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For the unit ball $B_X$ of the Banach space there are only two possibilities: sep$(B_X)= 1$, if $B_X$ is not separable, and sep$(B_X)=0$ if $B_X$ is separable. Indeed, if sep$(B_X)<1$ there are $\vare …

Locally convex spaces which satisfy the uniform bounded principle, i.e., every pointwise bounded family of continuous linear maps (with values in any normed space) is equicontinuous, are called barrel …

At least in the locally convex world, $\tau'$ is always finer than $\tau$: Given a continuous seminorm $p$ of $(X,\tau)$ the so-called local Banach space $X_p$ is the completion of $X/p^{-1}(\lbrace 0 …

I would like to add the following important case (which generalizes Bill Johnson's answer): An LS-spaces is defined as a countable inductive limit of Banach spaces $X_n\hookrightarrow X_{n+1}$ with co …

Yet another answer: EVERY incomplete Hausdorff (locally convex) topological vector space is a counterexample! It is due to the late Susanne Dierolf who proved (Manuscripta Math., 17(1):73–77, 1975) t …

There exists a norm on $Y$ making $L$ bounded if and only if the kernel of $L$ is closed. However, $\|y\|_Y = \inf\lbrace \|x\|: L(x)=y\rbrace$ is always a semi-norm (that is $\|y\|_Y$ may be $0$ for …

As suspected by Yemon Choi the question is very closely related to the approximation property: As can be seen e.g. in the book Tensor Norms and Operator Ideals of Defant and Floret (page 64 combined w …

For $\Omega=\mathbb N$ with the disrete $\sigma$-algebra measurability is no condition and $$\ell^\infty(\mathbb N)\hat\otimes_\varepsilon X \cong C(\beta\mathbb N)\hat\otimes_\varepsilon X\cong C(\be …

References are boring if the proof is simple: The open mapping theorem tells you that $L+M$ is closed if and only the canonical map $S:L\times M \to L+M$ $(x,y)\mapsto x-y$ is an isomorphism (where …

As mentioned in my comment, this is true for reflexive Banach spaces and the compactness game may generalize to other situations, e.g., if the Banach space is a dual space and the embedding in $\sigma …

As Gerald suggests, the answer to question 1) is yes. The answer to the second question is negative (if $E$ is infinite-dimensional) because the operator norm makes $B(E)$ a Banach space and, on any B …

Banach spaces where all weakly convergent sequences are norm convergent are said to have the Schur property. A classical Theorem of Schur says that $\ell^1(I)$ has the Schur property for every set $I$ …

For every countable subset $M\subseteq D$ set $X_M=\overline{\text{span}(M)}$ (with the norm of $X$). These spaces seem to satisfy your requirements. Am I missing something?

The isometry of $(X\otimes_\pi Y)^*$ and $X^*\otimes_\varepsilon Y^*$ for finite dimensional $Y$ is in 6.1 of the book Tensor Norms and Operator Ideals by Defant and Floret and the isometry of $(X\ot …

Väth only claims that the sequence $y_n$ is Cauchy in the space of measurable functions with the topology (or rather, uniformity) of convergence in measure which, for finite measures, is given by the …