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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

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

A different solution: Let $(x_k)$ be a sequence in $\displaystyle l^2(\mathbb{N})$. Put $\displaystyle S_n=\sum_{k=1}^n x_k^2$, with $S_0=0$. The sequence $S_n$ converge to a limit $S$. We may suppose that $x_k\geq 0$ for all $k$; hence we have $x_k=\sqrt{S_k-S_{k-1}}$. Put: $$T_n=\frac{1}{\sqrt{n}}\sum_{k=1}^n x_k=\frac{1}{\sqrt{n}}\sum_{k=1}^n ...


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

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 ...


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, 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 ...


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

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 $S\in M'$. $TSx=STx=SU|T|x$ ...


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

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

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 \\ ...


1

I would like to clarify my final question as descibed by the discussion with @Joel above (see the comments). Let $\mathbf{w}=(w_1,\ldots,w_n)^T$, $\mathbf{x}_i=(x_{i1},\ldots,x_{in})^T\in\mathbb{R}^n$, $i=1,\ldots,m$, and $A=\big(a_{ij}\big)_{i,j=1}^{n}$ an $n\times n$ symmetric positive definite real matrix. Let's suppose that we would like to minimize ...


1

Suppose that we have $\tilde f = f + h$ as described. Let $V = span( k(\cdot, x_i) )_{i=1,...,m}$. We can write any function $f \in \mathcal{F}$ as $f = f_s + f_p$ where $f_s \in V$ and $f_p \in V^{\perp}$. This means that $$\tilde f(x_i) = f(x_i) + h(x_i) = f_s(x_i) + f_p(x_i) + h(x_i)$$ but since $f_p \in V^{\perp}$ we have that $f_p(x_i) =0$. Which ...


1

Let's introduce the projections $P=S^*S$ (onto $(\ker S)^\perp$), and $Q=TT^*$ (onto $\operatorname{ran}T$). The condition of $ST$ being partial isometry is $ST=ST(ST)^*ST$ which simplifies to $ST=SQPT$. Rewrite this further as $S(I-QP)T=0$. Since $\ker S=\ker P$ and $\operatorname{ran} T = \operatorname{ran} Q$, the latter is equivalent to $P(I-QP)Q =0$. ...


1

Assume there is a non-zero $v \in \mathcal{D}(Q)\setminus \mathcal{D}(A)$, and let $K$ be the one-dimensional subspace spanned by $v$. Then $K\cap \mathcal{D}(Q)=K$, and $K$ is closed in $H$. The restriction of the form $Q$ to $K$ is finite-dimensional, and positive-semidefinite. $B$ is trivially defined, but $\mathcal{D}(B)\ne\mathcal{D}(A)\cap K=\{0\}$. ...



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