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It seems to me that, if $s(x)$ is strictly increasing then for $x$ between 0 and $s^{-1}(k)$, we have $0<s(x)<k$. Therefore: $$\int_0^{s^{-1}(k)}s(x)dx < \int_0^{s^{-1}(k)}kdx = k*s^{-1}(k) <= k*10$$ let $s_n(x)$, for $n\in\{1,2,..\}$ be a sequence of functions with: $$s_n(x) = k+(x-10)/n$$ Then $\int_0^{s^{-1}(k)}s(x)dx$ goes to 10*k when n ...


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If $x$ is analytic, \begin{align} T(t)x & = x+\int_{0}^{t}T(s)Ax ds \\ & = x + (s-t) T(s)Ax|_{s=0}^{t}-\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax -\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax +\left.\frac{t^{2}}{2!}A^{2}x\right|_{s=0}^{t}+\int_{0}^{t}\frac{(s-t)^{2}}{2!}T(s)A^{3}xds ...


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For $a\in A$, $$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$ Which shows that $\phi$ is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if $\{u_i\}$ is approximate unit of C*-algebra $A$, then $\|\phi\|=\lim \phi(u_i)$. Using this we have $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ...


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I think it is not true. The following should work. Take $g_n:=a_n = b_n \in C_c^\infty$ with $\text{supp} g_n \in [n,n+1]$ and $0\leq g_n \leq 1$ and $g_n = 1$ on $[n+\frac 14, n+\frac 34]$.


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No. Let $b$ be a smooth function with support of $b$ included in the unit ball. Assume that $\|b\|_{H^1}=1$. Then define $b_n$ to be translates of $b$: $$ b_n(x) = b(x + 2n e_1). $$ Then $b_n\rightharpoonup 0$ in $H^1$, but $b_n$ does not converge stronlgy to zero in $H^1$ or $L^2$. Set $a_n:=b_n$. Then $\|a_n\|_{H^1}\le 1$, $a_nb_n\in L^1$, but ...


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If $y_0$ is a local minimum, then there is $\delta>0$ so that $J(y) \ge J(y_0)$ whenever $||y-y_0||<\delta$. Now fix $h\neq 0$ and consider the function $\mathcal{J}: (- \epsilon, \epsilon) \to \mathbb{R}$, $\mathcal{J}(t) = J(y_0 + th)$. So if $\epsilon <\frac{\delta}{||h||}$, we have (for all $|t|<\epsilon$) $$||y_0 + th - y_0|| = |t|\cdot ...


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Suppose that $x=\langle x^{(j)}:j\in J\rangle\in (E^I)^J$, where each $x^{(j)}=\langle x_i^{(j)}:i\in I\rangle\in E^I$. Similarly, let $y=\langle y^{(j)}:j\in J\rangle\in (E^I)^J$, where each $y^{(j)}=\langle y_i^{(j)}:i\in I\rangle\in E^I$. Then we want to define $\mathfrak{W}$ so that $$\begin{align*} ...


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Let $a\in (-\delta, \delta)$. Then it is easy to see that $z\in B_1(0) \Rightarrow \frac{a}{\delta}z\in B_1(0)$ and $f(\frac{a}{\delta}z) =a.$ Therefore, $(-\delta,\delta)\subseteq f(B_1(0))$



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