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Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper A. Avilés, P. Koszmider, A Banach space in which every injective operator is surjective. Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074 the authors constructed an infinitely dimensional Banach space $... 11 No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take$X$to have bigger cardinality, for example,$X = \ell_\infty^*$. Then you have no chance of building such operators. Another possibility for counterexamples comes with non-separable reflexive spaces (or, ... 9 You ask for an instructive example, so I'll be long winded. Suppose$(x_n)$is basis for a normed space$(X,\|\cdot\|)$. The partial sum projections$S_n$are well defined but might be discontinuous. Define a new, larger, norm on$X$by$|x| := \sup_n \|S_n x\|$. Then$(x_n)$is a Schauder basis for$(X,|\cdot|)$. Note that this is the first step in the ... 8 In the generality stated, the question has a negative answer (which took me embarrassingly long to spot); the point is that when$T_1$and$T_3$are not assumed injective or surjective they give us very little traction on what$T_2$is, in contrast to the intuition one might have from the Five Lemma. The "silly" counterexample might be useful for other ... 6 The answer is no: you can even have$T_1=T_3=0$and$T_2$equal to the identity$id$on an infinite dimensional Banach space. Indeed, consider the following commutative diagram with exact rows:$$\begin{CD} 0@>>> 0 @>0>> X @>id>> X @>>> 0\\ &&@V0VV @VV{id}V @VV0V\\ 0@>>>X @>>id> X @>>0&... 5 Your post is awfully technical for MO, IMO. The lattice of closed ideals in$L(L_1)$contains at least a continuum of elements: https://arxiv.org/abs/1811.06571 5 This is well known but not trivial. It follows primarily from Theorem 2.c.13 of [LT]. The Theorem says (applied to this situation) if every operator$T:E\to \ell_2$is strictly singular, then every complemented subspace of$\ell_2\oplus E$is of the form (up to isomorphism)$X'\oplus E'$for some complemented subspace$E'$of$E$, and$X'$is either finite ... 5 Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density. Suppose that$X$fails to have a strictly convex renorming. Thus, there is no injective operator$T$from$X$into any space$Y$that is strictly convex, as if it were$\|x\|^\prime = \|x\| + \|Tx\...

4

Of course, Yemon was faster than me. But I want to emphasize that the point is very elementary linear algebra: The simple commutative diagram $\begin{CD} 0 @>>> \mathbb R @>f_1>> \mathbb R^2 @>f_2>> \mathbb R @>>> 0\\ @V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ 0 @>>> \mathbb R @>>f_1> \mathbb R^2 @>... 4 I don't know if this question is still interesting for someone, but in 2008 Argyron and Haydon constructed in "A hereditarily indecomposable L∞-space that solves the scalar-plus-compact problem. Acta Math. 206 (2011), no. 1, 1–54" a infinite-dimensional Banach space in which every operator is the sum of a compact operator and a multiple of the identity. ... 4 Yes. Using the injective property of$\ell_\infty$, get an operator$S:X\to \ell_\infty$that is compact on some infinite dimensional subspace$X_0$of$X$. Let$Q: X \to X/X_0$be the quotient map. Define$T:X\to \ell_\infty \oplus X/X_0$by$Tx = (Sx, Qx)$. With a slightly different argument you can replace$\ell_\infty$by any infinite dimensional Banach ... 3 Doesn't this already fail in finite dimensions? Take$X = \mathbb{R}^2$and let$A = B =$the closed region bounded by$y = \sqrt{x^2 + 1}$and$y = x + 1$, both for$x \geq 0$. The intersection with any ball is closed and therefore compact, but$A - B$is not closed: it contains points arbitrarily close to$(0, -1)$but does not contain that point itself. 3 Am I confused? The Hamel dimension of any infinite dimensional separable Banach space is$2^{\aleph_0}$, so they are all the same as vector spaces. So the underlying vector space of one of them can be the underlying vector space of any of them. 3 First, let me mention the result of Mackey: if$X$is an infinite-dimensional Banach space, then its Hamel dimension (= dimension as abstract vector space, over$\mathbf{R}$or$\mathbf{C}$) is$\ge c=2^{\aleph_0}$, with equality if$X$is separable. In the separable case,$X$has cardinal$\le c$so the upper bound is clear. The lower bound is trivial ... 2 In a recent note with J. Swaczyna (arXiv:2005.04873), we proved that assuming the existence of certain large cardinals (for example, the existence of a super-compact cardinal) that filter bases whose underlying filter is a projective subset of the Cantor set, have continuous coordinate functionals. The role of large cardinals is to make the heuristic proof I ... 2 This is not an answer, but I put it here to stop close votes that are apparently based on a misreading of the question. I do not know the answer to the OP's question, but I will point out that both hypotheses are necessary. For example, drop the hypothesis that$A$is convex. Then there is a counterexample with$A=B$; namely,$A=\bigcup_n \{(n+1)e_n, ne_n\}$... 2 The answer is negative. In$X :=\ell_2 \oplus_\infty \ell_1$let$x_n = (M_n \delta_n, 2 e_n)$and$y_n = (M_n \delta_n, e_n)$, where$(\delta_n)$is the unit vector basis for$\ell_2$and$(e_n)$is the unit vector basis for$\ell_1$and$M_n \uparrow \infty$fast. Let$B$be the convex hull of the$x_n$and$y_n$and let$A$be the closure of$B$. Then$...

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Let $\epsilon>0$ and let $(u_{i})_{i=1}^{n}$ be a $\delta$-net for $S_{U}(\delta>0$ will be specified later). For each $i$, choose $x^{*}_{i}\in S_{X^{*}}$ such that $\langle x^{*}_{i},x_{i}\rangle=1$. Let $V=\cap_{i=1}^{n}Ker(x^{*}_{i})$. Then $V$ is the required finite-codimensional subspace.

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The result (even for LF-spaces) is due to J. Dieudonné and L. Schwartz La dualité dans les espaces (F) et (LF), Annales de l’institut Fourier, tome 1 (1949), p. 61-101, propositions 2 and 4. (Proposition 2 says that the inductive limit topology induces on the ,,steps'' their original topologies, and proposition 4 says that bounded subsets of the inductive ...

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