# Tag Info

## New answers tagged functional-analysis

-1

The weak topology $\tau_{w}$ is the weakest topology for which the elements of $V^{\star}$ are continuous from $(X,\tau_{w})$ to the scalar field. For any $\phi \in V^{\star}$, the sets $\phi^{-1}B_{r}(0)$ must be open neighborhoods of $0$ for any $r > 0$. Finite intersections of such inverse images under elements of $V^{\star}$ will form a base of ...

0

The precise statement is the following: a set $A\subseteq X$ is in the weak topology iff $$\forall a\in A \exists \epsilon\in \mathbb{R}_{\geq0}\exists F\subseteq V^*: F\text{ is finite and }\{y\in X\mid\forall\phi\in F:|\phi(a-y)|<\epsilon\}\subseteq A.$$ It sounds like the key point you are missing is the implicit universal quantifier in the definition ...

0

let $(s_i)$ be a net in $A+B$ with $s_i=a_i+b_i$ which converges to $x$, since $B$ is compact then there is a convergent subnet $(b_{h(j)})$, let us say that $b$ is the limit. Then $a_{h(j)}=s_{h(j)}-b_{h(j)}$ converges to $x-b$. Since $A$ is closed, we have that $x-b\in A$, hence $x\in A+B$. We conclude that $A+B$ is closed.

0

Consider a net $\{a_\alpha + b_\alpha\}_{\alpha \in I}$ with $a_\alpha \in A$ and $b_\alpha \in B$ that converges to $c$. We need to show $c \in A + B$. Since $B$ is compact, there is a subnet $\{b_\beta\}_{\beta \in J}$ converging to some $b \in B$. Then $a_\beta = (a_\beta + b_\beta) - b_\beta$ converges to $c - b$, and since $A$ is closed this must be ...

0

The word interpolation generally used to mean "interpolation of a function". However the link refers to interpolation of "spaces" not "functions". It is known from Riesz–Thorin theorem that Fourier Transform transform element of one $L^p$ space to another $L^q$ space.

0

It is an easy consequence of U.B.P. applied to the normed space $X^*$ and the set of operators $S:={J(x_n); n\in N}$. Indeed, for every $f \in X^*$ there is a $C_f$ such that $|y(f)|\leq C_f$ for all $y\in S$. Applying the U.B.P we have that there is $C$ so that $\Vert y\Vert\leq C$ for all $y\in S$. But every y of S is of the form $y=J(x_n)$, and thus ...

0

Every Hausdorff topology on a real or complex finite-dimensional topological vector space is equivalent to the usual topology. So, without doing any checking: No, your topology cannot result in a TVS.

4

Scalar multiplication will not be continuous unless the topology on the scalar field $K$ is also discrete (or unless $X=0$). Indeed, if $x\in X$ is any nonzero vector, $a\mapsto a\cdot x$ is a continuous injection $K\to X$, as the restriction of the scalar multiplication map $K\times X\to X$ to the subspace $K\times\{x\}\cong K$. If $X$ has the discrete ...

3

The Cauchy-Kowalevski Theorem, together with Lewy's example, provides some food for thought. The CKT is the main result about local existence of a solution of an analytic first order system of PDEs. Roughly speaking, if all the coefficients, the force and the boundary datum are analytic at some point the PDE admits a local solution which is $C^{\infty}$ and ...

2

zhw points out a nice property of analytic functions: If $f,g$ are analytic on $(a,b)$ and $f(x_n)=g(x_n)$ for a sequence of distinct points converging to some some $x_0\in(a,b)$, then $f(x)=g(x)$ for all $x\in(a,b)$. This becomes false if we loosen the restriction of analyticity, as can be seen by considering the functions ...

0

Suppose $f\in L_{w}^{p}$. Fix $\lambda>0$ and define $f_{0,\lambda}:=f\chi_{\left|f\right|\geq\lambda}$ and $f_{1,\lambda}:=f-f_{0,\lambda}$. I claim that $f_{0,\lambda}\in L_{w}^{p_{0}}$ and $f_{1,\lambda}\in L_{w}^{p_{1}}$. Indeed, \begin{align*} ...

0

Well the answer was contained in the question linked by rschwieb (Prove the approximate identity from the unitization), I'll write it up here. If a C* Algebra $\mathscr{U}$ does not contain a unit then approximate identities will not be approximate identities in the C* Algebra $\mathscr{\tilde U}$ with a unit adjoined. Every approximate identity in a C* ...

0

We will first show that, given $x\in X$, there exists $z\in N$ s.t. $\|x-z\|_X=d(x-z,N)$ and $p(x)=p(x-z)$. Let $\{z_n\}\subset N$ s.t. $\|x-z_n\|_X\to d(x,N)$. Then $\infty>\|x\|_X+\sup_{n}\|x-z_n\|_X>\sup_{m}\|z_m\|_X$. Since $N$ is finite dimensional, there exists a subsequence $z_{n_k}$ and $z\in N$ s.t. $z_{n_k}\to z$. Therefore, ...

1

Is $e_0$ supposed to have norm 1? Notice that if we suppose that, we obtain $1=|g_i(e_0)|\leq \Vert g_i\Vert_E\Vert e_0\Vert=\Vert g_i\Vert$, hence $\Vert g_i\Vert_E =1$.

0

OK. So the whole thing was very simple. First note that both $\tilde{H}$ and $\tilde{L}$ commutes with $J$. Now $\tilde{H} + \imath J \tilde{L}$ \ge 0$is equivalent to $$\begin{bmatrix} \tilde{H} & -J \tilde{L} \\ J \tilde{L} & \tilde{H} \end{bmatrix} \ge 0.$$ Which is possible if and only if$\tilde{H} \geq - J ...

1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...

1

Given a random variable $X$ with cumulative distribution function $F_X$, we get a probability measure $m_X$ on the Borel $\sigma$-algebra of $\mathbb{R}$ which is generated by the equations $m_X((-\infty,x])=F_X(x)$ for each $x \in \mathbb{R}$. This $m_X$ is sometimes called a pushforward measure. Is this what you mean?

2

There are a few ways to define the span of $M$, what you wrote in your first paragraph is one of them. Taking all scalar multiples of elements of $M$ does not give the correct definition of the span of $M$: in $\mathbb{R}^2$, the set of all scalar multiples of $M=\{(1,0),(0,1)\}$ is the union of the $x$-axis with the $y$-axis which is not even a subspace. ...

-1

The distribution function simply is the function from reals to reals $f(x)=|\{X\leq x\}|$. The only function on $T$ involved is the original random variable $X$.

1

You are pretty much there. Since $f$ is continuous on $[0,M]$, it follows that $\lim_{\varepsilon\to0^+}\int_\varepsilon^Mf(x)\, dx=\int_0^Mf(x)\, dx$. EDIT: $f$ here is defined as you originally did before you edited the question: $$f(x):=\begin{cases} \frac{\varphi(x)-\varphi(0)}x&\text{if }x>0,\\ \varphi'(0)&\text{if }x=0. \end{cases}$$

4

For $0 < p < 1$, let $$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$ We endow it with the $p$-seminorm $$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$ and the metric $d_p(x,y) = s_p(x-y)$ derived from it. Note the difference from the case $p \geqslant 1$. If ...

0

The Heine-Borel theorem plays a crucial role in the development of both Riemann and Lebesgue integration on $\mathbb{R}^d$. Perhaps the most important result in Riemann integration is that continuous functions on closed intervals $[a,b]$ are integrable. The reason is that $[a,b]$ is compact by Heine-Borel, so continuous functions on it are uniformly ...

3

Daniel Fischer's comment says yes, since $\ell^1=c_0^*$. One can give an explicit projection from $(\ell^\infty)^*$ onto $\ell^1$ by saying $$P\Lambda=(\Lambda e_1,\Lambda e_2,,\dots).$$

0

Consider the function $f(x) = \alpha_1\|x\|_1 + \alpha_2\|x\|_2$ for some scalars $\alpha_1,\alpha_2 > 0$. It should be clear that $f$ defines a norm. Now show that $f$ satisfies the parallelogram law. It should be straightforward to do so, but let me know if you need any help.

1

Suppose that $\{e_1,e_2,...\}$ is a basis of the Banach spaces $M$. Let $M_n=\{e_1,e_2,...,e_n\}$. So $M_n$ is closed and $int(M_n)=\emptyset$. Given $x\in M$, since $\{e_1,e_2,...\}$ is a hamel basis of $M$ there exists $n$ such that $x=\sum_{j=1}^n\alpha_je_j$, so $x\in M_n$. This prove that $M=\cup_nM_n$. Then $M$ is a countable union of sets with empty ...

2

This looks suspiciously like one of the questions in my differential equations workshops this year. Anyway, let $z=||x-y||^2.$ Then $\frac{d}{dt}z=\frac{d}{dt}||x-y||^2$ which is equal to $$=2 \begin{pmatrix} x_1-y_1 \\ x_2-y_2 \end{pmatrix} \begin{pmatrix} -x_1+2x_2+y_1-2y_2\\ -2x_1-x_2+2y_1+y_2 \end{pmatrix} \\ ... 2 Of course every finite dimensional Banach space has a finite basis. As a consequence of Baires category theorem every infintedimensional Banach space cannot have a countable Hamel Basis, but very often a Schauder basis.There exist separable Banach spaces that don´t even have a Schauder basis so as was shown by Per Enflo. So You are right, the word ... 1 The counterexample doesn't hold true: Because \sup_{x\in[0,1]}x^{-\frac{1}{2}}=\infty it follows f\notin L^{\infty}([0,1]), so the pairing \langle f,f\rangle is not defined.For f\in L^p,g\in L^q, p^{-1}+q^{-1}=1 the dual pairing \langle f,g\rangle is always finite, because of Hölder's inequality. 1 "Scalar product" is a misnomer. It's actually a duality, that is, a bilinear map L^p\times L^q\to\mathbb F. 1 Short Answer: Sigma Additivity Explicitly: Suppose there is another measure Q such that Q(\{\omega\})=p_{\omega} for all \omega\in \Omega but P\neq Q. Therefore there is a subset A\subset \Omega such that$$P(A)\neq Q(A)\Rightarrow P(A)-Q(A)\neq 0.$$Note that A must be countable as a subset of a countable set. Hence the above above may be ... 0 Let x=(x_1,\dots,x_n)\in\mathbb{R}^n and ||x||_3^3=\sum_{i=1}^n|x_i|^3. Using the fact that \forall 1\leq i\leq n, |x_i|\leq||x||_{\infty}, we can chose m=n^{-\frac{1}{3}}. On the other hand:$$||x||_{\infty}^3=\left(\max_{1\leq i\leq n}|x_i|\right)^3=\max_{1\leq i\leq n}|x_i|^3\leq\sum_{i=1}^n|x_i|^3=||x||_3^3$$So you can chose M=1. 0 We have the following: Suppose z:=(x,y)\in R^2. First prove that | x |^3 + | y |^3\leq(| x | + | y |)^3. Therefore, 2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3. On the other hand, |x|^3+|y|^3\geq \Vert z \Vert_\infty^3. Thus, \Vert z \Vert_3 \geq \Vert z \Vert_\infty. Summing up: 2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 ... 2 It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, m and M. It is easiest to look at the case when ||(x,y)||_\infty=1. What does that equation tell you about x and y? You can use this to bound ||(x,y)||_3. In terms of the shape of ||.||_3, you really just want to sketch the curve ... 0 Using the Gram-Schmidt process, construct orthogonal vectors f_n from the A^nf. Note that \Vert f_n \Vert\le \Vert A^nf \Vert. From the Gram-Schmidt equations it is apparent that, on the orthonormal vectors \frac{f_n}{\Vert f_n\Vert}, A is a Jacobi matrix operator J with n,n+1 elements b_n := \frac{\Vert f_{n+1}\Vert}{\Vert f_n\Vert} and ... 0 So, as indicated in my earlier comments, I end up with the same expression as you involving \lambda and \delta; however, I couldn't arrive at the expression in the exercise statement. Here's my attempt at the optimization step. Define a function of \varphi=\varphi(\delta) by ... 1 The Spectral Mapping Theorem allows you to more easily compute the spectrum of some operators. If you know that you can write an operator A as A=f(a) for f\in hol(a) and a\in\mathcal{A}, where you already know the spectrum of a, you can compute the spectrum of A, since \sigma(A)=f(\sigma(a)). 0 The second and the third statement are basic facts in Sobolev Space theory. The first one is probably a corollary of the other two, but I think it can be proved directly. Anyway, Michel Willem's book on Functional Analysis, Birkhauser-Verlag, contains the proofs of these facts. You can also read Haim Brezis's book on functional analysis (Springer-Verlag) for ... 1 Consider the monomials$$f_n(x) = \frac{x^{n^2}}{n}.$$1 One classic example that can help you here is the set of functions of the form$$ f_n(x) = \sqrt{x^2 + 1/n} $$for n \in \Bbb N. Note that the functions f_n(x) form a Cauchy sequence, but their derivatives do not. 0 You can think of a linear functional on a vector space as giving you a coordinate in that space. A continuous linear functional is a coordinate that varies continuously as the vector varies continuously; one is not so much interested in the coordinates that are disconnected from the topology of the space. Representation of a linear functional gives the ... 2 The set S is convex and symmetric, i.e., -x \in S for all x\in S. If x is an interior point of S and B(x,\varepsilon)\subseteq S (where B(x,\varepsilon) is the ball around x with radius \varepsilon) you get B(0,\varepsilon/2)\subseteq S since for \|y\| <\varepsilon/2 you have y= \frac 12 (-x) +\frac 12 (x+2y) \in \frac 12 S + \frac ... 1 Look at this: http://planetmath.org/banachspacesofinfinitedimensiondonothaveacountablehamelbasis there is refference also. Example is any Banach space of infinite dimension . 1 (a) is false as the sequence (0,\dots,0,n,0,\dots) is in H. (b) is true due the continuity of inner product of l_2 and that the sequence (1/n) is in l_2. 0 I assume the correct definition of d_L is d_L(f,g) = \sum_i 2^{-(i+1)} d_Y(f(x_i), g(x_i)). Suppose \{f_n\} is a Cauchy sequence in L(X,Y). Hints: Show that for each i, \{f_n(x_i)\} is a Cauchy sequence in Y. Therefore... If f is going to be the limit of f_n, this tells you what f(x_i) has to be, for each i. If f is going to be ... 1$$ \begin{align} &d_\infty[T(x_1,x_2)-T(y_1,y_2)]\\ &=d_\infty\left[\left(\frac{x_1+2x_2}5-1,\frac{x_1-2x_2}7+1\right)-\left(\frac{y_1+2y_2}5-1,\frac{y_1-2y_2}7+1\right)\right]\\ &=d_\infty\left[\left(\frac{(x_1-y_1)+2(x_2-y_2)}5,\frac{(x_1-y_1)-2(x_2-y_2)}7\right)\right]\\ ...

1

Let's denote for convenience $\Delta x=y-x$, so that $y=x+\Delta x$, and consider $$\Delta T=T(x+\Delta x)-T(x)=(\Delta T_1, \Delta T_2)=\Bigl(\frac{\Delta x_1+2\Delta x_2}{5},\frac{\Delta x_1-2\Delta x_2}{7}\Bigr).$$ We need to estimate $$d_\infty(T(y),T(x))=\max\{|\Delta T_1|,|\Delta T_2|\}.$$ Let's do one component at at time $$|\Delta ... 0 In what follows below, I will denote the norm defined in (1) by \left\|\cdot\right\|_{\alpha,r;q}^{(1)} and the norm defined in (2) by \left\|\cdot\right\|_{\alpha,r;q}^{(2)}. So the algebraic identity that I was missing is the following lemma, which allows us to write the r^{th} difference [T(t)-I]^{r} as a linear combination of [T(2t)-I]^{r} and ... 1 The intersection of a countable dense subset M with the subspace Y can be empty. A possibility would be to project M onto Y (i.e., choosing p(x)\in Y such that \|p(x)-x\| \le \|y-x\| for all y\in Y) and then show that \lbrace p(x):x\in M\rbrace is dense. 0 Consider$$ f_n(x,y)=f(x,y)+ \frac{h(y)}{\sum_k h(k/n) g_n(k)}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right)$$with h_n(y) = ne^{-1/y} if y>0 and h_n(y) = 0 if y \leq 0 then it verifies:$$\sum_k f_n(x,k/n) g_n(k)=\sum f_n(x,k/n) g_n(k)+ \sum_k\frac{h_n(y/n)g_n(k)}{\sum_k h_n(k/n) ...

5

All normed linear spaces are metric spaces, and metric spaces are Hausdorff.

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