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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
11
votes
Accepted
Embedding the $\ 2^n+1$-point metric 1-space
Indeed no. If it exists, for each of our points $p_i$ consider the set $A_i$ homothetic to their convex hull with center $p_i$ and coefficient bit less than $1 /2$. By volume argument some two such se …
10
votes
Non-separable Banach space
Let $\varphi$ be a continuous function supported on $[0,1]$. Then continuum many combinations $\sum c_k \varphi(x+k)$, $c_k\in \{0,1\}$ are separated in our space.
9
votes
Accepted
Contact points for John's ellipsoid
Looks true. A necessary and sufficient condition for these points (let $E$ be a standard ball) is that the identity operator $I$ is a non-negative linear combination of projectors $P_i$ on lines throu …
8
votes
Accepted
If $K$ is a countable compact metric space is the set of extreme point of $Ba(C(K))$ countable?
It is true (if you mean real spaces, for complex spaces it is obviously false). Note that if a function $f$ is extremal, it takes only values $\pm 1$. Indeed, if $|f(a)|<1$, then choose a small ball $ …
6
votes
Explicit description of the closure of a given set
The closure is the set of uniformly continuous bounded functions. At first, each function $f$ in the closure is bounded, as you note, and it is uniformly continuous: If $\|f-f_n\|<\delta/3$ and $|f_n( …
5
votes
Accepted
$l^1$ versus $l^2$
Is it true? $l^1$ is a sum of finite-dimensional $l_n^1$ over $n=1,2,\dots$. In summands you have almost spherical sections of large dimensions by Dvoretzky theorem, this allows to change norm a bit s …
4
votes
Accepted
Finite-dimensional subspaces of $c_{0}$
For non-zero sequence $x=(x_1, x_2, \ldots)$ denote $L(x)=\min\{i:x_i\ne 0\}$ (leader of $x$). By Gauss elimination, $M$ contains a basis $(p_1, \ldots, p_d) $ with distinct leaders $m_1<m_2<\ldots<m_ …
3
votes
Accepted
$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$
There is a useful simple
Lemma. If $X\sim X\oplus X$, $Y\sim Y\oplus Y$, and each of $X,Y$ is isomorphic to a complemented subspace of another, then $X\sim Y$.
Proof. We have $X\sim Y\oplus A$, then …
3
votes
An elementary question on subspaces
Not always. Say, both subspaces may be dense, but have trivial intersection. This situation is generic: you may choose two countable dense subsets $A,B$ in a separable Banach space so that their union …
3
votes
Accepted
Compact restrictions of the inclusion of $J:L_\infty(0,1)\to L_1(0,1)$
Take the functions which are constant on the intervals $[1/(n+1),1/n)$ for all positive integers $n$.
3
votes
Regarding orthogonality in Banach space
Consider the plane generated by $x$ and $y$. Now draw a tangent line to the sphere $\{z:\|z\|=\|x\|\}$ at point $x$. It has the form $\{x+tw|t\in \mathbb{R} \} $, and $w$ may be chosen of the form $\a …
3
votes
Accepted
Vectors concentrated on one coordinate
Take $X=l^1$ and $Y$ the subspace of sequences with zero sum, $e_k$ standard. Then any $x\in Y$ with $\|x\|=1$ satisfies $|e_k(x)|\leqslant 1/2$ for all $k$.
3
votes
Accepted
Separable subspaces in dual spaces
Yes. Take a countable dense subset $A$ in $Y$. For each element $a\in A$ take a countable sequence $x_n$ in a unit ball of $X$ such that $a(x_n)$ tends to $\|a\|$. Let $Z$ be a closed span of all thes …
2
votes
Accepted
An inequality in product space $V$
It follows from the 1-dimensional case which you say is true: project everything to a randomly chosen line, apply 1d case and integrate.
2
votes
Integral estimate (inequality) with a Schwartz function
$\DeclareMathOperator\supp{supp}\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Bigabs[1]{\Bigl\lvert#1\Bigr\rvert}$
We have $$
\Bigabs{\int_{-\infty}^\infty \frac{f(x-t)}{t+iy}dt}=
\Bigabs{\int_{x-A}^{ …