New answers tagged

1

It can be done easier. Observe that \begin{equation*} \begin{split} \|x_n-x\|^2 & = \langle x_n-x, x_n-x\rangle = \langle x_n, x_n\rangle - \langle x, x_n\rangle - \langle x_n, x\rangle + \langle x, x\rangle \\ & = \|x_n\|^2 - \langle x, x_n\rangle - \langle x_n, x\rangle + \|x\|^2 \end{split} \end{equation*} From the hypothesis, $$\|x_n\| \to \|x\|,...


0

$$\|x_n-x_0\|^2 = | \langle x_n-x_0, x_n-x_0\rangle| = | \langle x_n, x_n-x_0\rangle| - \langle x_0, x_n-x_0\rangle| \leq | \langle x_n, x_n-x_0\rangle | + | \langle x_0, x_n-x_0\rangle|$$ Now because $\langle x_n, x_0\rangle \to \|x_0\|^2 \implies \lim_\limits{n\to \infty} | \langle x_n - x_0, x_0\rangle | = 0$ then the second term: $$ | \langle x_0, x_n-...


2

Let $X$ be a normed TVS. Fix $x\in X.$ Now, $\|f(x)\|\le \|f\|\cdot \|x\|$ so $\|\phi(x)f\|\le \|f\|\cdot \|x\|$ so $\tag1 \|\phi(x)\|\le \|x\|.$ On the other hand, the Hahn-Banach theorem gives us an $f\in X'$ such that $\|f\|=1$ and $\phi(x)f=\|f(x)\|=\|x\|$ so $\tag2\|x\|=\phi(x)f\le \|\phi(x)\|\cdot \|f\|=\|\phi(x)\|$ so $\phi$ is indeed an isometry.


1

We need to show that $\|\phi(x)\| = \|x\|$ for every $x\in X$. Fix $x \in X$ and notice that for every $f \in X'$ we have $$|\phi(x)(f)| = |f(x)| \le \|f\|\|x\|$$ which means $\|\phi(x)\| \le \|x\|$. To show the reverse inclusion, we need to find $f \in X'$ such that $\|f\|=1$ and $f(x)=\|x\|$. In this case we have $$\|\phi(x)\| \ge \frac{|\phi(x)(f)|}{\|f\|}...


0

Claim: Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset X$ and a linear operator $T: \mathrm{D}(T) \subset X \to X$, the vectors $T(\boldsymbol{e}_j)$ aren't always sufficient to describe the action of the operator on a general vector of $X$. Proof: Let $X = L^2([0, 2\pi])$ be the space of all square-integrable functions equipped with the Fourier ...


1

Note that $L \cap W=\{0\}$. Just define $f(v+\lambda x)=\lambda$. Then the kernel of $f$ is $L$.


1

The whole field of Analysis is highly dependent on the fact that we may approximate complicated things (think a vector $x$ in your space $X$) by simpler things (think truncated sums of the representation of $x$ in your Schauder basis). These approximations are only useful because the observations we want to make (e.g. $T(x)$) respect the approximations we ...


1

$$\|Px\|^2\le \|Px\|^2+\|Px-x\|^2=\|x\|^2$$ so $\|Px\|\le\|x\|$ which implies that $\|P\|\le 1$.


1

It is a consequence of the closed graph Theorem, hence a non-trivial fact, that, whenever a Banach space $X$ is the (algebraic) direct sum of two closed subspaces $E$ and $F$, then the natural projections $\pi_E$ and $\pi_F$ are continuous. Therefore, for every $x$ in $E$, the new norm $$ |||x||| = \|\pi_E(x)\| + \|\pi_F(x)\| $$ is equivalent to the ...


2

The proof you have written works. Denoting the inclusion by $i:V\to H$ you have for $f\in H$ and $v\in V$ that: $$|f(i(v))| ≤\|f\|_{H'} \cdot \|i(v)\|_H ≤ \|f\|_{H'}\cdot \|i\|_{L(V,H)}\cdot \|v\|_{V}$$ where you applied boundedness of $f$ and $i$ as linear maps to get that $v\mapsto f(i(v))$ is bounded as a linear map (with bound $\|f\|_{H'} \cdot \|i\|_{L(...


1

The notation you allude to is a very useful, but not rigorous notation used by physicists. With $|r\rangle$ it is meant the delta function at $r$ so $$ |r\rangle = \delta(r-x) $$ Now the above is not a function but a distribution. A distribution applied to a test function $\psi$ is written $$ \langle r , \psi\rangle $$ Now the above can be formally written ...


0

Consider the Hilbert space $\ell^2$ and let $(e_n)_{n=1}^\infty$ be the canonical vectors. Denote $e = \sum_{n=1}^\infty \frac1n e_n$ and let $B$ be an algebraic basis for $\ell^2$ containing the linearly independent set $S = \{e_n : n\in \Bbb{N}\} \cup \{e\}$. Consider $$x_n := \sum_{k=1}^n \frac1k e_k = \sum_{b \in B} \alpha_n^b b, \quad \alpha_n^b = \...


2

You are almost there. As you said we know that the unit ball is weak* compact, so by the EVT there exists $u$ in the closed unit ball so that $\inf_{\|x\|\leq 1}(f,x)=(f,u)$. We also note that $f\neq 0$ so $u\neq 0$. Set $\alpha=1/\|u\|$. We claim that $\inf_{\|x\|=1}(f,x)=(f,\alpha u)$. Assume not: Then there exists a $v$ on the unit sphere such that $(f,v)&...


0

One possible answer is that out of the $L_p$ spaces, the p-norm $\|\cdot\|_p$ satisfies the Parallelogram Law: $$2\|u\|^2+2\|v\|^2=\|u+v\|^2+\|u-v\|^2$$ only for $p=2$. It is a well-known result that by satisfying the above, one can define an inner product via: $$(u,v) := \frac{1}{2}\left(\|u\|^2+\|v\|^2-\|u-v\|^2\right).$$ Conversely, you can show that the ...


1

The answer is negative and here is a counter-example. Let us begin with the following: Lemma. If $T$ and $S$ are positive operators on a Hilbert space, with $0\leq T\leq S\leq 1$, and if $\xi $ is any vector such that $T\xi =\xi $, then $S\xi =\xi $, as well. Proof. We have $$ \|\xi \|^2 = \langle \xi , \xi \rangle = \langle T\xi , \xi \rangle \leq \...


2

By definition a spectral resolution on a Hilbert space $H$, defined on a measurable space $(X, \mathscr B)$, is a function $$ P:\mathscr B \mapsto B(H), $$ such that $P(E)$ is a projection for each $E$ in $\mathscr B$, $P(\emptyset) = 0$, and $P(X) = I$, whenever $\{E_n\}_{n\in \mathbb N}$ is a pairwise disjoint collection of measurable sets, one ...


0

HINT: Enough to get the spectrum of $T(e_n)= e_{n-1}$, where $(e_n)_{n\in \mathbb{Z}}$ is the orthnormal basis of $\mathcal{H}$. Now, $\mathcal{H}$ is isometric with $L^2(S^1)$ with the norm $$f\mapsto \frac{1}{2\pi} \int_{0}^{2\pi} |f(e^{it}|^2 dt $$ with $$e_n \mapsto \epsilon_n$$ with $\epsilon_n(t) = e^{i n t}$. (this is the inverse Fourier transform) ...


2

First, to show that $F$ is linear, $$ \langle F(\alpha x + \beta y),z\rangle=\langle \alpha x+\beta y,Fz\rangle \\ = \alpha \langle x,Fz\rangle+\beta\langle y,Fz\rangle \\ = \alpha\langle Fx,z\rangle+\beta\langle Fy,z\rangle \\ = \langle \alpha Fx + \beta Fy,z\rangle. $$ Because this holds for all $z$, then $F(\alpha x+\beta y)=\...


1

Let $f:H\to \mathbb R: f(x)=\|x\|.$ Then, if $x\neq 0,\ f$ is differentiable at $x$. Note that $f(x)\neq 0$ whenever $x\neq 0.$ Let $g:\mathbb R^+\to \mathbb R^+:g(x)=x^{-1}.$ If $x\neq 0,$ then $g$ is differentiable at $x$. By the chain rule, we have $$(g\circ f)'(x)h=g'(f(x))\circ f'(x)h$$ Calculating, we get $$(g\circ f)'(x)h=g'(f(x)(\langle\frac{x}{\|x\|...


1

You can apply the product rule on the relation $$1 = \|\cdot\| \cdot \frac1{\|\cdot\|}$$ to obtain $$D1(x_0) = \frac1{\|x_0\|}D(\|\cdot\|)(x_0) + \|x_0\|D\left(\frac1{\|\cdot\|}\right)(x_0)$$ $$0 = \frac{x_0}{\|x_0\|^2} + \|x\|D\left(\frac1{\|\cdot\|}\right)(x_0) \implies D\left(\frac1{\|\cdot\|}\right)(x_0) = -\frac{x_0}{\|x_0\|^3}.$$


1

There are two ways to solve this. The first way is to use the chain rule. This requires the knowledge that a chain rule holds also in Hilbert spaces for Fréchet derivatives and not just in $\mathbb R^n$. Or you could just prove the chain rule yourself. The second way is to guess the Fréchet derivative (you already have a conjecture) and then try to verify ...


2

We start out with the assumption that $K$ is Hilbert, which includes completeness in its definition. Therefore $K$ is already a closed convex subset of $H$ and step 1 is not needed. Just compose projection of $H$ onto $K$ with $T$ to get the desired map. $$H \stackrel{P_K}{\to} K\stackrel{T}{\to} X$$


2

Without more assumption, this may be false. If you add the hypothesis that $f\circ g_n$ uniformly converges to $f \circ g$, then it would be correct.


0

To see the equivalence, consider the following lemmas, the first of which is well-known. Lemma 1. $|\sum_i\pm a_i|\le 1\iff |\sum_it_ia_i|\le1,\quad \forall t_i\in[-1,1]$. Proof: $\Leftarrow$ is obvious. $\Rightarrow$: $|\sum_it_ia_i|\le\sum_i|a_i|=|\sum_i\pm a_i|\le1$ where a choice of signs is made so that $|a_i|=\pm a_i$. Lemma 2. $|\sum_i\langle u_i,w_i\...


4

No. For any reflexive Banach space $X$ consider $X \oplus X^*$. Then $(X \oplus X^*)^* = X^* \oplus X$, which is of course isomorphic to $X \oplus X^*$ but not necessarily a Hilbert space.


1

For showing that $F+M$ is dense you only have to show that any vector orthogonal to both $F$ and $M$ is $0$ and this is easy. For the second part your idea is good. Let $x=\sum\limits_{n=1}^{\infty} \frac 1 n e_{-n}$. Observe that $e_{-n}+ne_n, n \geq 1$ are orthogonal. This implies that any element of $M$ is of then form $\sum\limits_{n=1}^{\infty} a_n u_n$...


0

As mentioned in one of the comments, you need a construction as in $(3)$ for your space to be complete whereas your construction in $(1)$ is not necessarily complete, so they are not equivalent. However, if you take the Hilbert space completion of the space you defined in (1), this space will be the same as the space defined in (3). See the discussion here: ...


1

I suppose a precise statement of what the OP has in mind would be: Is there a basis $\{e_x\}_{x\in[0,1]}$ of $H$ such that, once any $f\in H$ is expanded in that basis, the coefficient of each $e_x$ is precisely $f(x)$? In other words $$ f= \sum_{x\in[0,1]} f(x)e_x. $$ The difficulty here is the fact that this will often have uncountably many nonzero ...


0

The problem is that integrals do not change when you change the integrand at a finite number of points (or, even on a set of measure $0$,) which makes it impossible for the integral inner product to give you a pointwise value for a function. You can retrieve an average value over a small interval with an integral such as $\frac{1}{2\delta}\int_{x-\delta}^{x+\...


1

I will denote $\mathcal B := (L(\mathcal H),\|\cdot\|_{\mathrm{op}})$, $\mathcal B_1 := (L(\mathcal H),\|\cdot\|_1)$, $\mathcal B_2 := (L(\mathcal H),\|\cdot\|_2)$. First of all, note that every operator $A$ with finite trace norm leads to a corresponding element of $\mathcal B^*$, since we can define $\phi_A(B) = \langle A,B \rangle$ (and make use of the ...


0

You are generating the Legendre polynomials with this procedures, i.e., if you take $\{1,x,x^2, x^3, ...\}$ and apply Gram-Schmidt on it you'll end up with Legendre polynomials.


0

Let $X,Y$ be the Hilbert space $\ell^2(\mathbb R)$ and $A$ defined by $(A \xi)_i = \frac{\xi_i}{i}$ for all $i \in \mathbb N$. The sequence $x_1=(1, 0, \dots, ), x_2=(1, 1, 0, \dots, ), x_2=(1, 1, 1, 0, \dots, ), \dots$ is in $\ell^2(\mathbb R)$. $\{A.x_n\}$ converges in $\ell^2(\mathbb R)$ to the harmonic sequence. However the harmonic sequence is not in $A[...


1

Let $X=Y$ have orthonormal basis $e_1, e_2, \dots$. Then let $A(e_k) = \frac{1}{k} e_k$. Every $e_k$ is in the image, so the image is dense. Also $$ x = \sum_{k=1}^\infty \frac{1}{k}e_k $$ is in $X$ because $\sum\frac{1}{k^2}$ converges. But this $x$ is not in the image of $A$, since the only choice $y$ with $A(y) = x$ would be $$ y = \sum_{k=1}^\infty ...


1

One criteria follows from the following. Theorem: If for $u \in H^k([0,\infty))$ with $k \leq 2,$ if we define $$ Eu(x) = \begin{cases} u(x) & \text{ if } x \geq 0, \\ u(-x)+3u(-x/3)-3u(-2x/3) & \text{ if } x < 0, \end{cases}$$ then $Eu \in H^k(\Bbb R),$ with comparable norms. It may be possible to use a simpler extension here, but I ...


1

The self-adjoint operators Kato was was dealing with were systems of finite numbers of particles. He was the first to prove that these Hamiltonians were essentially self-adjoint, which means that their closures are self-adjoint. As a consequence, the spectral theory of these operator may be written in terms of classical eigenfunction types of expansions ...


2

This easily follows from the following theorem: Let $(q_j)_{j \in J}$ be an increasing net of projections on the Hilbert space $H$. Then this net is strongly convergent to the projection of $H$ onto $$\overline{\bigcup_{j \in J} q_j(H)}$$ For a reference, see [1] theorem 4.1.2. Let us see how this implies the theorem: Given a finite subset $F \subseteq I$, ...


1

It cannot be extended uniquely, not even in the finite dimensional case. For example, take $\mathbb{R}^3$ and a basis $\{e_1,e_2,e_3\}$ of $\mathbb{R}^3$. Let $T_0(e_1)=e_1$ in $H_0=span(e_1)$. Then $T_1(e_2)=e_3,\ T_1(e_3)=e_2$ and $T_2=I$ ($I$ the identity). Then $T_1,\ T_2$ are two different extensions of $T$.


2

If $V$ is a vector space, $V^{\otimes k}$ denotes the tensor product of $k$ copies of $V$. For example $$V^{\otimes 2}=V\otimes V$$ and $$V^{\otimes 3}=V\otimes V\otimes V$$ Since the dimension of a tensor product is the product of the dimensions of the components $$\text{dim }V^{\otimes k}=(\text{dim }V)^k$$ In your case, the dimension of $\mathcal{H}$ will ...


1

The $q$ is given. The matrix with entries $q_{ij}$ is used to establish a property that $q$ satisfies.


0

The answer is no. Although $\mathcal T$ coincides with the WOT on bounded sets, as observed by the OP, these topologies do not coincide over the whole of $B(H)$. One very nice point of view from which to understand this question is the theory of duality. Suppose $B$ is a complex vector space (I am using the letter $B$ here because the application will be ...


0

As noted by Ian, my definition of $L^2$ is problematic because the inner product $\langle \tilde{X},\tilde{Y} \rangle \mapsto \int XYdP$ is not positive definite. This is because if $X=0$ $P$-a.s. and $X$ is not $\mathcal{C}$ measurable, then $\tilde{X}\neq\tilde{0}$ but $\langle \tilde{X},\tilde{X} \rangle =0$. We can alternatively proceed thus: Define $$...


1

The most thorough (but accessible) introduction to Hilbert spaces that I've seen is that given by Chapter 3 of Kreyszig's Introductory Functional Analysis with Applications. I find that his style is similar to that of Strang, so I think that if you like one you should like the other.


1

Let $\gamma \in \mathscr D(C)$ but not $\in \mathscr D(A)$. Then there is an $\alpha \in \mathscr D(A)$ such that $(A-i) \alpha = (C-i)\gamma$. Then $(C-i)(\alpha - \gamma) =0$, which implies $$((\alpha - \gamma),C(\alpha - \gamma))=i((\alpha - \gamma),(\alpha - \gamma))$$ which cannot hold for symmetric $C$.


2

This is the content of Riesz–Fischer theorem. As @Sten noted, we are talking about Schauder basis here (you allow infinite linear combination, not just finite ones). The uncountable "Hamel basis" (each function is a finite combination...) is much larger and impossible to write down explicitly.


1

Your statement (i) as it stands is wrong. To see this consider the map $$ L:\mathbb C^{2\times 2}\to \mathbb C^{2\times 2}\quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto \begin{pmatrix}1&0\\0&0\end{pmatrix}^*\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\0&0\end{pmatrix} $...


2

Claim: $A\psi=\lambda\psi$ for some $\psi\ne 0$ iff $$ P(\{\lambda\})\psi = \psi. $$ Proof: First assume that $A\psi=\lambda\psi$ for some $\psi\ne 0$. Then $$ 0=\|(A-\lambda I)\psi\|^2=\int_{-\infty}^{\infty}|\mu-\lambda|^2d\rho(\mu), $$ where $\rho(S)=\langle P(S)\psi,\psi\rangle= \|P(S)\psi\|^2$ is the measure associated with $\psi$. It ...


2

I'm not sure if there's a way to see this directly from the definition of $P_A$, but here is a proof using the resolvent, which is quite natural in view of the proof of the spectral theorem. With $R_A(z)=(A-zI)^{-1}$, we know that $$\langle\psi,R_A(z)\psi\rangle=\langle \psi,\frac{1}{z_0-z}\psi\rangle=\frac{1}{z_0-z}||\psi||^2$$ This is in turn, by the ...


1

It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not converge is a bit more difficult. The approach suggested by @Max, although somewhat sophisticated, is perhaps the easiest way to pin down all of the details. Here is a pedestrian way to describe this method. For each pair of vectors $(x, y)\in ...


1

Yes, the book is wrong. It was supposed to be $ub \in A$ for all $b \in B$. See also Farah's book Corollary 1.6.13.


2

I think you are using the left-shift instead of the right-shift. If $M$ is the one-dimensional subspace spanned by a non-zero sequence $(a_n)$ and if this is invariant under the right shift then $(0,a_1,a_2,...) =c (a_1,a_2,...)$ for some constant $c$. If $c=0$ we get $a_n=0$ for all $n$ immediately. Otherwise we have $0=ca_1,a_1=ca_2,...$ and these imply ...


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