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Let $\epsilon > 0$. Since $a_n$ is a bounded sequence, by compactness we can find a finite set of scalars $b_1, \dots, b_m$ such that for every $a_n$ there exists a $b_{k_n}$ with $|a_n - b_{k_n}| \ …

You've clarified in comments that $\le$ means that the operators $A_{n+1} - A_n$ are positivity preserving. So for any nonnegative $x \in \ell^1$, we have $A_1 x \le A_2 x \le \dots$ pointwise. Sinc …

According to Are $L^\infty(\Bbb R)$ and $L^2(\Bbb R)$ homeomorphic?, the map $f \mapsto \operatorname{sgn}(f) |f|^2$ is a homeomorphism from $L^2$ to $L^1$. It's clearly a bijection. Suppose $f_n \t …

For (1), take $X = c_0$ so that $X^* = \ell^1$, which is separable. Take $Y = \{y \in \ell^1 : \sum_i y(i) = 0\}$ which is a proper closed subspace. Since $Y$ contains all the elements of the form $ …

Matthew's answer reminded me of a fact that makes this easy: if $X$ is a normed space (say, over $\mathbb{R}$) and $f : X \to \mathbb{R}$ is a linear functional, then its kernel $\ker f$ is either clo …

I don't know whether you'll consider this "simple", but here is a proof. I distilled it from Eric Schechter's Handbook of Analysis and its Foundations, which has a proof of a more general statement a …

For a somewhat less explicit example, it suffices to find a continuous $g : X \to \mathbb{R}$ which is unbounded on the unit ball (then fix any nonzero $x_0 \in X$ and take $f(x) = g(x) x_0$). Pick a …

It depends on whether $\hat{X}$ spans $X$ (in the algebraic sense, i.e. finite linear combinations). If it does, then for every $x \in X$, we can write $x = a_1 x_1 + \dots + a_n x_n$ for some $x_1, \ …

I presume you want the operators $\phi_n$ to be continuous. Let $X$ be a separable Banach space (which necessarily has cardinality $\mathfrak{c}$). If such a sequence exists, then $X$ has the approx …

These are called Bochner spaces. Under mild assumptions (see Gerald's post), they are Banach spaces. It is sufficient to assume that $B$ is separable, or that $L^p(\Omega, B)$ is defined to include …

Let's write $H$ for $L_2(\mu)$, since its Hilbert space structure is all we care about. I assume $\gamma$ is a Borel measure. By duality, you can naturally identify $H$ with $H^*$, and $H^*$ is a se …

If $y_n$ does not converge weakly to $y$ then there is a weakly open neighborhood $U$ of $y$ and a subsequence $y_{n_k} \notin U$. By weak compactness this subsequence has a weak limit point $z \noti …

Update: Here is a ZFC (probably even ZF+DC) counterexample for Q1. It's from probability and kind of indirect, maybe someone will be able to find something shorter. Let $X = C_0([0,1])$, the space …

The answer to Q2 is No. I am grateful to Damian Sobota for drawing my attention to the following paper: Darst, R. B. On a theorem of Nikodym with applications to weak convergence and von Neuman …

Just to flesh out Bill's answer and comments thereon, we have the following facts. Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator. If $T$ has dense range then $T^*$ is injec …