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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.
7
votes
Accepted
Large ideally convex sets
Just take your favorite decreasing sequence $\lambda_k$ of positive numbers with sum $1$ and inductively construct the vectors $x_k\in C$ such that $\left \|\sum_{k=1}^n\lambda_k x_k-x\right\|\le\lamb …
6
votes
Accepted
Bounds on dimension of a subspace
Just use the Wirtinger's inequality that says that if $u=0$ at the center of an interval $I\subset \mathbb R$, then $\int_I|u|^2\le (|I|/\pi)^2\int_I|u'|^2$. Thus, if the dimension is $>n$, we can cre …
2
votes
Accepted
Boundedly complete bases
As to the question, the answer is "no". The simplest counterexample is the closure of finite support sequences in the norm $\sup_{j\ge 1}|a_j|+\sum_{j\ge 1}{|a_j-a_{j+1}|}$ (decaying to $0$ sequences …
6
votes
Accepted
Unconditionally convergent series in $\ell_2$ consisting of $\ell_1$-small vectors
If I understand correctly what you are asking, the answer is "certainly not".
Consider $k$ orthogonal vectors $v_j$ with $k$ coordinates $\pm 1$ (the Hadamard matrix). Now multiply them by $t$ and tak …
6
votes
Accepted
A map into a Hilbert space with prescribed orthogonality
Local compactness is not required but separability (and metrizability) seem essential for the construction below. Also, the function $L$ itself is more of a red herring: all we really need is the symm …
9
votes
Accepted
On sums of independent random variables in Banach spaces
Certainly not always. The most trivial example seems to be $X=\ell^\infty$, $\eta_n=\pm e_n$ (with probability $1/2$ for each sign), and $\xi_n$ being uniformly distributed on $\pm e_1,\dots,\pm e_N$ …
10
votes
Accepted
Multiple of identity plus compact
OK, let me try too. It is going to be a somewhat long story. WLOG, $\|T\|\le 1$.
Step 1: It is enough to show that for every finite-dimensional subspace $E$ and every $\delta>0$, there exists a unit …
1
vote
Accepted
Norming functionals for vectors in intersections
Let $X=\ell^1$, $Y=\ell^2$. Take $y_n^*=(1,1/n,1/n,\dots,1/n,0,0,\dots)$ with $\frac 1n$ repeated $n^3$ times. Looks like we are fried, or am I missing something in the setup?
6
votes
Accepted
Regular $p$-norm of a matrix
It is trivially $n^{1/p}$ for $p>2$ (and, therefore, $n^{1-\frac 1p}$ for $1<p<2$ because the norm of the adjoint operator in the dual space is the same as the norm of the operator in the space itsel …
8
votes
Operator norms of circulant matrices
The answer to Question 2 is "No". Note that the $2$-norm is just the maximum of $\left|\sum_{k=0}^{n-1}a_{k+1}z^k\right|$ over the $n$-th power roots of unity and the $1$-norm is just $\sum_k |a_k|$. …
5
votes
Continuity in Banach space for non-linear maps
For $X=\ell^1$, put $f(x)=\sum_{n\ge 1} nx_i^n$ (this one is even analytic if you understand that word in a not too restrictive sense).
6
votes
Projections in Banach spaces
I'm not sure if I like the ultrafilters, so I decided to find some elementary construction. I do not have much imagination for chains of commuting projections either, so let us consider the space of a …
8
votes
Banach-Mazur applied to a Hilbert space
By "explicit", I was really asking for a sequence of continuous functions...
Ah, OK. Take a Peano curve $(f,g)$ from $[-1,1]$ to $[-1,1]^2$. Now define inductively $F_1=f$, $F_{k+1}=F_k\circ g$. …