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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.

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 ...

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

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 $$... 2 Proof 2 is incomplete. Without the separability assumption, you are facing the fact that the closed unit ball is in general not metrizable in the weak topology, and thus you cannot deduce sequential compactness (the existence of [weakly] convergent subsequences) from the compactness. You could cite the Eberlein-Shmulian theorem to obtain the fact that the ... 2 Of course it is. You can prove it by convoluting C_c(\mathbb{R}^d) functions with a smooth and compactly supported approximation of unity (also called mollifier). This gives a smooth uniform approximation of the original function. The same argument shows that C_c(\mathbb{R}^d) is dense in L^p(\mathbb{R}^d) for 1\le p<\infty. 2 Yes, you can use Ascoli-Arzelà. Most easily seen here if you factor the map through L^\infty([0,1], \lVert\,\cdot\,\rVert_{L^\infty}):$$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], ...

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 $\pi$, ...

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 ...

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 ...

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 ... 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 ... 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 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 ... 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 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 $$(every element in C has norm at most 1/2). This is the most general possible result: let T=(1+\delta)I for some \delta>0. A subset C as desired satisfies ... 1 Try this: null(f) is a subspace of X whose codimension is 1. Now if it is not dense, then null(f) is closed (because its closure is a subspace containing null(f) and it is not X). Then you show that it implies continuity. For example, as f\neq 0 there is a y\in X such as f(y) = 1.$$ \{x\in X| |f(x)| = 1 \}= (y + null(f))\cup (-y + null(f)) ...

1

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 ...

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 ...

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 ...

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

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 ...

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

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

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 ...

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 $\{u_n\}_{n\in\mathbb N}\subset C_0(\mathbb R)$ 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 ...

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 $$(T_nx)_k = \begin{cases}x_k &, k \leqslant n\\ 0 &, k > n. ... 1 For (i) \implies (iii). For each N=1,2,3,\ldots, define F_{N}=\{ x : \sup_{n}\|T_{n}x\| \le N \}, and show every x is in some F_{N}. For (iii) \implies (ii). If \|T_{n}\|\le M for some M and all n,$$ \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

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 ...

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