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It is an old result of Victor Klee (answering a question of Banach) that a metrizable topological vector space (i.e., there is a translation invariant metric giving the topology) is a complete topolog …

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 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 …

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 …

A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet spac …

The answer to this question is YES -- but it is useless! In fact, a theorem of Valdivia (which you can find, e.g., as Theorem 6.5.8 in the book Barrelled Locally Convex Spaces of Bonet and Perez-Car …

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 …

Eric Wofsey's answer is very nice and simple. Nevertheless this might be interesting. Very old results of Kakutani [Concrete Representation of Abstract (M)-Spaces, Ann. of Math. 42 (1941)] show that f …

This counterexample is so simple so that I might overlook something stupid: Let $T:\ell_1\to c_0$ be the inclusion map. With the natural isomorphisms $\ell_1^*=\ell_\infty$ and $c_0^*=\ell^1$, the tra …

A colimit is an object $E$ together with morphisms $i_n:E_n\to E$ commuting with the inclusions $i_{n,m}:E_n\to E_m$ (i.e., $i_m\circ i_{n,m}=i_n$) such that, for every sequence of morphisms $f_n:E_n\ …

In general, there is no easy criterion. I recall the construction of two closed subspaces of a Banach space whose sum is not closed: Let $T:X\to Y$ be a linear map between Banach spaces with closed gr …

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 …

No. Let $I$ be an index set with the cardinality of the continuum. Endow $X=\mathbb R^I$ with the product topology. According to (a particular case of) the Hewitt-Marczewski-Pondiczery theorem (which …

It is well-known that a Hausdorff locally convex space is semi-reflexive (i.e., the canonical map into its bidual is surjective) if and only if every weakly closed bounded set is weakly compact. This …

Interpreting the question as Dmitri Pavlov I assume that $L^1_K$ is the space of $L^1$-functions with support in $K$ so that we have an inductive spectrum of Banach spaces (identifying almost everywhe …