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Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$. Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = ... 7 Assuming "polynomial" means "polynomial in$z$", i.e. of the form$\sum_{k=0}^n a_k z^k$, then no, they are not dense, and the function$\bar{z}$is not in their uniform closure. The high-level reason is that polynomials are holomorphic on the unit disk, holomorphic functions are closed under uniform convergence, and$\bar{z}$is not holomorphic. For a ... 7 Show that$T$and$S$are closed. Then apply the closed graph theorem to conclude that$T$and$S$are continuous. To show that$T$is closed, suppose$\{ x_{n} \}$converges to$x$and suppose$\{ Tx_{n} \}$converges to$y$, and show that$Tx=y$; to do this observe that $$(Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in ... 7 1. We first talk a bit about the underlying method of the proof. We want look for (I + \epsilon T)^{-1} as an infinite series, similar to the way we would expand 1/(1+x) into a power series for |x| < 1 over the real or complex field. We then show that our infinite series of operators converges. Along the way, we need to use the fact that \|TS\| ... 6 Let us assume that V is complete (otherwise, this is probably false). Note that finite dimensional normed vector spaces are always complete. Also, let us assume that T is not an arbitrary linear transformation, but a bounded linear transformation, which means that$$ \Vert T \Vert := \sup_{\Vert x \Vert \leq 1} \Vert Tx \Vert < \infty. \qquad ... 6 Consider the space of square summable sequences and the subspace of sequences whose only finitely many terms are different from zero. This subspace is dense and is not the whole space hence it cannot be closed. 6 If$a\in\ell^2$, then$a_n\to0$. Thus$1/a_n\not\to0$, which prevents it from being in$\ell^2$. 5 This is an inner product space and accordingly it satisfies the parallelogram law. Hence you cannot show that it is not a Hilbert space be showing it fails to satisfy the parallelogram law. The reason it is not a Hilbert space is that it is not complete: there are some Cauchy sequences in it that fail to converge. For example, consider the sequence whose ... 5 Everything boils down to just a few key facts: A non-vanishing continuous real function on the unit sphere (unit vectors)$S$in$\mathbb{C}^{n}$achieves its minimum (non-zero) value at some point. The Cauchy-Schwarz inequality for inner products,$|(x,y)| \le \|x\|\|y\|$, of which the following is a special case:$|\sum_j a_j b_j| \le (\sum_j ...

5

For your first question, I think you are misunderstanding what the theorems say. When you restrict your $\sigma$-weakly continuous functional to the unit ball, you don't get a wot functional on the whole space: so 4.6.4 does not apply. For your second question, here is an example: fix an orthonormal basis $\{e_n\}$ and let $$... 5 It doesn't. There's nothing to prevent you from defining a bounded linear operator on a proper subspace. There are also unbounded linear operators defined on the whole space (although you won't find explicit examples of this, because it requires some form of the Axiom of Choice). However, a closed linear operator defined on the whole space is bounded ... 5 First of all, we write down the fourier expansion of x(t):$$x(t) = \sum_{n=1}^\infty a_n \sin(n\pi t) + \sum_{n=1}^\infty b_n \cos (n\pi t)$$(the main point is that x(t) has no constant term, do you know why?). Thus$$\int_0^1 |x(t)|^2 dt = \sum_{n=1}^\infty a_n^2 + b_n^2$$and$$\int_0^1 |x'(t)|^2 dt = \pi ^2 \sum_{n=1}^\infty n^2(a_n^2 + ...

5

All of these functions are in the ball of radius $1$ centered at $0$. They don't converge uniformly to any continuous function, nor does any subsequence converge uniformly to any continuous function (why?). Hence the ball is not precompact. Conclude from there.

5

Since $\Omega$ is bounded, we have a continuous inclusion $$j \colon L^q(\Omega) \hookrightarrow L^p(\Omega).$$ Now we have the general fact that a continuous linear map $T\colon X \to Y$ between two normed spaces (it holds more generally for Hausdorff locally convex spaces) is also continuous if we endow both spaces with their respective weak topology. ...

4

A good example of an operator with one point in the spectrum is the integral operator $$Lf=\int_{0}^{t}f(u)\,du$$ defined on $X=L^{2}[0,1]$, or defined on $X=C[0,1]$. In both cases $\sigma(L)=\{0\}$. This operator is also compact but it has no eigenvalues, which makes it a good counterexample to remember when studying compact operators.

4

with something like $$d(x,y)=\sum_n2^{-n}|x(n)-y(n)|,$$ the collection of bounded sequences is separable, for instance the rational span of the "standard basis" $e_k(n)=\delta_{kn}$ is dense. as noted in the comments, the topology generated by this metric is that of pointwise convergence (i.e. bounded sequences as a subspace of $\mathbb{R}^{\mathbb{N}}$ ...

4

This just uses the definition of weak derivative. The trick is to notice that if $u,\phi$ are both in $L^2(\mathbb R)$ then $$\int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx = \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx \tag{\ast}.$$ If (2) holds, you can use the dominated convergence theorem and ($\ast$) to see that $$... 4 The Dominated Convergence Theorem works in L^p spaces, too: if the dominating function is L^p (with 1\le p < \infty) then the convergence is in L^p sense. To wit, assume that$$|f_n| \le g \in L^p,$$with p>1 (the case p=1 being the DCT). Then the pointwise limit f of f_n (which exists by assumption) is dominated by g and hence f\in ... 4 There is a direct and self-contained proof of HLS inequality in Analysis by Lieb and Loss, Theorem 4.3. It uses nothing but layer cake representation, Hölder's inequality, and clever manipulation of integrals. A bit too long to reproduce here, though. Also, the boundedness of Hardy-Littlewood maximal function is much more straightforward than the ... 4 Cyclic shift operator is the matrix$$A=\begin{bmatrix} 0&1 \\ &0&1 \\ & & \ddots \\ &&&&1\\ 1&&&&0 \end{bmatrix}$$It has order n, that is to say, A^n=I, and it is a unitary matrix. 4 Let H = L^2([0,1]) and let P be the set of polynomials in H. 4 Take$$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$for n = 1, 2, \dots and for some a_n. You'll get$$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$Then choose the sequance a_n so that the first sum is convergent, while the second one ... 4 You can do this in any II_1-factor. Note that Q_j is a II_1-factor. In any II_1-factor M you can always get a sequence of pairwise orthogonal projections that add to the identity (those could be the q_j in your setup). So now you want to embed M_n(\mathbb C)\hookrightarrow q_jMq_j. Since you are in a II_1-factor, you can divide q_j as a ... 4 In terms of difficulty, it really depends on the level of instruction. Complex and functional analysis at an introductory graduate level are roughly comparable, and it really comes down to the student's own preferences and the quality of the instructor. But complex is often taught at an undergraduate level as well, while functional is typically not. As to ... 4 p' is the conjugate exponent of p. Let (p-\epsilon)' be the conjugate exponent of p-\epsilon. Then$$ (p-\epsilon)'=p'+\frac{\epsilon}{(p-1-\epsilon)(p-1)}=p'+\delta,\quad\delta>0. $$Then, by Hölder's inequality:$$ ...

3

I can show the claim when the interval is the same for each integral, let's say it is $[-1,1]$, and if there is a restriction on the roots of $f(t)$. I thought about this for a while and figured I might as well post it. I'm slightly hopeful with some more thinking a full solution will come. First, suppose $f(t)=\prod (t-\alpha_i)$ with $|\alpha_i| > 3$. ...

3

Let us take a closer look at $\lVert\,\cdot\,\rVert_0$. Since for $x\in X$ we have $\hat{x} = \{ x-y : y \in Y\}$, we can write $$\lVert \hat{x}\rVert_0 = \inf_{z\in\hat{x}} \lVert z\rVert = \inf \{ \lVert x-y\rVert : y \in Y\}.\tag{1}$$ So we have $\lVert \hat{x}\rVert_0 = 0$ if and only if for every $\varepsilon > 0$ there exists a $y_\varepsilon \in ... 3 Surely when working in a locally convex space with a compact space the convex hull is already closed. No, that need not be. Consider$\ell^2(\mathbb{N})$, let$(e_n)_{n\in\mathbb{N}}$be the "standard basis". Define$v_n = \tfrac{1}{n+1}\cdot e_n$and $$M := \{ 0\} \cup \left\{v_n : n\in\mathbb{N}\right\}.$$$M$is compact, and the convex hull of$M$... 3 The sinc function is the inverse Fourier transform of the characteristic function$\chi_{[-1,1]}$, at least up to a constant:$\begin{align} \chi_{[-1,1]}^{\vee}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{ixs}ds \\ & = \left.\frac{1}{\sqrt{2\pi}}\frac{e^{isx}}{ix}\right|_{s=-1}^{1} \\ & = ... 3 The solution of your problem is an easy application of the proposition below: Theorem: LetA$be a nonempty subset of$\mathbb{R}$and$A$is bounded below. Then there is a$m\in\mathbb{R}$such that$m$is a lower bound for$A$and given any$\epsilon>0$there exists$a\in A$such that$m\leq a<m+\epsilon$If$d=\inf_{y\in M} \|x-y\|^2\$ then for ...

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