# Tag Info

10

Here's the way I like to think about it. I'll start with the finite dimensional space $\Bbb{R}^n$ because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity $z^Tx$ represents a linear functional on $\Bbb{R}^n$, that is a linear function which eats a vector and spits out a real number: $$... 4 The dual space is a space of linear functionals. If we want to define a norm on the dual space, we do what we always do to measure the "size" of a linear transformation: we use an operator norm. Alternatively, the dual norm of z is the matrix norm of the matrix z^T. 4 Let E_n, n\ge 1, be these nonempty closed convex sets. We want to pick a point x_n in each set so that they form a Cauchy sequence. If the diameter of E_n tends to zero as n\to\infty, the Cauchy property comes automatically. Otherwise, we have to make some intelligent choices of x_n. I present two versions; the second, pointed out by Daniel ... 4 If (X, \Sigma) and (Y, \tau) are measurable spaces, then a measure \lambda on (X \times Y, \Sigma \times \tau) is said to be a product measure if there exist measures \mu, \nu on (X, \Sigma), (Y, \tau) respectively, such that \lambda = \mu \times \nu. That is, for each A \in \Sigma, B \in \tau, we have \lambda(A \times B) = \mu(A) ... 4 It is known that there is an uncountable chain in \mathcal P(\mathbb N). (In fact, we have a chain of cardinality \mathfrak c.) See: Chain of length 2^{\aleph_0}in  (P(\mathbb{N}),\subseteq) Now if \mathcal C is such a chain and e_i is an orthonormal basis of \ell_2 we can take$$F(C):=\overline{\operatorname{span}\{e_i; i\in C\}}$$for any ... 3 Any Hilbert space is of course self dual. But this is only true if you use the inner product as the pairing. Note that g\in H^{-1} acts on f\in H^1_0 (formally) by g(f)=\int_U gf, where no derivatives appear. Often the dual space of a function space is considered to act by "a simple integral", which may not be the inner product (if we are not in ... 3 I assume that A is a C^\star-algebra and that A_0 is a subset of A? Then the generated C^\star-subalgebra is the norm-closure of the linear span of the multplicative closure of the \star-closure of A_0. That is, it is the norm-closure of the generated *-subalgebra \{\sum_i \lambda_i a^i_1 \dotsc a^i_n : \lambda_i \in \mathbb{C}, a^i_k \in ... 3 There are many possible conventions, but you can't get around the fact that a 2\pi will appear somewhere. In order to have a way of comparing different conventions easily, I wrote up restrictions on the various coefficients in the definitions and comparison between different choices. 3 Forming the convolution of the (scaled) bump function with indicator functions you get "qausi-indicator functions" in \mathscr D, in particular, there are \psi_n\in\mathscr D(\mathbb R) which are positive and equl to 1 on [-n,n]. It is then easy to see that the elements of E^{loc} are those distributions which, on every compact set, have the same ... 3 We start with a theorem: Theorem. Let H be a Hilbert space and T a compact self adjoint operator. Then there exist a Hilbert basis composed of eigenvectors of T. Proof. See Brezis chapter 6, in particular, theorem 6.11. Remark: Let (\cdot,\cdot) denote inner product. In the conditions of the theorem, if (Tu,u)> 0 for all u\neq 0 ... 3 Looking at examples helps... The function f(x)=\mathrm e^{-x^2} on the real line endowed with \mu the Lebesgue measure shows that f can be integrable while \{f\ne0\} has infinite measure. 2 One uses Zorn's lemma to have the existence of a maximal orthonormal set \mathscr{B} in R. It remains to see that \mathscr{B} is an orthonormal basis (not a basis in the algebraic sense, unless R is finite-dimensional), i.e. S := \overline{\operatorname{span} \mathscr{B}} is the whole space R. Clearly, the maximality of \mathscr{B} implies ... 2 Dual norm is a particular case of the support function, specifically it is the support function of the unit ball of the original norm. When the unit ball is smooth enough \|z\|_* is the Euclidean distance from the origin to the hyperplane with the normal vector z (of unit Euclidean length) tangent to the ball. The equation of this hyperplane is ... 2 You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional x^{\star} to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed ... 2 If t\gt0, then we want a\ge2|A| and$$ \gamma_r=a+ir[-1,1]\cup[a,-r]+ir\cup-r+ir[1,-1]\cup[-r,a]-ir $$If t\lt0, then we want a\le-2|A| and$$ \gamma_r=a+ir[1,-1]\cup[a,r]-ir\cup r+ir[-1,1]\cup[r,a]+ir $$In each case, \gamma_r is counterclockwise. When t\gt0, the finite part of \gamma_r is [a-i\infty,a+i\infty]. When t\lt0, the finite ... 2 As Nate's hint suggested, we can take$$(X, \Sigma, \mu) = (\mathbb{R}, \mathcal{P}(\mathbb{R}), 0 \text{ (the zero measure)})$$and$$(Y, \tau, \nu) = (\mathbb{R}, \mathcal{P}(\mathbb{R}), \mu \text{ (counting measure)})$$and writing X \times Y as the countable union \bigcup \limits_{n = 1}^{\infty} X \times Y_{n} (with Y = \bigcup \limits_{n = ... 2 Note that if \mathcal{H} is real then the notion of a unitary operator does not make any sense. However, you can do that with five orthogonal operators instead. See e.g. Theorem 4.3 in A. BĂ¶ttcher, A. Pietsch, Orthogonal and Skew-Symmetric Operators in Real Hilbert Space, Integral Equations and Operator Theory 74 (2012), 497-511. It should be added ... 2 Let f(s) = s^p for some p>1, then for every s\geq 0m we have$$f''(s) = \underbrace{p(p-1)}_{>0}\,\underbrace{s^{p-2}\vphantom{)}}_{\geq 0} \geq 0$$and thus f is convex on the non negative numbers. So for any x_1,x_2 \in X and t \in (0,1) we get$$\frac{1}{p}\|x_1t + (1-t)x_2 \|^p \overset{(1)}{\leq} \frac{1}{p}(t\| x_1\|+(1-t)\|x_2\|)^p ...

2

It seems that you are looking for interwining operators.

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 $$\sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^nx\rVert\leqslant \sup_{n_k\leqslant ... 2 Your interpretation in 1 and 2 is correct. Regarding the point I really dislike talking about the integral of a non-measurable function, even though the function equals a measurable one almost everywhere. I share your dislike. This is why I think that spaces X and Y should themselves be assumed complete. On a complete measure space, a function ... 2 Assume T=T^{\star}\in\mathcal{L}(H), where H is a complex Hilbert Space. Implication 1: Show (Tx,x) \ge 0 for all x \in H implies \sigma(T)\subseteq [0,\infty). To do this, assume that (Tx,x) \ge 0 for all x \in H, and let \lambda < 0. Then$$ 0 \le -\lambda(x,x) \le ((T-\lambda I)x,x) $$implies$$ |\lambda|\|x\|^{2} \le ...

2

It is of course true that norm-continuous linear maps are weakly continuous. This follows from the fact that the weak topology is the initial topology w.r.t. to all continuous linear functionals, i.e. $\sigma(Y,Y^*)$ is the coarsest topology on $Y$ such that all $f\in Y^*$ are continuous. Then, by abstract nonsense, a map $T:E \to (Y,\sigma(Y,Y^*))$ (where ...

2

The fact that $f$ is in $H^1$ implies that its derivative $Df$ is in $L^2$. The Fourier coefficients of $Df$ are (up to a constant) $\{n |f_n|\}$. The Parseval relationship then implies that $\{n |f_n|\} \in \ell_2$. Since $\{n |f_n|\} \in \ell_2$ the Cauchy-Schwartz inequality implies \begin{align*} \sum_n |f_n| &= \sum_n n^{-1} n |f_n| \\ ...

2

Yes. $x \rho_t(x)$ is smooth and compactly supported, and any such function $f$ is in $H^s$ for every $s$. The easiest way to see this may be this: Let $k$ be any integer larger than $s$. Verify from the definition that $H^k \subset H^s$. (Note that $(1+|\xi|^2)^s \le (1+|\xi|^2)^k$). Using the fact that the Fourier transform takes differentiation to ...

2

Look up the concept of an unbounded operator. These are exclusive to the infinite dimensional setting. The differences between finite dimensional linear algebra and analysis on infinite dimensional space are many and important. Basically, the entire game changes significantly. Of special importance to quantum mechanics, I would point out: In the finite ...

2

Here's a hint for getting a single smooth function: The function $y(x)=\sqrt{1+x^2}$ has properties similar to what you want, and so does $y(x)=x$. See if you can somehow combine these two behaviours.

2

This is true since if $g\ge0$ then: $$\int_{X}fgd\mu=\int_{X}fdg$$ where $dg(A)=\int_{A}gd\mu$. This is due to the Radon-Nikodym Theorem.

1

Yes, your proof is correct. Another way to use the completeness of $L^1$ is to set up an isomorphism between your space (usually denoted $AC$ or $W^{1,1}$) and $L^1\oplus \mathbb R$. Such an isomorphism is given by $$f\mapsto \left(f',\int_0^1 f\right)$$ This map is bounded and surjective. It also has trivial kernel. Being lazy, I'd just invoke open ...

1

Hint: Use the Uniform boundedness Principle to show that $T$ is bounded, and the use a)

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