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5

Fix $\varepsilon>0$ and $g\in L^{\infty}([0,1])$. If $T$ is surjective, then $g=T(f)$ for some $f\in L^1([0,1])$, indeed. By the continuity of $T$ at $f$, there exists some $\delta>0$ such that \begin{align*} \left.\begin{array}{ll}\bullet\,h\in ...


3

In fact, any infinite-dimensional Banach space $X$ has a compact subset that does not lie in any finite-dimensional linear subspace. Namely, define a sequence $x_n$ inductively such that $x_{n+1}$ is not in the linear span of $\{x_1, \ldots, x_n\}$ but $\|x_{n+1}\| < 1/n$. Then $\{0\} \cup \{x_1, x_2, \ldots\}$ is your set.


2

Let's first check that $P^\perp$ has invariant range. Let $a \in A$ and let $\eta \in [A\xi_1]^\perp = (A\xi_1)^\perp$. Then, for any $b \in A$, $$ \langle a\eta, b\xi_1 \rangle = \langle \eta, a^\ast b \xi_1 \rangle = 0, $$ where $a^\ast b \in A$ and hence $a^\ast b \xi_1 \in A\xi_1$ precisely because $A$ is a self-adjoint algebra of operators in $B(H)$, ...


2

The nearest point projection onto a closed subset $E\subset \mathbb R^n$ is single-valued if and only if $E$ is convex*. In this case, the Lipschitz constant is equal to $1$. If $E$ is not convex, there is at least one point $x\in \mathbb R^n$ for which $\min_{y\in E}\|x-y\|$ is attained at more than one point. We could try to discuss the continuity of ...


2

1: On a quick look, I don't immediately see how to show that $\psi$ is contractive in general. But note that when $A$ is unital so is $\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case. 2: When you prove that $\psi$ is positive, you don't need to use that $A$ is $A$, just that it is a ...


1

Try to adapt the following example: the functions $f_t(x) = \chi_{(-\infty,t]}(x)$ all belong to $L^\infty(\mathbb R)$, but satisfy $\|f_t - f_s\|_\infty = 1$ if $s \not= t$.


1

Let $(X_n)_{n=1}^\infty$ be a sequence of disjoint sets of strictly positive finite measure. Set $g_n = \mathbf{1}_{X_n}$. For each non-empty set $A\subset \mathbb{N}$ let $$f_A(x) = \sum_{n\in A}g_n(x).$$ Then $\|f_A - f_B\| = 1$ for distinct subsets $A,B\subseteq \mathbb{N}$. The power-set of $\mathbb{N}$ has cardinality continuum so $L_\infty(\mu)$ is ...


1

The fact that the range of $P$ is invariant under $A$ can be written as $PaP=aP$ for all $a\in A$. Now fix $a\in A$; since $a^*\in A$, $Pa^*P=a^*P$. Take adjoints and you get $PaP=Pa$. So we have shown that $Pa=aP$ for all $a\in A$.


1

I'm assuming a normed space $X$ based on your comments. If $X$ is an infinite-dimensional normed space, then there exists a sequence $\{ x_{n}\}_{n=1}^{\infty}$ of unit vectors such that $\|x_{m}-x_{n}\| \ge 1/2$ for all $n \ne m$. That's a consequence of the Riesz lemma. I think that answers your question.


1

Sure: the image of the unit ball in $\ell^2$ under the map that sends the standard basis element $e_n$ to $e_n/n$ (with $\ge 1$) is Hilbert-Schmidt, so a compact operator, etc. Indeed, any sequence going to $0$ replacing $1/n$ produces (pre-) compact image.


1

Build on the identity that you found: $$ f\left(\sum_{j=1}^{n}a_{j}e_{j}\right) = \sum_{j=1}^{n}a_{j}f(e_{j}) $$ The way you build on this is to choose $a_{j}$ so that $a_{j}f(e_{j})=|f(e_{j})|^{q}$. Then you get $$ \begin{align} \sum_{j=1}^{n}|f(e_{j})|^{q} & \le \|f\|\left(\sum_{j=1}^{n}|a_{j}|^p\right)^{1/p} \\ & = ...



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