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As noted by @Razieh Noori, if $f\in X^*$, then $f\circ T\in X^*$. Since $u_k\rightharpoonup u \Rightarrow \exists M:\|u_k\|\leq M\quad \forall k\quad$. Also $T_k\rightarrow T \Rightarrow$ for $\epsilon>0\quad\|T_k-T\|_{op}<\frac{\epsilon}{2 M \|f\|_*}$ for $k>K_1$. Then $|f\circ T_ku_k-f\circ T u|\leq |f\circ T_k u_k-f\circ T ...


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In my upload http://arxiv.org/abs/1509.01078 I tried to modify Sokal's idea in order to avoid the axiom of dependent choice and instead rely on the axiom of countable choice, which is strictly weaker than dependent choice. EDIT (proof sketch(EDIT2) as wished for): We first note that for a linear operator $T$, $$ \max\{\|T(x - y)\|, \|T(x + y)\|\} \ge 1/2 ...


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Note that the variable of integration is $t$, so, we can use the following steps. $$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt \Rightarrow\|x\|\int_{0}^{\infty}e^{-(\alpha-\omega)t}dt$$. Here we do a little substitution and let $v = (\alpha - \omega)t$ and so we have $\frac{dv}{dt}=(\alpha - \omega) \Rightarrow \frac{dv}{(\alpha-\omega)} = dt$. Upon ...


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Supposing that $\|x\|$ doesn't depend on $t$ and $R \in \mathbb{R}$: $$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt=\lim_{R \to \infty}\|x\|\int_{0}^{R}e^{-(\alpha-\omega)t}dt=\lim_{R \to \infty}\|x\|\frac{e^{-(\alpha-\omega)t}}{\omega - \alpha}\Bigg|_{0}^{R}=\lim_{R \to \infty}\|x\|\frac{e^{-(\alpha-\omega)R}-1}{\omega - ...


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I doubt that there is a systematic way besides the "educated guess". Looking at $ \sqrt{1 + 4x^2}$ and $1 + 4x$ we see that the former contains the functions $\sqrt{\cdot}$ and $(\cdot)^2$, and the latter does not. As squaring and the square root are inverses to each other, two of the functions in $g\circ k \circ f$ should be $\sqrt{\cdot}$ and $(\cdot)^2$. ...


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Question 1. The quickest way to embed $S_1$ into a $C(K)$-space with $K$ countable (hence scattered) is to go via this construction. Let $L=\{-1,0,1\}^\mathbb{N}$. This is a compact Hausdorff space homeomorphic to the Cantor set. Define $$K = L / \{(\xi_n)_{n=1}^\infty\colon \text{support of }(\xi_n)_{n=1}^\infty \text{ is admissible}\}.$$ Note that $K$ ...



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