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In an $L^1(\mu)$ space, we can tell when two elements are disjoint (that is, have disjoint support): $f_1, f_2$ are disjoint if and only if $\|f_1\pm f_2\| = \|f_1\|+\|f_2\|$. I claim that if $L_1(\m …

Space $X = l_1$ is separable, therefore has an equivalent norm which is strictly convex. Let $Y$ be the space with that norm. Now every subspace of $Y$ is strictly convex, and so it remains to show …

The answer to the first question is NO. Even among norms on $\mathbb R^2$, the only ones that have this amazing property (any isometry defined on a finite set extends to an isometry defined on the wh …

The answer is sometimes "no", even when $Y$ is one-dimensional. Let's say the scalars are $\mathbb R$; the same argument will work for $\mathbb C$. Let $\omega_1$ be the least uncountable ordina …

The support of any $L_1$ function is $\sigma$-finite. A countable union of $\sigma$-finite sets is $\sigma$-finite. So a sequence of $L_1$ functions is supported by a $\sigma$-finite set. Now you a …

Bill is correct: $J(\omega_1)$ is not of codimension $1$ in its bidual. The remarks after Proposition 3 say: (if $\eta$ is infinite) then $J(\eta)^{**}$ is isometric to $\widetilde{J}(\eta+1)$, and t …

Yes, the James space. This is a good question, and R. C. James is rightly praised for this example. MR0044024 (13,356d) James, Robert C. A non-reflexive Banach space isometric with its second …

Every metric space of nonmeasurable cardinality is realcompact. [1] 15.24. Thus, if there are no measurable cardinals, then every metric space is realcompact As you noted...To get a Banach space t …

Let $X$ be the space of real polynomials, normed as functions in $C[a,b]$. Here we want $0 < a < b$ fixed. Now define $T \colon X \to X$ so that $T(x^n) = 0$ if $n$ is even and $T(x^n) = nx^n$ if $n …

The classical reference is: M. M. Day, Normed linear spaces. Springer-Verlag, 1958 (reprinted: Academic Press, 1962)

Let's build a "fat Cantor set". Start with $A_0 = [0,1]$ with measure $\alpha_0=1$. Then remove a short open interval centered at $1/2$, leaving a set $A_1 \subset A_0$ of measure $\alpha_1 < \alpha_ …

Alternatively (or maybe it is the same...?) in Solovay's model (which has DC, so I don't have to tell you what I mean by "continuous") where every subset of R has the property of Baire, it follows tha …

Let $X$ be a separable Banach space. Any linear subspace with the property of Baire (in particular, any linear subspace that is a Borel set) is closed. According to the Axiom of Choice, if $X$ is al …

Counterexample. $E = L^2[0,1]$ and $\gamma \colon [0,1] \to E$ defined by $\gamma(x) = 1_{[0,x]}$, the characteristic function of interval $[0,x]$. Then $\gamma$ is continuous, in fact $\|\gamma(x) - …

If K, L are countable compact metric spaces, then the isomorphism of C(K) and C(L) depends on the Cantor-Bendixson rank ... If the first empty derived set of K is K^(n) and the first empty derived set …