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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 …

Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space …

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 …

With the definitions in the OP, this is false. It is OK if the Banach space $B$ is separable and $(\Omega,\mathcal F, P)$ is an arbitrary probability space. It is OK if the Banach space $B$ is arbit …

Let $f \colon \Omega \to E$ be your function. $\mu$ is a measure on $\Omega$. Assume, for every $x^* \in E^*$, the composition $x^* \circ f$ is $\mu$-integrable. The Dunford integral in general lie …

Take $\mathcal X = L^1(\Omega,\mu)$ where $\Omega = \{1,2,3,\dots\}$ and measure $\mu(\{k\}) = p_k$ with $p_k > 0$, $\sum p_k = 1$. The norm in $\mathcal X$ is $$ \|f\|_{\mathcal X} = \sum_k |f(k)|p_ …

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 …

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 …

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 …

A particular case of a Banach algebra is the one-dimensional case $\mathbb C$, right? And according to Pietro, if you can prove this in that case, you get the general case by saying "functional calcul …

No. Let $(X, \cal A, \mu)$ be $[0,1]$ with Lebesgue measure. Let $E = L^2[0,1]$ with inner product $\langle \alpha,\beta\rangle := \int \alpha(t)\overline{\beta(t)}\;dt$. Define $f : [0,1] \to L^2[0,1 …

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_ …

Find many worked-out examples, all the important ones, in Dunford & Schwartz, volume I. See the top row of the table beginning on page 374.

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) - …

An example. $X = [0,1]$ with the usual metric. $\mathcal C[0,1] = \mathcal C_b[0,1] = \mathcal C_0[0,1]$. $\mathcal C[0,1]^* = \mathcal M[0,1]$. Let $\mu_n$ be the unit point-mass at $1/n$ and $\mu …