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Strictly speaking, the norm of a Banach space is part of its structure, and two equivalent norms give two different Banach spaces. Since an isomorphism should preserve the whole structure, norm includ …

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 …

A bump function on a Banach space $X$ is a non-zero function with bounded support. Existence of a bump function of a given smooth regularity has, of course, strong immediate consequences --by translat …

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 …

Just to chat, an easier counterexample is $X:=L^p(\mathbb{R})$ for $0\le p<1$, a complete metric separable TVS. The identity map can't be approximated by finite rank continuous linear operators, for …

I think it is true (even in the more general setting of second countable topological spaces). Let $\{A_n\}_{n\in\mathbb{N}}$ be a countable basis of the topology of $X$. For ${n\in\mathbb{N}}$ let $\ …

In other words, if $A$ is the infinitesimal generator of $T$, the mild solution of the abstract inhomogeneous Cauchy problem $$\begin{cases}\dot v =A v +f\\v(0)=0\end{cases}$$ needs not to be $W^{1,1 …

Actually I can't see the role of reflexivity for an answer to the question as it is, unless further properties on $f$ are assumed. In general, to start with the obvious case: if $f$ is a contraction …

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 …

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 …

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 …

True. For any $n\in \mathbb{N}$ consider the inclusion to the $n$-th coordinate $j _ n : X\to c _ 0(X)$ which is right inverse to the evaluation at $n$, so that $(j _ n x)(n)= x$, for any $x\in X$. …

I'd like to mention the following, even though it is just a reformulation: For $V\subset X $ and $I\subset X$ a closed linear subspace, the sum $V+I$ is closed in $X$ if and only if $\pi(V)$ is close …

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 …

For $t\in I$ let $p_t:Y^I\to Y$ denote the projection on the $t$-th coordinate of the product, that is the evaluation map $x\mapsto x(t)$, and let $p^\intercal_t:Y^*\to (Y^I)^*$ be its transpose oper …