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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.

13 votes
2 answers
749 views

Smooth Urysohn's lemma on Fréchet spaces

Let $V$ be a Fréchet topological vector space. Let $K_0$ and $K_1$ be two closed subsets which are disjoint. I wish to show the existence of a Fréchet-smooth function $f:V\to [0,1]$ whose restriction …
André Henriques's user avatar
6 votes
2 answers
200 views

holomorphy in infinite dimensions (holomorphic families of operators)

Let $X$ be a Banach space (over $\mathbb C$), and let $\mathcal L(X)$ be its algebra of bounded linear operators. Let $U\subset \mathbb C^N$ be an open subset, and $f:U\to \mathcal L(X)$ a function th …
André Henriques's user avatar
13 votes
1 answer
704 views

Trace-class operator satisfies $\sum |\lambda_n|<\infty$?

Here's an "exercise" which I thought should be easy, but which I find myself unable to do. Let $V$ be a Banach space. Recall that an operator $f:V\to V$ is trace-class if it is in the image of the n …
André Henriques's user avatar
6 votes
1 answer
212 views

Nice S¹-action implies existence of unconditional basis?

Let $V$ be a Banach space equipped with a continuous linear action of $S^1$ (meaning, the map $S^1\times V\to V$ is continuous). Assume that all the eigenspaces of the $S^1$-action are finite dimensio …
André Henriques's user avatar
9 votes
2 answers
336 views

Does $End(V)$ remember $V$, where $V$ is a locally convex space?

Let $V$ be a locally convex topological vector space over $\mathbb C$, and let $A=\mathrm{End}(V)$ be its algebra of continuous linear endomorphisms (viewed just as a $\mathbb{C}$-algebra, not as a to …
André Henriques's user avatar
15 votes
0 answers
258 views

Spaces locally modelled on $L^2(\mathbb R)$

In this recent question, I learned that any two separable Banach spaces are homeomorphic. Based on some readings, I'm guessing that $L^2(\mathbb R)$ is homeomorphic to $\prod_{n=1}^{\infty} (0,1)$ (in …
André Henriques's user avatar
33 votes
1 answer
1k views

Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$?

Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$? More generally, are there instances of surprising homeomorphisms between non-isomorphic Banach spaces?
André Henriques's user avatar