# Tag Info

9

Your condition implies that $X$ is isomorphic to a Hilbert space with isomorphism constant at most $M^2$. The distance condition implies that both type 2 constant and cotype 2 constant of $X$ is bounded by $M$. By Kwapien theorem the Banach-Mazur distance of $X$ to a Hilbert space is bounded by type 2 constant times cotype 2 constant. As a reference see e.g.,...

6

If $f_0\ne0$ or if $f_0=0$ and the $f_1,\dotsc,f_n$ are not linearly independent, then the answer is trivial: In this case there is another functional, say $f_1$, which is in the span of the remaining ones, say $f_1=\sum_{k\ne1}\lambda_kf_k$ with $\lambda_0\ne0$: Extend the $f_k$ to $F_k$ $(k=2,\dotsc,n)$ and put $F_1=\sum_{k\ne1}\lambda_kF_k$. In the ...

5

Assuming $P$ first countable, the standard contraction principle and elementary bounds are sufficient to conclude. You do not need higher regularity: Let $X$, $Y$ be Banach spaces, $P$ a topological space, $f:X\times P\to Y$. Assume that i. $f:X\times P\to Y$ is continuous; ii. $f(\cdot,p):X\to Y$ is differentiable, for any $p\in\ P$. iii. $f(x_0,p_0)=0$ ...

4

If $X$ is not HI, then there exist a decomposable closed subspace $Y\subseteq X$. Let $Y=Y_1\oplus Y_2$ be a decomposition, let $Z_1\subseteq Y_1$ and $Z_2\subseteq Y_2$ be separable closed subspaces, and let $Z=Z_1\oplus Z_2$. Clearly $Z$ is a separable decomposable closed subspace of $X$.

4

No. $A = \ell_\infty$ is a dual Banach algebra. Every separable complemented subspace of $A$ is finite-dimensional, so there is no way to exhaust $A$ by nicely complemented separable subspaces.

4

I am inclined to say yes. This is because reflexive Banach spaces have projectional resolutions of the identity, which are increasing ordinal-indexed nets of contractive, commuting projections that exhaust the whole space and satisfy certain compatibility conditions. From this, it follows that separable subspaces of reflexive Banach space are contained in 1-...

2

While Pietro Majer has already provided an argument and answered the questions; here is a Paper reference for a more general parameter dependent implicit function Theorem (also has Differential dependence on parameters) https://arxiv.org/abs/math/0303320 Glöckners Paper was published in the Israel Journal of math some years later. Addendum (from my comment ...

2

On $L^2(R)$, consider the densely defined operator $u\mapsto \int u\,dx$, defined on $L^2\cap L^1$. This operator is neither closed nor closable. If you want the operator defined on all of X, with X incomplete, just take X to be $L^2\cap L^1$ with the $L^2$ norm.

2

If $T$ is everywhere defined and closed, then it is continuous. See this. If you mean densely-defined $T$, then it's not so hard. The (strong) derivative operator $d/dx$ defined on (a dense subset of) $L^2(-1,1)$ is closable (its closure is the weak derivative), but not closed. To see that it is not closed, Let $u_n(x)$ be a sequence of functions converging ...

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