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2

They are tight, in the sense that we have $\text{trace}(AB) = \lambda_{max}(A)\; \text{trace}(B)$ if $A = I$. Similarly in the second one if $B=I$.


2

If I understand the question correctly, if we have any linear operator $A$ for which the exponential $\exp(A)$ is meaningful, then $A$ commutes with $-A$, and so $$ \exp(A) \exp(-A) = \exp(A + (-A)) = \exp(\mbox{the zero operator}) = I. $$ The inverse of $(I + B)^{-1}$ (again, if the latter is defined) is $(I+B)$.


2

Short answer: Bernstein. First note that since $\widehat{S_0u}$ has compact support $S_0u$ is smooth, in fact real-analytic. So we forget about $S_0u$, at least for now. Say that dyadic block $2^{j-1}\leq|\xi|\leq 2^{j+1}$ is $A_j$. There exists a $C^\infty_c$ function which equals $i\xi$ on $A_0$, hence there exists a Schwarz function $F$ with $$\hat F(\...


1

Let $h'(x) := \max(f'(x), g'(x))$. $h'$ is continuous. Get a primitive function $h$ with $h(0) = \max(f(0), g(0))$. This should do it.


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Sometimes the map $(\cdot, \cdot) : X \times X^* \to \mathbb{R}$ defined via $$(x,y) := y(x)$$ is denoted as duality map. In my opinion, this notion is in particular used if one has a space $Y$ which is isometric to $X^*$, i.e., for the map $$(x,y) := (I\,y)(x),$$ where $I : Y \to X^*$ is an isometric isomorphism. However, typically one might use acute ...


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Yes, you should add $\Delta t \to 0$ to be explicit, but in general, $o(\Delta t)$ is understood to hold for $\Delta t \to 0$, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of $y$ being differentiable at $t$, iff $$ y(t + \Delta t) = y(...


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It is often very difficult to calculate. A point $y\in X$ is in the weak closure if you can not enclose it in a weak neighborhood disjoint from $S$, i.e. if for every $\epsilon>0$ and linear functionals $\ell_1,\ldots,\ell_k\in X'$ the set $N=\{ x\in X : |\ell_i(y-x)|<\epsilon, i=1,\ldots,k \}$ intersects $S$. Note that $k$ must be finite. In finite ...


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If one of the two sets is bounded, the statement is true. One even gets the stronger result you mentioned. For unbounded sets it is false. I do not know if there are really simple counterexamples (e.g. twodimensional?), but I would try something like: With $n=3$, set $A = \{x \colon x_2 = x_3 = 0\}$ and $$B= epi f = \{x \colon x_3 \geq f(x_1,x_2) \}.$$ If ...



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