x_{n} = 1/n$ and $M = \{y \in H \mid \langle x,y\rangle = 0\}$. Then $M \subsetneqq H$ is closed and one can ...


1

To prove boundedness of $S\otimes T$, it suffices to prove it when $S$ and $T$ are unitaries (since the unitaries span $\mathcal{B}(H)$. Now if $$ z = \sum_{i=1}^n x_i\otimes y_i $$ where $\{y_1, y_2,\ldots, y_n\}$ are orthogonal. So $$ \|(S\otimes T)(z)\|^2 = \|\sum_{i=1}^n S(x_i)\otimes T(y_i)\|^2 $$ Since $\{y_i\}$ are orthogonal $$ = \sum_{i=1}^n ...


1

Branimir had posted an answer as a comment: 2. refines 1. because the representation constructed in 1. is, in general, a direct sum of infinitely many GNS representations, i.e., representations constructed from states on $\mathcal A$, whilst 2. gives you a single GNS representation that does the job.


1

No continuity is needed. Only linearity, positivity, and traciality. We can write $I=P+Q$ for two infinite projections, both equivalent to $I$. Explicitly, for matrix units $E_{kj} $ we can take $$ P=\sum E_{2n-1,2n-1},\ \ \ Q=\sum E_{2n,2n},\ \ \ V=\sum E_{2n,2n-1}, \ \ \ W=\sum E_{2n-1,n}. $$ Then $V^*V=P$, $VV^*=Q$, $W^*W=I$, $WW^*=P$. So $$ ...


1

For the constants: prove that the subspace is complete. In an Hilbert space, this is equivalent to being closed, and in this case it is easier. For the null integrals subspace: show that $\int f_n\to \int f$.


1

The remaining step that $\lVert U_v(f_0) - f_0\lVert_2 \to 0$ when $v \to 0$ holds for $f_0 \in C_c^{\infty}$ follows from two observations $f_0$ is uniformly continuous because it has compact support On a space $X$ with finite measure we have $\lVert f \rVert_2 \leq \mu(X)^{1/2} \lVert f \rVert_\infty$. Fix $\varepsilon > 0$. Since $f_0$ is uniformly ...


1

Try the following, if you want to avoid any Fourier series methods: first throw in the function $1$ into your set $\{\sin{kx}\}$. The resulting linear span is a subalgebra of $C(T,\mathbb{R})$, where $T$ is the one-dimensional torus. Then apply Stone-Weierstrass to obtain density in the $C$-norm. Now use the fact that $C(T,\mathbb{R})$ is dense in $L^2(T, ...


1

Hint: Let $L(x,y) = \lim_{n \to \infty} \langle y, T_n x \rangle$. You want to define $T$ so that $\langle y, T x \rangle = L(x,y)$. The Principle of Uniform Boundedness will be useful.


1

First of all, your requirements imply that for each $x \in H$, the sequence $\iota(T_n x)$ of bounded linear functionals, where $\iota$ denotes the canonical embedding of $H$ into the dual $H'$ by $\iota(x)(y) := \langle y, x\rangle$ is pointwise bounded. By the uniform boundedness principle, this implies that $\Vert T_n x \Vert$ is a bounded sequence for ...


1

Let $N=\{ (Ax,x) : \|x\|=1\}$. Suppose $\lambda \notin N^{c}$ so that there exists $\delta$ such that the following holds whenever $\|x\|=1$: $$ 0 < \delta \le |(Ax,x)-\lambda|=|((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\| = \|(A-\lambda I)x\| $$ Then $\|(A-\lambda I)x\| \ge \delta \|x\|$ for all $x$. (The same holds for ...


1

Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of $A$ is $$ (\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ ...



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