Search Results

For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations. Let $X$ be a dense, countably generated linear subspace of the separabl …

In general, no. Let $T:X\to Y$ be a bounded linear operator between infinite dimensional Banach spaces, as assumed in the question. Assume further $N:=\ker T$ is a separable subspace of both infinite …

Yes, it is true. This inverse function theorem is in a sense half-way between the Lipschitz and the $C^1$ setting. To be more precise let me review some classic results. (Invertibility of Lipschitz p …

As already recalled, a kernel of any non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension as a consequence of the existence of Hamel basis. That said, …

If $(X,\Sigma, \mu)$ is semi-finite (it has no infinite atoms) and non-$\sigma$-finite, by transfinite induction there is a family $\{E_\alpha\}_{\alpha \in \omega_1}$ of disjoint measurable sets of …

I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and onl …

1) "Other examples of $T$ aside from the class of homogeneous functions". Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always mea …

For dual spaces, there is an important characterization: $X^*$ has the Radon-Nikodym property if and only if $X$ is Asplund (its separable subspaces have separable duals). Of course, $\ell_1$ is not …

A funny isometry invariant to distinguish these normed spaces is: The space of spheres of radius $2$ in the unit sphere of $(X,\|\cdot\|_X)$ which are maximal by inclusion, as described below. It turn …

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm. Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets: $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$, $B$, the …

I would say, monomials are not a Schauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implie …

Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual. So $\el …

For a counterexample, let $Y$ be a Banach space with a closed, non-complemented subspace $Z\subset Y$. Consider the restriction of the sum operation ${\bf +}:Y\times Y\to Y$ to the space $X:=Y\times …

Recall the canonical isometric isomorphism $ (X/N )^*\sim N^\perp$, for a Banach space $X$, and a closed linear subspace $N$ of $X $. Also, a linear subspace $G$ of $X^*$ is weakly-star closed …

A simple reason, among many others: for instance, you would like to know whether a certain functional is bounded below, because you are looking for a minimizer of it, and you can prove some inequality …