# Tag Info

1

Hint: define $f: {\mathbb R}^2 \to {\mathbb R}$ such that $f(x,y) = 0$ unless $x^2 < y < 2 x^2$. Note that the intersection of any line through the origin with the exceptional set $A = \{(x,y): x^2 < y < 2 x^2\}$ misses some interval around the origin, so what $f$ does in $A$ does not affect the Gâteaux derivative at the origin.

1

I will answer your last question ("Can someone give me a good definition of weakly compact (in terms of sequences and boundedness)?"), and I will try to edit the post later to see if I can answer the rest. Let us denote by $X^*$ the space of all linear continuous functional on $X$, a normed vector space. Then, the weak topology $\sigma(X,X^*)$ on $X$ is the ...

0

You can view these excellent answers (especially the longest one, who gives you every kind of references) Obviuosly all credits go to the authors of the answers linked above :)

2

If $null(f)$ is not dense in $X$, you can find $x\in X$ and $r>0$ such that $B(x,r)\cap null(f)=\varnothing$. if $y\in X$ is such that $|f(y)|\geq|f(x)|$, then for some $\alpha$ with $|\alpha|\leq 1$ we have $f(\alpha y)=f(x)$, so $x-\alpha y\in null(f)$, hence $x-\alpha y\not\in B(x,r)$, so $\Vert y\Vert\geq\Vert\alpha y\Vert\geq r$. What this just said ...

1

Try this: $null(f)$ is a subspace of $X$ whose codimension is 1. Now if it is not dense, then $null(f)$ is closed (because its closure is a subspace containing $null(f)$ and it is not $X$). Then you show that it implies continuity. For example, as $f\neq 0$ there is a $y\in X$ such as $f(y) = 1$. $$\{x\in X| |f(x)| = 1 \}= (y + null(f))\cup (-y + null(f)) ... 1 This result is true for any bounded operator with \|T\|\leq1. Fix x\in X with \|x\|\leq1. Put$$ C=\tfrac12\,\overline{\text{conv}\,\{T^nx:\ n\in\mathbb N\}}\subsetneq B_X $$(every element in C has norm at most 1/2). This is the most general possible result: let T=(1+\delta)I for some \delta>0. A subset C as desired satisfies ... 2 Fix a basis v_1,\ldots,v_m \in V_k. Then the map \sum \alpha_jv_j\to\left(\sum |\alpha_j|^2\right)^{1/2} defines a norm on V_k, and this norm is induced by the inner product \langle \sum \alpha_jv_j,\sum \beta_jv_j\rangle = \sum \alpha_j\overline{\beta_j}. In a finite-dimensional space all norms are equivalent, so the identity map becomes a ... 0 A compact operator has a closed range iff it has a finite dimensional range. Without loss of generality,we can assume that the range of A is closed,otherwise we can consider the restriction of A to E where E=T−1(F),In all cases the theorem assures that F is finite dimensional. to prove the theorem,consider the canonical map associated with A and the fact ... 2 No, in general you cannot expect this. Simply consider (X,\|\cdot\|)=(\mathbb{R},|\cdot|), and set x=0. Then the inequality reads$$|y|^3 \leq \frac{1}{2} |y|.$$Obviously, this is in general not true for y>1/\sqrt{2}. Yes. The equality$$\begin{align*} \|x\|^2 x - \|y\|^2 y &= (x-y)\|x\|^2 + y (\|x\|^2-\|y\|^2) \\ &= (x-y)\|x\|^2 + y ...

1

By the Lebesgue differentiation theorem, $$(Df)^\prime(t)=\frac{d}{dt} \int_0^t f(s)ds=f(t)$$ for a.e. $t\in[0,1]$. Therefore, $$\langle Df, Dg\rangle_{C^\prime}=\int_0^1 (Df)^\prime(t) (Dg)^\prime(t) dt = \int_0^1 f(t) g(t) dt = \langle f,g\rangle_{L^2[0,1]}$$ so $D$ preserves the inner product. By the characterization of absolutely continuous functions by ...

0

The inclusion $R\supseteq {N^*}^\perp$ follows from a more general fact: If $T:V\to W$ is a bounded linear operator (without assuming closed range), then $\overline{R}={N^*}^\perp$. The inclusion $R\subseteq {N^*}^\perp$ is easy, and since ${N^*}^\perp$ is closed, you get $\overline{R}={N^*}^\perp$. for the converse, take $w\notin \overline{R}$. ...

4

Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

3

Yes. The isomorphism is given by: $$\begin{array}{llll} \varphi:&C^0[a,b]&\longrightarrow&C^0[0,1]\\ &f(x)&\longmapsto&(b-a)f(a+(b-a)x) \end{array}$$

2

Let $X=\ell^2(\mathbb{N})$, let $Y_1=X$, and let $Y_2=\{(0,x_2,x_3,\ldots)\in\ell^2\}$ (the subspace of sequences whose first entry is $0$). Then $Y_1\cong Y_2$ - specifically, the right-shift map $S_r:Y_1\to Y_2$ is an isomorphism - but $X/Y_1\cong 0\not\cong \mathbb{R}\cong X/Y_2$. A similar trick will show for many sorts of mathematical objects that ...

0

The idea is good. I prefer to think in terms of lower bound $m(T)=\inf\{\|Tx\|:\|x\|=1 \}$, which is the same thing as $\|T^{-1}\|^{-1}$ when $T$ is invertible, but does not require us to consider the inverse explicitly. The definition of lower bound implies that $|m(T)-m(S)|\le \|T-S\|$ for any two operators $T,S$. Therefore, $T_{n_k}\to T$ implies ...

2

I think there is something to be fixed with the notions of "equality or the same " and "essential equality or isomorph to" and "isomorphic and isometric to" 1) equality is a "defined" as an axiom ( extension axiom, X=Y iff they have the same elements i.e. $X\supset Y, X \subset Y$) 2) Isomorphic equality between two Banach space means there exists an ...

1

The answer that I was searching is that $Y$ has to be closed to make $X/Y$ actually a normed vector space. I somehow did not find this answer satisfying. Let's consider an example. Let $f:X\to \mathbb R$ be a discontinuous linear functional on $X$ (over real scalars). The space $Y=\ker f$ has codimension $1$. So, $X/Y$ is isomorphic to $\mathbb R$, ...

0

What are the hypothesis on $K$? Observe that a strongly exposed point is a denting point and a denting point is an extreme point. So, for instance take the set $K$ as the unit ball of $c_0$. This set is convex and closed but it has no extreme point. Hence, $K$ can not be the closed convex hull of any kind of extreme points of $K$. Items 1 and 2 can be ...

0

If a Banach space $(X, \|\cdot\|$ is a Hilbert space, then the norm satisfies the "parallelogram identity" $$\|x+y\|^2+\|x-y\|^2=2\big(\|x\|^2+\|y\|^2\big).$$ But the norm of $C[0,1]$ does not satisfy such an identity: for $f=1$ and $g=x$, $$\|f\|=1,\,\,\|g\|=1,\,\,\|f+g\|=2,\,\,\|f-g\|=1,$$ and hence $$... 0 Banach spaces that are not a Hilbert space are, among many others, L^{p}(Rⁿ,dⁿx) for p∈[1,∞),p≠2. Linear functionals on such spaces can be written as an integral similar to the Hilbert space inner product but in general the functional cannot be associated with an element of the space itself. But there exists the notion of a semi-scalar product which was ... 0 The space C[0,1] of continuous functions f:[0,1]\to\mathbb R with the supremum norm is an example of a Banach space which is not a Hilbert space. We need to check that the parallelogram law is not satisfied. Take f(x)=x, x\in[0,1], and g(x)=1, x\in[0,1]. Then 2(\|f\|_\infty^2+\|g\|_\infty^2)=4, but \|f+g\|_\infty^2+\|f-g\|_\infty^2=5. 1 What follows is not a complete answer. First, a general fact. Let E be a Banach space (say real for simplicity) and let \phi be a continuous linear functional on E. Set K:=\ker(\phi), and let also f\in E. Then one can find g\in K such that \Vert f-g\Vert={\rm dist}(f,G) if and only if the linear functional \phi attains its norm, which means ... 1 For (i) \implies (iii). For each N=1,2,3,\ldots, define F_{N}=\{ x : \sup_{n}\|T_{n}x\| \le N \}, and show every x is in some F_{N}. For (iii) \implies (ii). If \|T_{n}\|\le M for some M and all n,$$ \begin{align} \|T_{n}x-T_{n+k}x\| & \le \|T_{n}x-T_{n}y\|+\|T_{n}y-T_{n+k}y\|+\|T_{n+k}y-T_{n+k}x\| \\ & \le ...

1

First of all, in order to have $(iii)$ be equivalent to $(i)$ and $(ii)$, we need that $Y$ is complete (a Banach space) too. As a counterexample when $Y$ is not complete, consider $X = \ell^p$, and $Y$ the subspace of sequences with only finitely many nonzero terms. Then let $$(T_nx)_k = \begin{cases}x_k &, k \leqslant n\\ 0 &, k > n. ... 0 For first point, use that any vector in a pre-Banach space can be normalized to unity. 2 Proof 2 is incomplete. Without the separability assumption, you are facing the fact that the closed unit ball is in general not metrizable in the weak topology, and thus you cannot deduce sequential compactness (the existence of [weakly] convergent subsequences) from the compactness. You could cite the Eberlein-Shmulian theorem to obtain the fact that the ... 0 Given any x\in E, there exists a rank-operator T with Tx=x (use Hahn-Banach to construct a bounded functional f with f(x)=1, and then define Ty=f(y)x). Then$$ S_\alpha x=S_\alpha Tx\to Tx=x. $$This shows part 1. For part 2, let X\subset E be compact. Fix \varepsilon>0. Then there exist x_1,\ldots,x_n such that the balls of radius ... 3 In your proof, you should start with a Cauchy sequence \{(x_n,y_n)\}\subset X\times Y and show that this sequence is convergent in X\times Y. First observe that if \{(x_n,y_n)\} is a Cauchy sequence in X\times Y, then both \{x_n\} and \{y_n\} are Cauchy sequences in X and Y respectively, since$$ ...

1

The correct statement is that a Banach space such all of its closed subspaces are complemented is isomorphic to a Hilbert space. It was proven by Lindestrauss and Tzafriri in 1971. The proof is not very hard but not trivial either.

1

Let $\{u_n\}_{n\in\mathbb N}\subset C_0(\mathbb R)$ be a Cauchy sequence, i.e., for every $\varepsilon>0$, there exists an $N=N(\varepsilon)>0$, such that $$m,n\ge N\quad\Longrightarrow\quad \|u_m-u_n\|_\infty=\sup_{x\in\mathbb R}\lvert u_m(x)-u_n(x)\rvert<\varepsilon.$$ We shall show that there exists a $u\in C_0(\mathbb R)$, such that ...

0

For any $f \in C(\mathbb{R})$, the $L^{\infty}$ norm of $f$ is the same as the $C(\mathbb{R})$ norm of $f$. So the closure of $C(\mathbb{R})$ in $L^{\infty}(\mathbb{R})$ consists of all functions which are equal a.e. to continuous functions. Choose any function in $L^{\infty}$ which is not equal a.e. to a continuous function, and that function cannot be the ...

5

The simplest counterexample is the non-zero constant function $$f(x)=1.$$ If $g\in C_0^1(\mathbb R)$, then $\lim_{|x|\to\infty}g(x)=0$, and hence $$\lim_{x\to\infty}|f(x)-g(x)|=\lim_{x\to\infty}|1-g(x)|=1.$$ Thus $$\|f-g\|_\infty=\sup_{x\in\mathbb R} |f(x)-g(x)|=1,$$ and therefore $f$ can not be approximated by $C_0^1$ functions.

1

It suffices to prove that $B_X(0,1)$ is totally bounded in $Y$; that is, for every $\epsilon>0$ it admits a finite $\epsilon$-net (in the norm of $Y$). Pick linear functionals $\lambda_1,\dots,\lambda_n\in X^*$ such that $$\|x\|_Y\le \frac{\epsilon}{3}\|x\|_X+\max_{1\le i\le n} |\lambda_i(x)|\qquad \forall\ x\in X$$ The image of $B_X(0,1)$ under the map ...

1

The answer is "no". In $c_0$ and $\ell_p$, $1<p<\infty$, the sequence of the standard unit vectors provides a counterexample. One can find counterexamples in a large class of spaces: Let $X$ be a Banach space that lacks the Schur property. ($X$ is said to have the Schur property if every weakly convergent sequence in $X$ is norm convergent. Note ...

1

Example. Let $X=L^2[0,2\pi]$ and $f_n(x)=\sin nx$. Then $f_n\to 0$ weakly, due to Riemann-Lebesgue Lemma, but $\|f_n\|=\sqrt{\pi}$. On the other hand, if $\|f_n\|\to 0$, then $f_n\to 0$, strongly. Note. If the case of Hilbert spaces, if $f_n\to f$ weakly, then $$(f_n-f,f_n-f)=(f_n,f_n)+(f,f)-(f_n,f)-(f,f_n).$$ Clearly $(f_n,f),\, (f,f_n)\to (f,f)$ but ...

4

Bessaga showed something stronger, but only for Hilbert spaces. Generalization to certain Banach spaces (i.e., those which are linearly injectable into some $c_0(\Gamma)$) was given by Dobrowolski: In particular, in 1966 C. Bessaga [1] proved that every infinite-dimensional Hilbert space $H$ is $C^\infty$ diffeomorphic to its unit sphere. The key to prove ...

1

Let $\Phi:B(H,K)\rightarrow S(H,K)$ be the morphism you defined. Given $\phi\in B(H,K)$, you can verify that $\Vert\phi\Vert=\sup\left\{|\langle\phi(x),y\rangle|:x\in H, y\in K, \Vert x\Vert\leq 1,\Vert y\Vert\leq 1\right\}=\Vert\langle\phi(\cdot),\cdot\rangle\Vert=\Vert\Phi(\phi)\Vert$, so $\Phi$ is an isometry. Also, it follows (almost directly) from Riesz ...

1

Let's prove that every finite-dimensional subspace of a topological vector space is closed. Suppose $M$ is the linear span of linearly independent vectors $x_1,\dots,x_n$, and we have a convergent sequence of the form $v_k = \sum_{i=1}^n c_{ik} x_i$. Consider two cases: There is a constant $C$ such that $|c_{ik}|\le C$ for all $i,k$. Then we can choose a ...

0

To allow functions from an open subspace of $X$ makes the definition more general, and thus applicable for cases where the given function could not be extended in a differentiable way to the whole space (simplest exaple is $x\mapsto 1/x$ which is defined only on an open subset of $X=\Bbb R$, and is differentiable there). If a linear function $U\to Y$ is ...

2

Of course it is. You can prove it by convoluting $C_c(\mathbb{R}^d)$ functions with a smooth and compactly supported approximation of unity (also called mollifier). This gives a smooth uniform approximation of the original function. The same argument shows that $C_c(\mathbb{R}^d)$ is dense in $L^p(\mathbb{R}^d)$ for $1\le p<\infty$.

1

Yes, take for example the Haar basis consisting of functions of the form $h_I=\chi_{I_l}-\chi_{I_r}$ where $I$ is a standard dyadic interval and $I_l$, $I_r$ its left, respectively right child intervals and $\chi_I$ are (depending on your needs, properly normalized) characteristic functions. This constitues a countable (Schauder) basis for $L^p([0,1])$ when ...

0

Apparently, $X$ is a dense subspace of $\bar X$, in order to define its dual. Clearly, if $\ell\in X^*$, then $\ell$ extends uniquely, by a standard density argument to an $\bar\ell\in \bar X^*$, and clearly $\|\bar\ell\|=\|\ell\|$. Inversely, if $\bar\ell\in\bar X^*$, then its restriction to $X$ is a bounded linear functional on $X$.

1

But I don't understand $C=\overline{\operatorname{conv}}T(C)$? $\operatorname{conv} A$ is the convex hull of $A$, and $\overline{\operatorname{conv}} A$ denotes the closed convex hull of $A$, that is, $\overline{\operatorname{conv} A}$. So the requirement is that $C$ be equal to the closure of the convex hull of its $T$-image. and why we need $X$ be ...

1

I suppose we can assume the fact that $\overline{S}\colon E_1/\ker S \to E_2$ is a well-defined bijective linear mapping as known. That is pure linear algebra, and has nothing to do with the topologies. Now suppose that $E_1$ and $E_2$ are normed spaces, and $S$ is continuous. Then $\ker S$ is a closed subspace of $E_1$, and hence $E_1/\ker S$ is a normed ...

2

It follows for example with the closed range theorem. $\pi \colon X \to X/Y$ has closed range (it is surjective, by definition of $X/Y$), so $\pi^\ast \colon (X/Y)^\ast \to X^\ast$ has closed range, and thus $\pi^{\ast\ast}\colon X^{\ast\ast} \to (X/Y)^{\ast\ast}$ has closed range. Since $\pi^\ast$ is injective (that follows from the surjectivity of $\pi$, ...

0

If you have all the $\left(\mu_n\right)_{n=0}^\infty$, you can constrain the moment generating function and optimize your $\int \sqrt{f_c(x)} dx$. You should make sure the integral exists, that would determine the underlying vector space...

0

This is true even in operator space setting. See proposition 7.2 in Introduction to operator space theory. G. Pisier

2

Yes, you can use Ascoli-Arzelà. Most easily seen here if you factor the map through $L^\infty([0,1], \lVert\,\cdot\,\rVert_{L^\infty})$: C^1([0,1],\lVert\,\cdot\,\rVert) \underbrace{\hookrightarrow}_{\text{compact}} L^\infty([0,1], \lVert\,\cdot\,\rVert_{L^\infty}) \underbrace{\hookrightarrow}_{\text{continuous}} L^1([0,1], ...

0

Hint: Note that $||f(x)-f(y)|| = ||A(x-y)|| \leq ||A||_{op}||x-y||$ where $||A||_{op}$ is the operator norm of $A$. So the map will be a contraction for $||A||_{op} <1$. Now you just have to find an expression for the operator norm of $A$.

0

I think it might not even be true it's equivalent to a norm like that. My approach would be to show if $(f_n,g_n,h_n)$ is Cauchy in this norm, then $(f'_n + g_n+a h_n)$ , $(h'_n-af_n)$ and $(g'_n)$ converge to some $L^2$ functions, by completeness of $L^2$. Now you should be able to write all 6 of $f_n,g_n,h_n,f'_n,g'_n,h'_n$ in terms of these three to get ...

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