# Tag Info

10

For $p>1$, the set is not closed. To see this, note that $$x_n = (1/n, \dots ,1/n,0\dots)$$ is an element of $S$ (the number $1/n$ appears $n$ times). Now, we have $$\Vert x_n \Vert_{\ell^p} = 1/n \cdot n^{1/p} = n ^{1/p -1}\to 0,$$ for $p>1$. But $0$ is not an element of the set. EDIT: Here is more on the general principle involved: Your set ...

8

The set is path-connected. Given $A$ and $B$, construct a path from $A$ to $B$ by starting with the straight line segment from $A$ to $B$, with uniform speed; whereever this is in the space, the path is given by the line segment; wherever it is not, this is because of a specific disk; define the path to map the interval that would have been mapped to a chord ...

6

This might be a counter example where I replace open balls by open rectangles. let consider the set {$1/n$}... if $n$ is odd thne consider the open rectangles $(1/n,1/n+1) \times (2k, 2k+2)$ s.t $k \in \mathbb{Z}$ ...and if $n$ is even then consider open rectangles $(1/n,1/n+1) \times (2k+1, 2k+3)$ s.t $k \in \mathbb{Z}$... now my claim is that if I ...

6

Please pay attention: you end up with the statement $x'(x-y)=0$ for all $x' \in X'$. This implies $x-y=0$, as a corollary of the Hahn-Banach theorem.

6

If $B = H$, we can of course conclude it. Otherwise, choose an $x \in H \setminus B$, and consider $A = x^\perp$. Then $A^\perp = \operatorname{span} \{x\}$, hence $A^\perp \cap B = \{0\}$, but $A \neq H$.

6

If $a=0$ or $b=0$ or $c=0$ then it is obvious. Let $a,b,c\neq 0$. Divide by $b^pc^q$ and get: $$\frac{a^pa^q}{b^pc^q}+1\leq (\frac{a}{b}+1)^p(\frac{a}{c}+1)^q$$ Now substitute $x=\frac{a}{b}> 0$ and $y=\frac{a}{c}>0$. We have to prove $$x^py^q+1\leq (x+1)^p(y+1)^q$$ Note that $\frac{1}{\frac{1}{p}}+\frac{1}{\frac{1}{q}}=1$ and use Holder's ...

6

I don't believe that Hausdorff is used; the result is true for compact $K$. They state the result for compact Hausdorff $K$ just because nobody cares about $C(K)$ for non-Hausdorff compact $K$.

6

Any bijection on a discrete space is automatically a homeomorphism. A nontrivial vector space over nondiscrete numbers with discrete topology fails to be a TVS. This gives you a counterexample as follows: Given a nontrivial vector space over the rational, real or complex numbers. Endow it with the discrete topology. Then translation by any vector and ...

6

To answer your question: in the several decades that I've been reading math papers, I haven't ever seen definitions like this that I can recall. My reading's mostly been in topology, however, so I can't speak to other areas of mathematics. But there's probably a good reason I haven't seen it: When you define a class this way, it may well be empty, or ...

5

Of course if you can plug in $$x(t) = \begin{cases} 1 & \text{if } t\in [-1, 0] \\ -1 & \text{if } t\in (0,1]\end{cases}$$ Then this function satifies $f(x) = 2$. But this $x$ is not continuous. However you can approximate this by $x_n \in C([-1,1])$, where $$x_n (t) = \begin{cases} 1 & \text{if } t\in [-1, -\frac 1n)\\ -1 & \text{if } ... 5 Use the word "the" here; you are referencing the unique space that is the span of your basis. 5 It's not even differentiable. What happens if you fix b and let a vary from one side of b to the other? It is piecewise linear though. If one cuts the plane along the diagonal x=y then in the left/upper component the function is just y, while in the right/lower component the function is x. 4 Consider f_1 = \chi_{[0,1]}, f_2 = \chi_{[0,0.5]}, f_3 = \chi_{[0.5,1]}, f_4 = \chi_{[0,0.25]}, f_5 = \chi_{[0.25,0.5]}, f_6 = \chi_{[0.5,0.75]}, f_7 = \chi_{[0.75,1]}.... f_n converges to 0 in L^1 but does not converge pointwise anywhere in [0,1]. 4 The "real" definition for weak convergence is: u_n\rightharpoonup u in H^1(\Omega) if for each bounded linear functional f on H^1(\Omega): \langle f,u_n\rangle\rightarrow \langle f,u\rangle as n\to\infty. We want to show that both deffinitons are equivalent. 1) Note that for arbitrary functions g_0,g_1,...,g_n\in L^2(\Omega) the functional ... 4 Real inner products are bilinear (that is, linear in each argument) and symmetric. Therefore$$\begin{align*} \langle x+y,x-y\rangle&=\langle x+y,x\rangle-\langle x+y,y\rangle\\ &=\langle x,x\rangle+\langle y,x\rangle-\langle x,y\rangle-\langle y,y\rangle\\ &=\langle x,x\rangle +\langle x,y\rangle-\langle x,y\rangle -\langle y,y\rangle\\ ...

4

This is a fairly typical notation for function space-valued things. The idea is that for each fixed time $t$, we have a function of a particular type with respect to $x$. So here the notation $\mathcal{D}^\prime_t(I;\mathcal{S}(\mathbb{R}^n))$ would mean distributions of the kind $u(t,x)$ such $u$ is a distribution with respect to $t$, but for each fixed ...

4

Hint. $A_k u$ is by definition the map $x \mapsto u(x + 1/k)$ (where $u \colon \mathbf R \to \mathbf R$ is a given measurable, $p$-integrable map). You only have a sequence of maps $A_k \colon L^p(\def\R{\mathbf R}\R) \to L^p(\R)$ here, neither the domain nor its subsets consist of sequences, $L^p(\R)$ is the space of $p$-integrable maps $\R \to \R$. Do not ...

4

Let's use the closed graph theorem. We want to prove that: $${\rm gr}(P) = \{ (x,P(x)) \in \Omega \times \Omega \mid x \in \Omega \}$$is closed. Let $((x_n, P(x_n)))_{n \geq 1}$ be a sequence in the graph such that $x_n \to x$ and $P(x_n) \to y$. Since $(P(x_n))_{n \geq 1}$ is a sequence in the range of $P$, which we assume closed, we have that $y$ is in ...

4

I think you have two questions: If $f\in L^p$, how does $f$ induce a tempered distribution $T_f$? Why is the Fourier transform of a tempered distribution tempered? Here are the answers: Tempered distributions are the continuous dual of the Schwarz class $\mathcal{S}$. This means that a tempered distribution is a continuous linear functional $T\colon ... 4 One of the nice things about Functional Analysis is that you can generally reduce to the scalar case by applying a linear functional to everything, rearranging scalar integrals, and then pulling the functional back outside. Then, knowing that you have enough functionals to separate points allows you to remove the functional from both sides of the resulting ... 4 Let$x$be non equal to zero. By the Hahn Banach theorem, the functional defined by$\phi(x)=1$on$\mathbb K.x$can be extended in a functional$\phi'$defined on all$X$and we have$\phi'(x)\neq 0$. 4 The space will be separable if and only if$\Sigma$has only finitely many sets of positive measure (by ignoring null sets, from here on out I will just refer to$\Sigma$as being "finite" or "infinite"). If$\Sigma$is finite then there exist pairwise disjoint nonempty sets$A_1,\ldots,A_n\in\Sigma$such that$\bigcup_iA_i=X$and $$\text{for all ... 4 The proof of linear independence is the same as for integer powers. Let 0\le a_1<a_2<\dots<a_n and suppose that$$ \sum_{k=1}^n\lambda_k\,x^{a_k}=0,\quad x\in\mathbb{R}. $$Divide by x^{a_1}, take derivatives and multiply by x to obtain$$ \sum_{k=2}^n\lambda_k(a_k-a_1)\,x^{a_k-a_1}=0,\quad x\in\mathbb{R}. $$Repeat n-1 times to get ... 4 So you want to prove that if \ker(f) is dense, then f is discontinuous, or equivalently, if f is continuous, then \ker(f) is not dense. Suppose f is continuous. Then \ker(f), which is the preimage of the closed set \{0\}, is closed in X by continuity. If it were dense, then \ker(f)=X, so f would be the zero functional. Contradiction. 4 Assume \| f \|_\infty=\infty. Let M>0. Define A_M=\{ |f| \geq M \}. Then \mu(A_M)>0. Take p so large that \mu(A_M)^{1/p} \geq 1/2, then \| f \|_p \geq (\mu(A_M) M^p)^{1/p} \geq M/2. Since M was arbitrary, f is not uniformly bounded in L^p, and your result follows by contraposition. This is essentially the argument suggested by John ... 3 You should prove that the E_{ij} within K(\ell^2(I)) satisfy the universal property. This proves existence and uniqueness of C^\ast(X;R) in one swoop. Obviously the E_{ij} are linearly independent, so whenever you have elements x_{ij} \in A satisfying the relations, you can define a linear map by sending E_{ij} \mapsto x_{ij}. It is easy to see ... 3 Let A=I+T^{\star}T. Then A=A^{\star} and$$ \|x\|^{2}\le \|x\|^{2}+\|Tx\|^{2}=(Ax,x) \le \|Ax\|\|x\| \\ \implies \|x\| \le \|Ax\|. $$So A is injective. The range of A is dense because \mathcal{R}(A)^{\perp}=\mathcal{N}(A^{\star}) = \mathcal{N}(A)=\{0\}. To see that the range is closed, suppose \{ Ax_n \} converges ... 3 Let \chi_A be the characteristic function of the set A. Now observe that \| \chi_A - \chi_B\|_{L^p}^p = \int|\chi_A - \chi_B|^p = \int_{A\setminus B}1 + \int_{B\setminus A}1 = \mu(A\setminus B) + \mu(B\setminus A) = \rho_{\triangle}(A,B) Now consider the function \varphi: \Sigma \to L^p(X, \Sigma,\mu ) defined as A \mapsto \chi_A. This is an ... 3 Take a test function \phi with support in [-R,R]. Then your operator applied to \phi becomes$$\sum _{n,d}\frac{\phi(n/d)}{(nd)^\sigma}\le \|\phi\|_{\infty}\sum_{n,d:1\le n\le Rd}\frac{1}{(nd)^\sigma}.$$It easy to check that the sum$\sum_{n,d:1\le n\le Rd}\frac{1}{(nd)^\sigma}$is finite for$\sigma>1$. Hence, by definition, your operator is a ... 3 Let$X$be uncountable and$\mu$the counting measure on$X$. Then for all$p \in (1, \infty)$,$L^p(X, \mu)\$ is reflexive but not separable.

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