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0

As Frank provided a counterexample the inclusion in general doesn't hold. In fact if your measure is finite the inclusion is vice versa because via Hölder you get \[ \|f\|_q \leq \mu(X)^{\frac{1}{p} -\frac{1}{q}} \cdot \|f\|_p.\] Maybe you are talking about the sequence spaces $\ell^p$ because there the inclusion holds as every function which is in ...


2

Let $f(x):= x^{-1/p} \mathbf{1}_{(1,\infty)}$. Then $f^q$ is integrable because $\frac{q}{p} > 1$. But $f^p = \frac{1}{x}$ (on $(1,\infty)$) is not. That is, $f \in L^q(\mathbb{R})$ but $f \notin L^p(\mathbb{R}) $.


0

Yes, it is. Let $M=\{x\in X: y^*(x)=0 \text{ for all }y^*\in Y^*\}$. This is sometimes denoted ${}^\perp Y^*$ and called the pre-annihilator of $Y^*$. Every $y^*\in Y^*$ induces a linear functional on $X/M$ in a natural way: $y^*(x+M)=y^*(x)$ is well-defined. Conversely, if $\phi$ is a linear functional on $X/M$, then its composition with the projection ...


0

Your set $A_\sim$ is the projection onto the second coordinate of the set $$ \{(X,Y)\in \mathrm{SB}^2 \mid (X,Y)\in A\times \mathrm{SB} \text{ and } X\sim Y\}, $$ and $A\times \mathrm{SB}$ is the preimage of $A$ by the first projection, so both sets are analytic.


0

First note that the way you are defining $T$ does not give you guarantee that $T$ will be linear. To make sure that $T$ is linear you need $\{x_n \mid n \in \mathbb{N}\}$ and $\{y_n \mid n \in \mathbb{N}\}$ to be linearly independent, which you can assume without loss of generality. Because if they are not linearly independent then you can choose a maximal ...


1

"Naturally identified" leaves enough wiggle room to identify an infinite subset $X$ of $\mathbb N$ with the sequence in $\mathcal N$ that enumerates the elements of $X$ in increasing order. The set of all these increasing sequnces is a closed subset of $\mathcal N$.


1

Let $U$ be a subspace of a Banach space $V$. Suppose $x\in U$ is an interior point of $U$, which means that $$ B(x,\varepsilon)=\{y:\|x-y\|<\varepsilon\}\subseteq U $$ for some $\varepsilon>0$. Since $U$ is a subspace, also $$ B(0,\varepsilon)=-x+B(x,\varepsilon) $$ is contained in $U$. Now, for every $v\in V$, $v\ne0$, the vector $$ ...


1

This might seem redundant but also seems correct to me. If we take the sequence $(S_n)$ to be the constant sequence of $\textbf{Identity operators}$ then this sequence converges strongly to the Identity operator. Since $(T_n)$ converges to $T$ weakly and in this case the sequence $T_nS_n$ is nothing but the sequence $(T_n)$, which clearly fails to converge ...


1

Hint: Try to prove the following result $\textbf{Result}:$ If for a sequence $(x_n)$ in a Banach space $X$, the sequence $(f(x_n))$ is bounded for all $f\in X^*$, then the sequence $(\|x_n\|)$ is bounded. Now suppose that you have proved the result. Suppose $(T_n)$ converges to $T$ in weak operator topology, i.e., \begin{align*} |f(T_nx)-f(Tx)| ...


2

The category of Banach spaces is locally $\aleph_1$-presentable.


2

$$\langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle\forall \alpha\\ \langle x,x\rangle+2\alpha\langle x,y\rangle+\alpha^2\langle y,y\rangle\geq\langle x,x\rangle\forall\alpha\\ 2\alpha \langle x,y\rangle+\alpha^2\langle y,y\rangle \geq0\forall\alpha$$ But the left-hand side equals zero if $\alpha=0$ and if $\alpha=-2\langle x,y\rangle/\langle ...


0

I just did |T'(X)| = |2014 cos(2014x)| < sup (2014 cos(2014x)) and it was not smaller than 1 so T is not a contraction is that right?


1

I think that you should add $\alpha < 1$ in your statement. So, you just have to show that there exists at least one $x\in[0, 2\pi]$ such that this inqequality doesn't hold. Do you have any ideas? Just take the derivative of the $\sin(2014x)$.


1

First of all you should understand that what is meant when one says that "$\textit{Not all normed linear spaces are inner product spaces.}$" Here is the explanation. An $\textit{inner product space}$ is a vector space with an inner product defined on it. An inner product on $X$ defines a norm on $X$ given by \begin{equation} \|x\| = \sqrt{\langle x,x ...


4

The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ...


0

There are at most $2^{\aleph_0}$ subsets of a standard Borel space, or of a separable metric space. The same bound applies to coanalytic subsets. But usually, you have a lot more of non Borel subsets (for instance, in the case your family ${G}$ is uncountable).


0

Here's a proof that $T$ is continuous in the strong operator topology. As noted in the question, for each $x\in X$, the map $t\mapsto T_tx$ is right-continuous under the weak topology, and norm-bounded on each bounded interval. To complete the proof given in the question for an arbitrary Banach space, we just require the following lemma. Lemma: Let $X$ ...


-1

every finite dimensional Banach space is a reflexive bot it need not be uniformly convex .$X=\mathbb{R}^n , n\geq2 $with the norm $\Vat x\Vat_{1}=\sigma {i=1}^\n\vert x{i}\vert


2

Yes, it follows from the following general theorem (see Dixmier "$C^*$-algebras and representations", 2.10.2): Every irreducible representation $\rho$ of a $C^*$-subalgebra $C$ of a $C^*$-algebra $A$ can be continued to an irreducible representation $\pi$ of $A$ in a possibly larger Hilbert space. You can also apply the Lemma 2.10.1 from Dixmier directly. ...


0

Well, it is enough to show that the unit ball is compact with respect to some locally convex topology, that is weaker than the norm topology. See this article. For example, if $X$ is a reflexive space with a basis, then the unit ball of $\mathscr B(X)$, the space of bounded linear operators on $X$, furnished with the weak operator topology is compact. This ...


3

You assumed that $T_\lambda$ is surjective (when applying bounded inverse theorem). It is not true. In fact a small modification of your second argument shows precisely that $T_\lambda$ cannot be surjective.


0

What about the following generalisation (it works, but is it useful?) ? Let's deal with polynomials of the form $P(A,B,z)=A-Bz$ first and let's give a modified definition to the "generalised resolvent set" $\rho(A,B)$: \begin{equation} \left\{z\in \mathbb{C}\left| \right. \left(A-zB\right)^{-1}\text{,}B\left(A-zB \right)^{-1}\text{exist on a dense (common) ...


2

If $\epsilon>0$, choose a step function $\tau$ such that $\sup_{x\in[a,b]}\|f(x)-\tau(x)\|<\frac\epsilon4$. Now, since $\tau$ is a step function, there is a partition $\{I_k\}_{k=1}^n$ of $[a,b]$ such that $\tau$ is constant on each interval $I_k$. Let $M$ denote the set of endpoints of the intervals $I_k$. This set is finite. Now, suppose that $x\not ...


0

Let $\{x_n\}_{n\in \mathbb{N}}$ be a set with dense linear span and $T:X\rightarrow Y$ be an isomorphism. Put $y_n=Tx_n$. Then $\{y_n\}_{n\in \mathbb{N}}$ has dense linear span in $Y$. Now let $(a_n)$ be a sequence such that $\sum_{n=1}^\infty a_n x_n$ exists. For every $N$ the following inequality is true: $$\|\sum_{n=1}^N a_n x_n\|=\|T^{-1}\sum_{n=1}^N a_n ...


2

Q1. Non-zero constant functions DO NOT BELONG to $W_0^{1,2}(\Omega)$. Q2. The space $W_0^{1,2}(\Omega)$ is defined as the «completion» of $C_0^\infty(\Omega)$ with norm $$ \|u\|_{W^{1,2}_0(\Omega)}=\left(\int_\Omega\big(u^2(x)+\big|\nabla u(x)\big|^2\big)\,dx\right)^{1/2}. $$ Thus, indeed, every element of $W_0^{1,2}(\Omega)$ can be ...


1

In terms of getting a proof done, it really is just Cauchy-Schwarz: $$\left | \int_0^t f(s) ds \right | = (f,1)_{L^2} \leq \| f \|_{L^2} \| 1 \|_{L^2} = t^{1/2} \left ( \int_0^t f(s)^2 ds \right )^{1/2}.$$ Cauchy-Schwarz can be proven in the abstract setting of a general inner product space, and then it follows in your case from the fact that the $L^2$ ...


0

You can characterize $W^{1,2}_0$ as "the 'compactly supported' $L^2$ functions that have $L^2$ weak derivatives." Since the weak derivatives of a constant function are all zero, hence $L^2$, the answer to your question is the answer to, "are constant functions compactly supported $L^2$?" This is never true unless the constant is zero.


0

Just take the union of $0$ and a sequence of disjoint closed annuli around $0$ of decreasing sizes. It's still closed and absorbing, but not convex, and it includes no ball around 0. Edit: the above is wrong. The definition of absorbing set is equivalent to be a neighbor of 0 in every intersection with a one-dimensional subspace. An example of a radial set ...


0

Let $y \in V \setminus W$ with $\def\norm#1{\left\|#1\right\|}\norm y = 1$. Then, as $W$ is closed $d := \def\dist{\mathop{\rm dist}}\dist(y, W) > 0$. Choose $\delta > 0$ such that $\frac{d}{d+ \delta} > 1-\epsilon$. There is some $w \in W$ with $\norm{y-w} < d + \delta$. We let $v := \frac{y-w}{\norm{y-w}}$. Then $\norm v = 1$ and $$ \dist(v,W) ...


1

if $x \in V$ is non-zero then $||x|| \gt 0$, so set: $$ x' = \frac{x}{||x||} $$ it is now evident that $||x'|| = 1$


1

Let $(g_{n})$ be a sequence in $K$ satisfying $\left\|f-g_{n}\right\|_{p}\rightarrow\delta:=\inf_{g\in K}\left\|f-g\right\|_{p}$. It follows from the triangle inequality that the $g_{n}$ are bounded in norm, and therefore lie in a closed, convex subset $E\subset K$, which is weakly compact by Alaoglu's theorem. Whence $(g_{n})$ has a subset ...


2

For $L^1$, take $K$ to be the set of functions for which $\int h \mathrm{d}x = 0$. Let $f$ be the characteristic function of the interval $[0,1]$. Note $$ 1 = \int h - f \mathrm{d}x \leq \|h - f\|_1 $$ Note that this minimum is achieved for any $h$ of the form: $h = +1$ on a measure 1/2 subset of $[0,1]$, $h = -1$ on its complement in $[0,1]$, and $h = 0$ ...


0

It means closed under isomorphism.


1

Hint: $$\|f\|^2 = \sum_{k=1}^n \|f^{(k)}\|_2^2 \ge \|f^{(k)}\|_2^2$$


1

One example might be $M(K)$, the space of all regular Borel measures on $K$ of finite variation, where $K$ is compact space. This space arises as the dual of $C(K)$.


3

No, it cannot. For any $x_1, \ldots, x_n \in X$, the linear span $V$ of $x_1, \ldots, x_n$ is finite-dimensional, so it is not all of $X$, and by Hahn-Banach there is a nonzero continuous linear functional $f$ such that $f = 0$ on $V$. Take $y \in X$ with $0 < \|y\| < r_0$ so that $f(y) > r \|f\|$. Now $$\|y - x_j\| \ge \|f\|^{-1} |f(y - x_j)| = ...


0

That $I - A$ is both injective and surjective follow from the fact that it is invertible. Indeed, let $B:X \to X$ be any invertible operator; then we have an operator $B^{-1}$ such that $BB^{-1} = B^{-1}B = I. \tag{1}$ To see that (1) implies $B$ is injective, suppose that $Bx_1 = Bx_2 \tag{2}$ for some $x_1, x_2 \in X$. Then from (1) $x_1 = Ix_1 = ...


2

Hint Prove that $Id+A+A^2+..+A^n+...$ defines an operator which is the inverse of $Id-A$.


1

You know that $B(X)$, the space of bounded linear operators on $X$ is a Banach space. Consider the series $$ \sum_{n=0}^{\infty} A^n $$ Since $\|A\| < 1$, it converges absolutely, and thus converges in $B(X)$ to an operator $B$. Now check that $B(I-A) = (I-A)B = I$ and so $(I-A)$ must be surjective.


0

Given $\varepsilon > 0$, there is $n_0 \in \mathbb{N}$ such that: $$ m,n \geq n_0 \implies \Vert f_m - f_n \Vert < \frac{\varepsilon}{4}$$ Then, we have: $|f_m(0) - f_n(0)|< \varepsilon /4$ $\sup \left\{ \frac{|(f_m-f_n)(t)-(f_m-f_n)(s)|}{|t-s|^p}: t\neq s, \quad t,s\in [0,1] \right\} < \varepsilon /4$ Given $t\in]0,1]$, we have for all ...


1

In fact all normed spaces are subspaces of some function spaces. This could be the reason why functional analysis have its name.


1

$\mathbb{R}$ is an infinite dimensional vector spaces over $\mathbb{Q}$, but it is not an Hilbert space. You can see that all the axioms for a vector space are verified if you define the sum of two ''vectors" as the usual sum of real numbers and the product for a scalar $q \in \mathbb{Q}$ as the usual product. This space has an infinite dimensional Hamel ...


1

Nevermind, I answered it myself using an elementary argument. Thanks anyway! EDIT: Here is the argument, in brief. Consider the real Banach spaces $\widetilde{X}=X\oplus_{\ell_1}X$ and $\widetilde{Y}=Y\oplus_{\ell_1}Y$. It is a simple exercise to show that $T\oplus 0$ and $0\oplus T$ both lie in $\mathcal{FSS}(\widetilde{X},\widetilde{Y})$, and hence so ...


1

Relative directions of elements of vector spaces can be expressed via the scalar product: If we have a (pre-)Hilbert space $\cal H$ with a scalar product $\langle \cdot, \cdot \rangle$, then we say that two elements $f,g \in \cal H$ are orthogonal (i.e. their relative angle is 90°) if $\langle f, g \rangle = 0$. If you take a (say, finite) set of vectors in ...


0

In my opinion direction can be defined only fo Hilbert spaces ( sice these are the spaces where we cand define an angle). If $V$ is an Hilbert space than, given two vectors $v,w$ we can say that they have the same direction ad are equi-oriented iff $ \langle v,w\rangle=||v||\cdot||w||$ and have the same direction but are opposite iff $ \langle ...


2

Well we already seem to have a structure that encompasses length and direction in the notions of $\mathbb{R}^2$ and $\mathbb{R}^n$. Describe a vector in $\mathbb{R}^2$ by $(r,\theta)$ where $r$ is the length of the vector and $\theta$ is the direction of the vector. In $\mathbb{R}^n$ we can do the exact same thing, just using the $n$-tuple ...


2

Note: Here are two examples of partial converses which could be convenient. Partial converse in section 27 of A Hilbert Space Problem Book by Paul R. Halmos: Every bounded subset of a Hilbert Space is weakly bounded. We can read in section $27$: Uniform boundedness The celebrated principle of uniform boundedness (true for all Banach ...


1

Let $(u,v)\neq(0,0)$. $D_{(0,0)}f(u,v)=\displaystyle\lim_{\epsilon\to0}\dfrac{f\big((0,0)+\epsilon((u,v))\big)-f(0,0)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{f(\epsilon u,\epsilon v)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{\epsilon^4 u^3v}{\epsilon(\epsilon^4 u^4+\epsilon^2 ...


1

Being able to extend an isometry to an isometry is rare. I don't think any nontrivial Banach space has this property. Here is a proof for $\ell^1$ and $\ell^\infty$, the two spaces of main interest to you. In both cases $e_1,e_2,\dots$ is the standard set of $0-1$ vectors, e.g. $e_2=(0,1,0,0,\dots)$. The case of $\ell^1$ Let $A=\{0,e_1,e_2\}$. Define ...



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