# Tag Info

13

You need to check if the functions are independent, as you said. A way to go about this, which that ties it in with things you likely know is to evaluate it at several points, as you did for $x=0$. You get one condition for $x=0$. You get another condition for $x=1$ and still another one for $x=2$. Each will allow more than one solution, but they'll ...

13

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. ...

11

Write $$\alpha e^x + \beta e^{2x} + \gamma e^{3x} = 0$$ You can go ahead and cancel out a positive number like $e^x$ so: $$\alpha + \beta e^{x} + \gamma e^{2x} = 0$$ Suppose you have some solution for this with $\alpha$, $\beta$, $\gamma$ not all zero. Then, as you say $$\alpha + \beta + \gamma = 0\qquad \qquad (1)$$ Because this must be true at $x = 0$ but ...

6

Hint: let $e^x=y$, $e^{2x}=y^2$, $e^{3x}=y^3$ you have: $\alpha y +\beta y^2+ \gamma y^3=0$ where the $0$ at RHS is the zero polynomial. Now: when a polynomial is the zero polynomial? In general: The $0$ at RHS is the neutral element for the sum of functions in the vector space, not simply the number $0$ and this means that it is the function ...

6

Ah, but how do you know $||f||=<f,f>$ defines a norm? That requires a proof too! Suppose $f\ne 0$. Then we can find $x \in [0,1]$ such that $f(x) \ne 0$. So $f(x)>0$ or $f(x)<0$. In either case, $f(x)^{2}>0$. But since $f$ is continuous at $x$, for any $\epsilon>0$ we can choose $\delta$ sufficiently small that if $|y-x|<\delta$, ...

6

Let $\{a_n\}$ be a sequence of positive real numbers that increases to $1$, with the property that the sequence of products $$a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$ converges to a positive value. It's not hard to write down a specific example. Let $\cal H = \ell^2(\mathbf R)$ and define $T : \cal H \to \cal H$ by $$T(x_1,x_2,x_3,\ldots) = (0, ... 6 There's a general framework that the Fourier transform fits into using Pontryagin duality and studying the characters of a locally compact abelian group, such as \mathbb{R}. The characters of \mathbb{R} are exactly the maps x \mapsto e^{itx}, which is where the complex factor comes from. This has all sorts of wonderful consequences, like the fact that ... 5 Take T_k x = k^2 x_1 -k x_2. Then for any x \neq 0 we see that |T_kx| \to \infty. However, T_k ({1 \over \sqrt{1 + k^2}}(1,k)) = 0 for all k. 5 As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) \mathbb R^\infty=\bigcup_n \mathbb R^n, where \mathbb R^n has its usual topology and is included in \mathbb ... 5 Consider f:\mathbb R\to\mathbb R  defined by f (x)=1. The set \{1\}  is compact, but f^{-1}(\{1\})=\mathbb R  is not. 5 Let us assume that we are working with sequence spaces \ell^p and \ell^q. Since the$$p=q=2$$case is taken care of already, assume$$p>2>q.$$Since \ell^p\subset\ell^q for q\le p as here, both sides of the desired inequality make sense. Start the indices of the sequences at zero, because we feel like it. So, an example to stretch things: ... 5 I'll assume S not empty. Since the function f\colon S\to\mathbb{R}, f(p)=d(p_0,p) is continuous, when S is compact its image is compact, hence closed and bounded; therefore the image of f contains its minimum. If S is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider X=\{0\}\cup (1,2], with ... 5 Hint: Use Wronskian and show that the Wronskian-Determinant does not vansish. 5 Just consider e.g.$$f(x) := \begin{cases} \frac{1}{\sqrt{x-5}}, & x \in (5,6), \\ 0, & \text{otherwise}. \end{cases}$$Then f \in L^1((1,\infty)), but f \notin L^2((1,\infty). 4 Since K is compact, 0\in\sigma(K) (because K is not invertible). Because K is compact, any element of its spectrum has to be an eigenvalue. Now suppose that \lambda\ne0, and that Ku=\lambda u. This, with your concrete k, looks like$$ \lambda u(x)=\int_0^x y\,u(y)\,dy-x\,\int_1^x u(y)\,dy. $$Since u is integrable (otherwise Ku makes no ... 4 This is not a complete answer. Rather, it is a suggestion on how to reduce the problem to a possibly simpler one. For each c\in(-\pi/2,\pi/2), multiply each of your two identities by e^{i c k - \epsilon k^2/2} with \epsilon>0, integrate in k\in\mathbb R, let \epsilon\to0, and then subtract one of the resulting identities from the other. ... 4 Notice that for each vector x, one has$$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$Thus$$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$4 You have to prove$$ \forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\Leftrightarrow\alpha,\beta,\gamma=0, $$but I think the quantifier applies only to the part on the left side of the \Leftrightarrow, like this:$$ \left(\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\right) \Leftrightarrow\,\alpha,\beta,\gamma=0. $$So ... 4 You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of x. If you consider D: \mathcal{F} \rightarrow \mathcal{F}, the derivative operator, is an endomorphism in \mathcal F (i.e. a linear map from ... 4 We know that for \|A\| < 1, (I-A)^{-1} is well-defined (prove this yourself if you have not done so yet) so we can talk about the inverse. Thus:$$ I = (I-A)(I-A)^{-1}.$$Here is a hint:$$ 1 = \|I\| = \|(I-A)(I-A)^{-1}\|.$$Try doing some basic norm manipulations to this. You need to increase the norm, not decrease it since you want a lower ... 4 A function is a machine, a machine that assigns to any value of some set X a unique element f(x) beloging to some set Y. So the proper notation would be f:X\rightarrow Y:x\mapsto f(x). You read it as follows: "f is a function from X to Y which assigns to any x\in X a value f(x) in Y". When you write f you refer to the function, when you ... 4 I’d attack it much more directly. HINT: Suppose that a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty. Then a has a subsequence \langle a_{n_k}:k\in\Bbb Z^+\rangle such that |a_{n_k}|\ge k for each k\in\Bbb Z^+. For each k\in\Bbb Z^+ let$$x_{n_k}=\frac1{ka_{n_k}}\;,and let all the other terms of x be 0. Show that x\in\ell_1, ... 4 This holds for first countable or sequential spaces. It need not hold, even within functional analysis (the dual of L^\infty([0,1]) does not obey it, IIRC). For first countable spaces there is a simple theorem: A \subseteq X is closed iff for all sequences (x_n) that converge to x, if all x_n are from A, then also x \in A. I.e. A is closed ... 4 A dense set D of X is such that the closure of D equals X. Or equivalently, every non-empty open set contains a pont of D. So the points of D are in a sense "close" to all points of X, we can "approximate" points of X by points in D. The name separable is somewhat unlucky (what can be separated, exactly?). It probably has an historic ... 4 You have \begin{align} a_1\sin(x+b_1)+a_2\sin(x+b_2)&=a_1\sin x\cos b_1+a_1\cos x\sin b_1+a_2\sin x\cos b_2+a_2\cos x\sin b_2\\ &=(a_1\cos b_1+a_2\cos b_2)\sin x+(a_1\sin b_1+a_2\sin b_2)\cos x. \end{align} So it is enough to show that \alpha\sin x+\beta\cos x\in M_1. For this, note that we can always find t with ...

3

An alternate approach is induction on $n=\dim(W)$. The base case $n=0$ is clear, so the hard part is the induction step. For this, it's enough to prove the following result: if $M$ is a closed subspace of $V$ and $x\in V,x\not\in M$, then $M+\mathbb{C}x$ is also a closed subspace. Indeed, by the Hahn-Banach theorem there is a continuous linear functional ...

3

Let $A \subseteq B(H)$ be any AW*-algebra that is not a von Neumann algebra. (Actually, we don't need a full AW*-algebra, see below.) Let $t \in A$ be any operator. By definition of AW*-algebra, every right-annihilator is generated by a projection. In particular, the right-annihilator of the singleton set $\{t\}$ is generated by a projection $q \in A$. ...

3

You can do it when the Banach space $E$ is separable (i.e. $\overline {\{x_n\}}=E$ for some sequence). Here a sketch of the way following the "Cours de Theorie de l'Approximation" of Professor P.J.Laurent (Grenoble University). Let $V$ a vectorial subspace of $E$ and $f\in V^*$ with $$||f||=\sup_{||x||\le 1;\space x\in V} |f(x)|$$ We have to show $\exists ... 3 This is studied in potential theory: the function$u$is the Newtonian potential of$f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where$K(x)=c_n|x|^{2-n}$for$n\ne 2$and$K(x)=c_2\log|x|$for$n=2$. In dimensions$n\ge 3$the kernel$K$decays at infinity, so$u(x)\to 0$as$|x|\to\infty$in this case, provided$f$is reasonable (integrable and ... 3 The$\ell^p$spaces are a special case of the$L^p$spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on$\mathbb{N}\$.

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