# Tag Info

## Hot answers tagged functional-analysis

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 $$\hat F(\... 1 Let h'(x) := \max(f'(x), g'(x)). h' is continuous. Get a primitive function h with h(0) = \max(f(0), g(0)). This should do it. 1 Sometimes the map (\cdot, \cdot) : X \times X^* \to \mathbb{R} defined via$$(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),$$where I : Y \to X^* is an isometric isomorphism. However, typically one might use acute ... 1 Yes, you should add \Delta t \to 0 to be explicit, but in general, o(\Delta t) is understood to hold for \Delta t \to 0, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of y being differentiable at t, iff$$ y(t + \Delta t) = y(...

1

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 ...

1

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible