# Tag Info

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.

3

The Banach function spaces may be a better choice for your question. You can see here for the definition of a Banach function space or this book Function Spaces, Volume 1 (chapter 6). Answer to your question: The first thing is to introduce the definition of the Riesz-Fischer property. We say that a normed linear space $(X,\left\|\cdot\right\|)$ has ...

3

Complementing the very well detailed answer of Zev Chonoles, I think you've got it in the opposite direction. $T$ as is defined in the title is the mapping $x\mapsto T(x)=x^T$, such that $T(x)(y)=x^Ty$ is a linear transformation of the vector $y$. This means that the domain of $T$ is $(\mathbb R^n, \Vert \cdot \Vert_\infty)$ and its codomain is the dual ...

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 ...

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 ...

1

Sketch of proof. Suppose you have a Cauchy sequence $\{f_n\}\subset C_b^k(U)$, with respect to $\|\cdot\|_k$. Then, for every $\lvert\alpha\rvert\le k$, the sequence $\{D^\alpha f_n\}$ is uniformly Cauchy in $U$ and hence uniformly convergent to a continuous and bounded function, say $f^\alpha\in C_b(U)$ - All the $f^\alpha$'s are for the moment only ...

1

If $V$ is not separable, then $L^p(X,\mu,V)$ is not either. Take $\{v_j\}$ an uncountable set in $V$ without a limit point and $f_j\in L^p(X,\mu,V)$, such that $f_j(x)=\varphi_j(x)v_j$, where $\varphi_j\ne 0$ scalar. Clearly, there is no limit point in $\{f_j\}$. If $X$ is separable, then again it is not certain that $L^p(X,\mu,\mathbb R)$ is separable. It ...

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 $f(y) = 1$. $$\{x\in X| |f(x)| = 1 \}= (y + null(f))\cup (-y + null(f)) ... 1 A pair (V,\|\cdot\|) denotes a vector space V over \mathbb{R}, together with a norm function \|\cdot\|:V\to\mathbb{R}. Thus, (\mathbb{R}^n,\|\cdot\|_1) and (\mathbb{R}^n,\|\cdot\|_\infty) mean "the vector space \mathbb{R}^n equipped with the L^1 norm", and "the vector space \mathbb{R}^n equipped with the L^\infty norm", respectively. ... 1 Let X, Y be Banach spaces and U\subset X open. A function u\colon U\to Y is said to be differentiable at a\in U if there exists a linear operator Du_a\colon X\to Y (the differential of u at a) such that$$ u(a+h)=u(a)+Du_a(h)+o(h). $$If u is continuous, then Du_a is also continuous. If u is continuous and differentiable at very point ... 1 If the space X is banach it is an easy consequences of the open mappig theorem. Anyway with the norm induced topology over  X  you in fact are resizing a ball so it is a ball again and it is open by definition of the topology. So the map sends open ball in open ball therefore it is open. This reasoning heavily rely on the "absolute omogeneity" of the ... 1 It is true if you assume that T is self-adjoint (i.e. symmetric), meaning that$$ (Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1} $$and assuming that$$ |(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2}  Note that your inequality holds even for $T=-2I$, and thus we NEED to assume these two additional things: $(1)$ and $(2)$. So ...

1

The answer is yes. For every normed space $X$ we can define as $\alpha$, the cardinality of the set of minimum cardinality which is dense in $X$. Clearly, $\alpha\ge\aleph_0$. For every such set $D$, described in the question, we shall show that $\lvert D\rvert=\alpha$. It is not hard to see that the set $E=\bigcup_{n\in\mathbb N} 2^{-n}D$ is dense in ...

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