# Tag Info

32

Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of "nice" functions on real values - e.g. bounded and having continuous derivatives. The action of operators $L^h$ on this space has a semigroup structure: $$L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = ... 30 The essential idea of many transforms is to change the basis in the space of functions with the hope that in the new basis the problem will simplify. Let me give a finite-dimensional example. Suppose we have a 2\times2 matrix A and we want to compute A^{1000}. Direct approach would not be very wise. However, if we first diagonalize A as PA_dP^{-1} ... 14 There is also a proof by "amplification" which is really cool and easy to remember, imho. Also the exact same trick works to prove Hölder's inequality and is generally a very important principle for improving inequalities. It goes like this: We start out with$$\langle a-b,a-b\rangle\ge 0$$for a,b in your inner product space, and a\not=0, b\not=0. ... 13 Recall the Pythagorean theorem: If u_1, \cdots u_n are pairwise orthogonal, then$$ \| u_1 + \cdots u_n \|^2 = \|u_1\|^2 + \cdots + \| u_n \|^2.$$I want to use this to tell us something about two non-zero vectors u and v, but they aren't necessarily orthogonal. So consider the projection of u onto the plane of vectors orthogonal to v:$$w = u - ...

13

$\Bbb R[x]$, the polynomials in one variable. All the continuous functions from $\Bbb R$ to itself. All the differentiable functions from $\Bbb R$ to itself. Generally we can talk about other families of functions which are closed under addition and scalar multiplication. All the infinite sequences over $\Bbb R$. And many many others.

12

If $V$ is a Banach space we call $V'$ the dual space (see continuous dual space on wikipedia), i.e. the space of linear continuous functionals $\xi \colon V\to \mathbb R$. Then it is well known that there exists a natural injection $$J \colon V \to V''$$ defined by $$J(v)(\xi) = \xi(v)$$ for all $\xi \in V'$. We know that $J$ is an isometry, in ...

12

\begin{align} e^{d/dx} x^n & = \left(1+\frac{d}{dx} + \frac{(d/dx)^2}{2}+ \frac{(d/dx)^3}{6}+\frac{(d/dx)^4}{24}+\cdots+\frac{(d/dx)^k}{k!}+\cdots\right) x^n \\[10pt] & = x^n + nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3}+\cdots \\[10pt] & \phantom{{}={}}\cdots+\frac{n(n-1)\cdots(n-k+1)}{k!}x^{n-k}+\cdots+0+0+0+\cdots ... 12 I submit this counterexample which, in my opinion, proves that the statement is false. In the vein of this MathOverflow post by Gerald Edgar, let V denote the real vector space of all polynomials of one variable and let\lVert P\rVert=\max_{x\in[0, 1]} \lvert P(x)\rvert,\qquad \forall P\in V.$$Moreover, let$$W=\{a_0+a_2x^2+a_4x^4+\dots+a_{2k}x^{2k}\ ...

12

Take $$f\left(x\right)=\begin{cases} x^{-1/2}, & x\in\left(0,1\right]\\ 0, & \text{otherwise} \end{cases}$$ This function is obviously in $L^1$; note also that $f\ast f$ is $0$ on $(-\infty,0] \cup [2,+\infty)$, $\pi$ on $(0,1]$, and it decays from $\pi$ to $0$ continuously on $[1,2]$. Therefore, $f \ast f(x)$ is everywhere defined, in the sense ...

11

Hints: $\mathcal O(n)$ is the continuous inverse image of a closed set… There exists a rather obvious bound for every element in $\mathcal O(n)$ and thus this set is bounded… Now Heine-Borel and we're done. Added: Thanks to the comments by Berci I think some confusion may happen here, since the orthogonal group is not the inverse image of $\,\{1,-1\}\,$ ...

11

The reason is that we would like to define a norm by the following formula: $$\|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}.$$ Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for $p\geq 1$. See more details in Chapter $6$ of Folland's book: Real Analysis.

10

To answer your 4 questions briefly: Your $\alpha$ is an operator from a space of real valued functions to itself. You probably require the functions to be bijective in some sense (in order for $f^{-1}$ to even make sense), perhaps with both domain and codomain a subset of the reals. By the recursive relation, each $f_n(x)$ is also bijective in the same ...

10

To answer the question in the title, there is the following general theorem: The natural map $I \colon L^\infty(X) \to (L^1(X))^\ast$ is an isometric injection if and only if $(X,\Sigma,\mu)$ is semifinite. an isometric isomorphism if and only if $(X,\Sigma,\mu)$ is localizable. You already covered point 1 in your question (the isometry ...

10

The sequence $K \supset f(K) \supset f(f(K)) \supset \cdots$ is a nested sequence of compact sets. Then their intersection is non-empty. Denote the intersection by $A$. Then since $$A =\cap f^{(n)}(K)$$ it is easy to show that $f(A)=A$. Note there is a typo in the problem, the inequality can only hold for $x \neq y$. Since $A$ is non-empty and ...

10

For me, the simplest example would be $A=M_2(\mathbb{C})$ equipped with the operator norm induced by any $\ell^p$ norm ($1\leq p\leq +\infty$, $p\neq 2$) on $\mathbb{C}^2$. In short: $A$ is naturally a $*$-algebra and an isometric embedding into $B(H)$ for the latter norms is necessarily a $*$-homomorphism, turning $A$ into a $C^*$-algebra. And a ...

9

This is a variant of the open mapping theorem. If we consider the identity map $i$ on $X$ as a linear mapping from $(X, \|\cdot\|_2)$ to $(X, \|\cdot\|_1)$, then your condition says that $i$ is continuous. Then by the theorem $i$ is open and hence a homeomorphism, so its inverse is also continuous and the norms are equivalent.

9

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2. I tried to solve the exercise, but I'm not quite sure that I understood it. Let ...

9

Suppose that $R^2=T$. Then $\ker R\subseteq\ker T=\Bbb Re_0$, where $e_0=\langle 1,0,0,\ldots\rangle$. Clearly $\ker R$ is non-trivial, so $\ker R=\ker T$. Moreover, $T$ is surjective, so $R$ must also be surjective. In particular, $e_0=Rx$ for some $x\in\ell^1(\Bbb N)\setminus\ker T$, and therefore $R^2x=Re_0=0\ne Tx$. Now suppose that $R^2=S$, and let ...

9

We can finish the argument as follows. (Note: We'll assume that the limit in question exists for $f$ and establish that it's equal to $f(1)$. Technically, we should prove that this limit exists as Peter Tamaroff notes below (thanks!). A minor modification of the following argument simultaneously establishes the existence of the limit and its value but we'll ...

9

What Lax shows is that there is a finitely additive, rotationally invariant set function $m \colon P(S^1) \to [0,1]$ on the circle such that $m(S^1) = 1$: This implies in particular that it is impossible to decompose $S^1$ paradoxically into a disjoint union of finitely many pieces $A_1,\dots,A_n$ in such a way that $S^1$ can be written as disjoint union ...

8

There is such an induced map. However it is in the other direction. If $h:X\to Y$ is a continuous map between topological spaces, then for each $f\in C(Y)$, we have $f\circ h\in C(X)$ since compositions of continuos maps are themselves continuous. This induces a map $$f\in C(Y)\xrightarrow{h'}f\circ h\in C(X).$$ This is the ...

8

We have to consider the problem over the completion $\overline X$ of $X$. But that does not hurt our reasoning: if $A$ is not invertible, neither will its extension to $\overline X$ be. Assuming that $\|A_n^{-1}\|\leq M$ for all $n$, we have $$\|A_n^{-1}-A_m^{-1}\|=\|A_n^{-1}(A_m-A_n)A_m^{-1}\|\leq\,M^2\,\|A_m-A_n\|.$$ As $\{A_n\}$ is Cauchy, we deduce ...

8

In short the list of reasons is: 1) The modern trend in math today is to develop non-commutative analogues of well known theories. Vaguely speaking operator spaces are normed spaces over "non-commutative" scalars, in fact over matricies. 2) There was several long standing problems, that was solved via methods of operator space theory. As soon as a problem ...

8

Some important examples are furnished by integral kernels. Basically if $k:\mathbb{R}^2 \to \mathbb{R}$ is some function then we can define an operator $K$ on a function space by setting $$(Kf)(x) = \int_\mathbb{R} \! k(x,y) f(y) \, dy.$$ Of course one has to be careful here that one restricts the functions $f$ and $k$ so that everything makes sense. One way ...

8

This is the operator-valued version of the fact that the map $$x\mapsto \frac{1-ix}{1+ix}$$ transforms the real line onto the unit circle. In operator terms, the real line becomes the set of self-adjoint operators. And the unit circle is the group of unitary operators. So, let $T$ be self-adjoint. Then $(I\pm iT)^*=(I\mp iT)$, which implies ...

8

Maybe a good point to start is this useful corollary of Baire Cathegory Theorem the cardinality of an Hamel base of a Banach Space can be finite or uncountable. It can't be countable The proof is a delightful application of Baire theorem. Now to give an explicit example, we can consider the space $\ell^2$ which has the standard base $M:=$ $\lbrace ... 8 Consider the collection$I$of all isolated points of$X$. (By the Baire Category Theorem$I$is nonempty, but that is somewhat immaterial for the moment.) Note that$I$is then a discrete subspace of$X$. If$I$were not dense, then$U = X \setminus \overline{I}$is a nonempty (open) set without isolated points. From here we can construct in the usual ... 8 In finite dimension, the invariant subspaces of$T$are exactly the nullspaces of the operators that commute with$T$. So there is no such example in finite dimension$\geq 2$. If$T$is normal, it admits a handful of non-injective commuting projections (=reducing subspaces). So this is a question about non-normal operators in infinite dimension. As ... 8 Here is a more elementary method than you proposed: First, note that if$f$is continuous on$[0,1]$, then it is necessarily bounded on$[0,1]$; say$\lvert f(x)\rvert\leq M$for all$x\in[0,1]$. If we define$\delta_n:=\frac{1}{\sqrt{n}}$, then$\$ \left\lvert n\int_0^{1-\delta_n}f(x)x^n\,dx\right\rvert\leq ...

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