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7

It does not hold in general. Recall two basic facts: for any subspace $E$ of a Hilbert space, we have that $E^{\perp}$ is closed; $E^{\perp\perp} = \bar{E}$, the closure of $E$. In particular, if $E$ is closed then $E^{\perp\perp}=E$. Now in The direct sum of two closed subspace is closed? (Hilbert space), you can find an example of two closed ...


4

While I can't satisfyingly answer your question about the $L_1$ norm, the $L_2$ norm has an easy answer. Remember that a signal is generally, physically speaking, some sort of electronic oscillation in voltage (or current, if you prefer to think of it that way). It is well known that instantaneous power of an electronic signal is given by $$ \frac{dE}{dt} = ...


4

It is known that there is an uncountable chain in $\mathcal P(\mathbb N)$. (In fact, we have a chain of cardinality $\mathfrak c$.) See: Chain of length $2^{\aleph_0}$in $ (P(\mathbb{N}),\subseteq)$ Now if $\mathcal C$ is such a chain and $e_i$ is an orthonormal basis of $\ell_2$ we can take $$F(C):=\overline{\operatorname{span}\{e_i; i\in C\}}$$ for any ...


4

In general, the inequality $$\mathbb{E}\|X-Y\|^2 \leq \mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2$$ does not hold. Simply consider arbitrary $X$, $Z := \frac{1}{2} X$ and $Y = \frac{1}{4}X$. Then $$\mathbb{E}\|X-Y\|^2 = \frac{9}{16} \mathbb{E}\|X\|^2$$ whereas $$\mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2 = \left( \frac{1}{4} + \frac{1}{16} \right) ...


3

The space $\mathcal{C}_{0}$ is not a Hilbert space. Consider, for example, a sequence of functions which is zero on $(-\infty,-1/n)$, goes linearly from 0 to 1 on $[-1/n,1/n]$, is one on $(1/n,1)$, and goes linearly from 1 to 0 on $[1,2]$, and is 0 on $[2,\infty]$. This sequence is Cauchy, but it cannot converge to a continuous function. In the completion of ...


3

Consider the Heat semigroup on $L^2(\mathbb{R}^n)$: for $t>0$, $$U(t)g=K(t)\ast g$$ where $$K(t)=(4\pi t)^{-n/2}e^{-|x|/4t}$$ is the heat kernel. $U(t)$ is a strongly continuous semigroupo (of contractions). Let's show that it cannot be extended to a group ($U(t)$, $t\in\mathbb{R}$) on $L^2$. If such a group eventually exists, it should satisfy ...


3

The simplest case I know is the translation $T(t)$ semigroup on $L^{2}[0,\infty)$ defined by $(T(t)f)(x)=f(x+t)$. There's no way to invert once you've lost part of the function.


3

Your proof is correct. But with almost no extra work you can prove a better result: considered as a subset of $\ell^2$, the space $\mathbb R^\infty$ is dense. (Consequently, not closed, hence not complete in the $l^2$ metric.) The proof goes just as yours, but without specifying $x$. Just let $x$ be any element of $l^2$, then define $$x_n(k)= \begin{cases} ...


3

Every bounded linear operator on a complex Banach space has non-empty spectrum $\sigma(B)$. A simple way to argue this is to assume the contrary, and conclude that $(B-\lambda I)^{-1}$ is an entire function of $\lambda$ which vanishes at $\infty$; then Liouville's theorem gives the contradiction that $(B-\lambda I)^{-1}\equiv 0$ for all ...


2

Yes, correct and also true in reflexive Banach spaces. Kakutani's theorem (at least the direction you use in the proof) is also a special case of Banach-Alaoglu theorem. In History of Banach Spaces and Linear Operators, Albrecht Pietsch remarks ... the weak* compactness theorem is an elementary corollary of Tychonoff's theorem. Therefore priority ...


2

Note that if $\mathcal{H}$ is real then the notion of a unitary operator does not make any sense. However, you can do that with five orthogonal operators instead. See e.g. Theorem 4.3 in A. Böttcher, A. Pietsch, Orthogonal and Skew-Symmetric Operators in Real Hilbert Space, Integral Equations and Operator Theory 74 (2012), 497-511. It should be added ...


2

Well... you just have to show $||u_n - u|| \to 0$. But $||u_n-u||^2 = ||u_n||^2 - <u_n,u> -<u,u_n> + ||u||^2$ Now, by assumption the first term converges to $||u||^2$, the second and third term converge (by assumption of weak convergence) to $<u,u>=||u||^2$.


2

First observation: by the uniform boundedness principle, the sequence $(\lVert T^n\rVert)_{n\geqslant 1}$ is bounded. Let $x\in H$. Using compactness of $T$ and the fact that $T^nx\to 0$ weakly, we can find $n_k\uparrow \infty$ such that $\lVert T^{n_k}x\rVert\to 0$. Since $$\sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^nx\rVert\leqslant \sup_{n_k\leqslant ...


2

Assume that $A$ is bounded. Take $x_n\in A, y_n\in B$ such that $\lim||x_n-y_n||=e$. Since $A$ is bounded - sequence $x_n$ is bounded and so is $y_n$. Thus using Banach-Alaoglu theorem one can choose subsequences $x_{n_k},y_{n_k}$ converging in the weak topology to $x,y$. Now since closed and convex sets are weakly closed (Mazur theorem) one gets that $x\in ...


2

If $k_1,\dots,k_p$ are kernels with RKHSs $\mathcal{H}_i$, then $k=\sum_i k_i$ is indeed a positive definite kernel again without further assumptions. The relation $\lVert f\rVert_{\mathcal{H}}^2 = \sum_{i=1}^{p}\lVert f\rVert_{\mathcal{H}_i}^2$ holds automatically for all functions that are in the intersection of all participating RKHSs. There is no ...


2

Reflexive spaces interest mathematicians because they have a lot of nice properties: Unit ball is weakly compact, so you can exploit compactness to prove exitence of fixed points, convergent subsequences and etc. Reflexive spaces are characterized by the property that weak and weak* topology coincide. You can forget about weak topology and work with much ...


2

Hint: Try proving the contrapositive. If $T$ is not continuous, show that it maps some bounded sequence $\{x_n\}$ to a sequence $\{T x_n\}$ with $\|T x_n\| \to \infty$. Passing to a subsequence and rescaling, we can even assume $\|x_n\| \to 0$, and in particular $x_n \to 0$ weakly. Every weakly convergent sequence is bounded. (This comes from the ...


2

No such basis exists. What you call a "linear algebra basis" is more commonly called a Hamel basis in this setting. If $H$ is any infinite-dimensional Hilbert space, then it does not have a Hamel basis whose elements are pairwise orthogonal. Let $B \subset H$ be an infinite set whose elements are pairwise orthogonal; we will show $B$ does not span $H$. ...


2

An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of ...


1

My answer: only 2 is true. It is not closed as $\lim e_n$ does not exist. It is bounded as $||e_n||=1$ It is not compact as one can not find a finite sub-cover containing S. Neither does it contain a convergent subsequence as $\lim e_{n_k} $ does not exist.


1

Every pair of orthonormal vectors $e_i\neq e_j$ has distance precisely $d(e_i,e_j)=\sqrt{2}$. So $S$ is closed as it contains only isolated points, it is bounded as all its elements have norm one, it is not compact as it contains infinitely many isolated points, it contains a convergent subsequence the constant ones?!


1

The subspace $S_{n}$ spanned by $\{ e^{2\pi i kx}\}_{k=-n}^{n}$ is invariant under $A$, and $A$ is bounded on $S_{n}$ because its a matrix. However, $A$ is not bounded on the union of these nested subspaces. That puts a limit on any Zorn's lemma argument. I suppose you could find a maximal subspace on which $A$ is bounded by a given fixed constant $M$.


1

The explanation of Omnomnomnom of the $L_2$ norm is very good alreay. One possible interpretation of the $L_1$ norm, or rather the ration between $L_1$ and the $L_2$ norm is that it can be used as a measure of sparsity or compressability of a signal. Therefore it has often been used as a regularization term in optimization problems. Today this is a ...


1

No, you don't need another tool. Parallelogram law is good enough: One has $$4=\|x+y\|^2+\|x-y\|^2\ge \|x+y\|^2+\epsilon^2.$$ So $$\epsilon^2\le 4-\|x+y\|^2=(2-\|x+y\|)(2+\|x+y\|).$$ Can you find the constant $C$ so that $1-\|\frac12(x+y)\|\ge C\epsilon^2$ from here?


1

As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator ...


1

Notice that $v_h=Au_h$ for some (actually a unique) $u_h$. Using the assumption with $u:=u_k-u_j$, we obtain after an application of Cauchy-Schwarz inequality that $(u_k)_{k\geqslant 1}$ is Cauchy. For density, notice that if $v$ is orthogonal to the range of $A$, then $(Av,v)=0$. For the norm inequality, use $(A(A^{-1}u),A^{-1}u)\geqslant\alpha ...


1

It is not a subspace. If your operator is positive definite, then $F$ is empty. More generally, your $F$ cannot contain $h=0$, so it will not be a subspace for any choice of an operator, selfadjoint or not. Even if you replace $<$ with $\leq$, it will still not be a subspace in general: consider the operator $T=\begin{bmatrix} ...


1

I'll give a couple important properties of reflexive Banach spaces, one more geometric and one more operator-theoretic (though there are of course many more). First of all, given any closed convex subset $A$ of a reflexive Banach space $X$ as well as any $x \in X$, there exists a point $a \in A$ which minimizes the distance $\inf_{a \in A} \| x- a \|$ from ...


1

$||AB-BA||=||AB-A_{\alpha}B+A_{\alpha}B-BA|| \leq ||AB-A_{\alpha}B||+||A_{\alpha}B-BA||$. But $A_{\alpha}B=BA_{\alpha}$, and $||AB-A_{\alpha}B|| \leq ||B||||A-A_{\alpha}||$. So $S'$ is closed in the norm topology as well. A "high-level" explanation can be given: $S'$ is a von-neumann algebra, and every von-neumann algebra is a $C^*$ algebra- and so $S'$ ...


1

This is just an intuition for the much simpler case of finite dimension. Let $A$ be any matrix and $v$ an associated eigenvector, i.e. $Av = \lambda v$. Then $(A-I)v = Av-v = (\lambda-1)v$, and thus $\sigma(A-I)= \sigma(A)-1$. And now an attempt for infinite dimension, suppose that $(A-i\lambda I)$ is not invertible for every $\lambda \in (-\infty, 0]$ but ...



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