Tag Info

New answers tagged

0

I'm no expert on this, so I might say stupid things, but lets have a go. Let me assume that $A$ is unital (with unit $\mathbb{1}$). The answer to your first question is positive, since $$\sigma(F)=\bigcup_{\omega\in\Omega} \sigma(F(\omega)).$$ Indeed, the inverse $h$ (if it exists) of the function $\lambda I-F$ satisfies $h(\lambda I-F)=(\lambda I-F)h\equiv ...


0

Let $n=1$, fix $k$, and consider the $k$ vectors \begin{eqnarray} v_1 &=& (-1,1,0,0,0,0,\ldots) \\ v_2 &=& (0,-1,1,0,0,0,\ldots) \\ v_3 &=& (0,0,-1,1,0,0,\ldots) \\ v_4 &=& (0,0,0,-1,1,0,\ldots) \\ \vdots & \vdots & \vdots \end{eqnarray} Here, each $v_i\in \mathbb{R}^d$ where $d>k$. Let $v=v_1+\cdots+v_k$. Then ...


4

In fact, any infinite-dimensional Banach space $X$ has a compact subset that does not lie in any finite-dimensional linear subspace. Namely, define a sequence $x_n$ inductively such that $x_{n+1}$ is not in the linear span of $\{x_1, \ldots, x_n\}$ but $\|x_{n+1}\| < 1/n$. Then $\{0\} \cup \{x_1, x_2, \ldots\}$ is your set.


2

Sure: the image of the unit ball in $\ell^2$ under the map that sends the standard basis element $e_n$ to $e_n/n$ (with $\ge 1$) is Hilbert-Schmidt, so a compact operator, etc. Indeed, any sequence going to $0$ replacing $1/n$ produces (pre-) compact image.


1

I'm assuming a normed space $X$ based on your comments. If $X$ is an infinite-dimensional normed space, then there exists a sequence $\{ x_{n}\}_{n=1}^{\infty}$ of unit vectors such that $\|x_{m}-x_{n}\| \ge 1/2$ for all $n \ne m$. That's a consequence of the Riesz lemma. I think that answers your question.


6

We have the following characterization of adjoint operators: Suppose that $X$ and $Y$ are normed spaces. If $T:X\rightarrow Y$ is a bounded linear operator, then $T^*$ is weak*-weak* continuous. Conversely, if $S$ is a weak*-weak* continuous linear operator from $Y^*$ to $X^*$, then there is a bounded linear operator $T:X\rightarrow Y$ such that $T^*=S$. ...


0

Consider a real monomial: $x^\alpha$ Now poles give a restriction of the form $\alpha>\alpha_0$ while decay rates give a restriction of the form $\alpha<\alpha_0$. This can be used to glue a desired example, e.g. if $$f(x):=x^{\alpha_0}\chi_{[-1,1]}+x^{\alpha_\infty}\chi_{(-\infty,-1)\cup(1,\infty)}$$ then the integrability is restricted to ...


1

There exists a discontinuous functional $f:X\to \mathbb{R}$ take $p(x) =|f(x)|$ . Since $f$ is discontinuous then $\mbox{Ker}(f)$ is dense in $X.$ Take any $v\in X\setminus \mbox{Ker}(f)$ and construct absolutely convergent (with respect to norm) series such that $v=\sum_{k=1}^{\infty} v_k$ then you have $$p(v)>\sum_{k=1}^{\infty} p(v_k) =0.$$


1

Consider the space $X = C_b[0,\infty)$ of bounded continuous functions on $[0,\infty)$ endowed with the uniform norm $$\|f\|_{\infty} := \sup_{x \geq 0} |f(x)|.$$ Define $$T_t f(x) := f(x+t),\qquad x \geq 0, t \geq 0.$$ Obviously, $(T_t)_{t \geq 0}$ has the semigroup property (i.e. $T_t T_s = T_{t+s}$) and $\|T_t\| \leq 1$. This means that $(T_t)_{t \geq ...


0

Without loss of generality, assume that we are dealing with a probability space $(\Omega,\Sigma,\mu)$. Let's call a function F-integrable if it satisfies the definition. Every simple function in a Hausdorff locally convex topological vector space is F-integrable. Proof: Let $f$ be simple and $V$ be a neigborhood of $0$. Let $n=|f(\Omega)|$ and let $U$ be a ...


8

Here is an outline for proving your statement. $\ell^p$ is a space of (some) functions $\mathbb N\to\mathbb R$, and so is $L^p$ (with the choices indicated by OP). (I chose to work with the reals, but this choice is irrelevant.) The counting measure gives nonzero measure to all nonempty sets, so each equivalence class in $L^p$ only consists of one ...


2

It can be done fairly easily in any infinite dimensional normed space $X$. For a proof, see Lemma 1.4.22 in Robert E. Megginson's An Introduction to Banach Space Theory.


0

Take for example $V = \ell^2(\Bbb{N})$ and $a_n = 1/n \cdot \delta_n$ with $(\delta_n)_m =1$ for $n=m$ and $0$ otherwise. Verification of the claimed properties is an exercise.


1

To show $X = \bigcup_{k=1}^{\infty} kB_1$ we must check both $\subset$ and $\supset$ relations between these. $\subset$ : for every $x\in X$ there is $k$ such that $x\in kB_1$. E.g., $k$ could be any integer greater than $\|x\|$. $\supset$ : the Banach space $X$ is our Universe here; no elements from outside of it enter the proof. The unit ball ...


2

Simply use the Riesz lemma as $i(X)$ is norm-closed in $X^{**}$.


2

Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$. Set $$ g_{k,\ell}(x)=\frac{1}{k} g(\ell x). $$ It is not hard to see that $$ g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}. $$ Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior. Let ...


3

I assume you mean $C[0,1]$ with sup-norm. Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$. Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$. Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a ...


1

In fact, $\alpha\beta$ is optimal. To prove it, let $f,g:[0,1]\to\mathbb{R}$ with $f(x)=x^\alpha$ and $g(x)=x^\beta$. Note that $f\circ g\in C^{\alpha\beta}$, however, for all $\gamma>\alpha\beta$, $f\circ g$ does not belong to $C^\gamma$.


1

As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator ...


1

Putting measures on infinite-dimensional spaces is hard and an interesting problem of current research. There is, for the same reasons you point out, no translation-invariant measure -- no analogue of Lebesgue measure -- on an infinite dimensional space. Probability is a good source of examples for such measures: Wiener measure is the prototypical example of ...


0

These are the books I recommend: A short course on Banach space theory. N. L. Carothers. A friendly introduction into geometry of Banach spaces An Introduction to Banach Space Theory Graduate Texts in Mathematics. Robert E. Megginson. A more academic, but still very basic exposition. Topics in Banach space theory. F. Albiac, N. Kalton. Though this is still ...


1

You observe correctly that many results about series of real or complex numbers generalize to Banach spaces. Since $\mathbb R$ and $\mathbb C$ are examples of Banach spaces it should not be surprising that some (or even a lot) of what is taught about real or complex analysis generalizes to general Banach spaces. Moreover, in many cases the true nature of a ...


2

If $t\gt0$, then we want $a\ge2|A|$ and $$ \gamma_r=a+ir[-1,1]\cup[a,-r]+ir\cup-r+ir[1,-1]\cup[-r,a]-ir $$ If $t\lt0$, then we want $a\le-2|A|$ and $$ \gamma_r=a+ir[1,-1]\cup[a,r]-ir\cup r+ir[-1,1]\cup[r,a]+ir $$ In each case, $\gamma_r$ is counterclockwise. When $t\gt0$, the finite part of $\gamma_r$ is $[a-i\infty,a+i\infty]$. When $t\lt0$, the finite ...


2

You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional $x^{\star}$ to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed ...


2

Let $f(s) = s^p$ for some $p>1$, then for every $s\geq 0$m we have $$f''(s) = \underbrace{p(p-1)}_{>0}\,\underbrace{s^{p-2}\vphantom{)}}_{\geq 0} \geq 0$$ and thus $f$ is convex on the non negative numbers. So for any $x_1,x_2 \in X$ and $t \in (0,1)$ we get $$\frac{1}{p}\|x_1t + (1-t)x_2 \|^p \overset{(1)}{\leq} \frac{1}{p}(t\| x_1\|+(1-t)\|x_2\|)^p ...


1

If the open unit ball were totally bounded, then so would its closure, which is the closed unit ball. The closed unit ball, in turn, is complete (as a closed subset of a Banach space). Hence, the closed unit ball would be compact, which it can be shown it is not, given that the space is infinite-dimensional.


5

You are right: it is not totally bounded. Riesz's lemma directly leads to an infinite uniformly separated subset of unit ball, as the Wikipedia article shows.


0

The two properties are equivalent as long as you only consider sequences: $RR\implies KK$ because on the unit sphere, by definition, all norms are 1, therefore the norm obviously converges. For the other direction, just scale your sequences, then you get $$ \frac{f_n}{\|f_n\|}\rightarrow \frac{f}{\|f\|} $$ if the norms converge, this converges strongly ...


0

The idea to show that $d(0,C)=1$ holds, is to use a sequence of suitably chosen characteristic functions. Define the sequence $u_n$ by $$ u_n = \frac{2n^2}{2n-1}\chi_{[1-1/n,1]}(x), $$ where $\chi_{[1-1/n,1]}$, is the characteristic function of the interval $[[1-1/n,1]$. Then $u_n \ge0$. $$ T(u_n) = \frac{2n^2}{2n-1}\int_{1-1/n}^1 x dx = ...


0

Hint: Since $T$ is finite rank, you can pick an orthonormal basis $g_1,\ldots,g_n$ of the range of $T$, $$ Tv=\sum_{j=1}^n \langle Tv,g_j\rangle g_j. $$ Show that $\mbox{ker } T\supset \mbox{span }\{T^*g_1,\ldots,T^*g_n\}^\perp$. Pick and orthonormal basis of $\mbox{span }\{g_1,\ldots,g_n,T^*g_1,\ldots,T^*g_n\}$, say $e_1,\ldots,e_m$. Show that $m\leq 2n$. ...


8

We know that, $L_\infty$ is not separable, so neither does its dual $L_\infty^*$. It is remains to recall that $L_1$ is separable.


0

It seems you are thinking of the Fredholm alternative, which ensures that the index of $I+K$ is zero whenever $K$ is compact.


2

Except in the trivial case of the zero space, being Banach prevents pathological examples such as you link to. More precisely, in a Banach space with at least one nonzero vector, we cannot have $B(x,R)\subseteq B(y,r)$ with $R>r$. You can see this by considering the two balls restricted to a line that contains $x$ and $y$, with the induced metric. (In ...


2

Notice that $\lVert f_h\rVert_{C^1}\leqslant 2$ if we consider $h\geqslant 1$, hence the sequence $(f_h/2)_{h\geqslant 1}$ has all its terms in the closed unit ball of $C^1([0,1])$.


0

At face value, this never has any meaning unless $X$ is a field containing real numbers as a subfield. But, of course, as you mentioned, one can assign meaning to things like adding $c$ to a function, by defining $c$ not to be an element of $\mathbb{R}$, but an element of the set of constant functions. This is always possible, for example, if $X$ is a ...


1

It's easy to show that the inverse of an isometry between normed spaces is an isometry. Since an isometry is continuous, you're done.


0

If you want to show that an isometric isomorphism is continuous with continuous inverse, then the way you start is "Given an isometric isomorphism $T$, a point $x$ and an $\epsilon >0$, ..." and then you use the isometric property of $T$ and the triangle inequality to find your $\delta$, both for $T$ and its inverse.


1

A uniformly isolated set is closed: Let $(x_n)$ be a sequence in $A$ and suppose $x_n \to x \in X$. Choose $N \in \mathbb N$ such that $d(x_n, x) < \frac{r}3$ for $n \ge N$, hence $d(x_n, x_m) < \frac{2r}3$ for $n \ge m$. As the points of $A$ are at least $r$ apart, $(x_n)$ must be constant for $n \ge N$, so there is an $a \in A$ such that $x_n = a$ ...


1

As you write $A \cap r\cdot B_{X^*}$ is weak$^*$ly metrizable as a subspace of the metrizable space $r\cdot B_{X^*}$, now let $x_i^* \in A \cap r \cdot B_{X^*}$ be any net converging weakly$^*$ to $x^* \in X^*$. As $rB_{X^*}$ is weakly$^*$ closed (the norm is weakly$^*$ lower semicontinuous), we have $x^* \in rB_{X^*}$. Now, as $rB_{X^*}$ is metrizable in ...


-1

Hint. It is enough to show that one can approximate non-negatve functions by rational linear combinations of the bump functions$F_u$ you constructed, for every function is a difference of non-negative functions. So let $f:X\to\mathbb R$ be continuous and non-negative, and suppose that $f$ is not identically zero. let $\mathcal G$ be the set of all finite ...


3

Your proof is correct. But with almost no extra work you can prove a better result: considered as a subset of $\ell^2$, the space $\mathbb R^\infty$ is dense. (Consequently, not closed, hence not complete in the $l^2$ metric.) The proof goes just as yours, but without specifying $x$. Just let $x$ be any element of $l^2$, then define $$x_n(k)= \begin{cases} ...


1

The subspace $S_{n}$ spanned by $\{ e^{2\pi i kx}\}_{k=-n}^{n}$ is invariant under $A$, and $A$ is bounded on $S_{n}$ because its a matrix. However, $A$ is not bounded on the union of these nested subspaces. That puts a limit on any Zorn's lemma argument. I suppose you could find a maximal subspace on which $A$ is bounded by a given fixed constant $M$.


2

Reflexive spaces interest mathematicians because they have a lot of nice properties: Unit ball is weakly compact, so you can exploit compactness to prove exitence of fixed points, convergent subsequences and etc. Reflexive spaces are characterized by the property that weak and weak* topology coincide. You can forget about weak topology and work with much ...


1

I'll give a couple important properties of reflexive Banach spaces, one more geometric and one more operator-theoretic (though there are of course many more). First of all, given any closed convex subset $A$ of a reflexive Banach space $X$ as well as any $x \in X$, there exists a point $a \in A$ which minimizes the distance $\inf_{a \in A} \| x- a \|$ from ...


3

I will give you a hint: Let us start with the definition, i.e. take sequences $(x_n)_n$ in $M$ and $(y_n)_n$ in $N$ with $x_n + y_n \to z$ for some $z \in X$. We want to show $z \in M+N$. Now, let us write $\Gamma : X \to X/M \oplus X/N, x \mapsto (x+M) \oplus (x+N)$. We have $$ \Gamma(x_n) = (0, x_n +N) = (0, x_n+y_n+N) \to (0, z + N). $$ Why/how does ...


0

By the Holder inequality, $\lVert Af \rVert \leq ||\sum_{m \geq 1} \alpha_{mn} ||_{\infty} ||f||_1$. By (a) and the definition of sup norm, $M=||\sum_{m \geq 1} \alpha_{mn} ||_{\infty}$. Thus, $A$ is a bounded operator with operator norm at most $M$. Can you get the second part now?


3

The simplest case I know is the translation $T(t)$ semigroup on $L^{2}[0,\infty)$ defined by $(T(t)f)(x)=f(x+t)$. There's no way to invert once you've lost part of the function.


0

Nothing is gone wrong. The limit function is: $$ f(x) = x^2 + \sum_{i=1}^{+\infty}\frac{x^{2i+2}}{2^i(i+1)!} = 2\left(e^{x^2/2}-1\right).$$


3

Consider the Heat semigroup on $L^2(\mathbb{R}^n)$: for $t>0$, $$U(t)g=K(t)\ast g$$ where $$K(t)=(4\pi t)^{-n/2}e^{-|x|/4t}$$ is the heat kernel. $U(t)$ is a strongly continuous semigroupo (of contractions). Let's show that it cannot be extended to a group ($U(t)$, $t\in\mathbb{R}$) on $L^2$. If such a group eventually exists, it should satisfy ...


1

Assume that $(A-\lambda I)^{-1}$ is defined on all of $X$ and is bounded. Then $(A-\lambda I)^{-1}$ is closed. The graph of $A-\lambda I$ and the graph of $(A-\lambda I)^{-1}$ are transposes of each other in $X\times X$, which guarantees that $A$ is closed. So you can deduce that $A$ is closed under those circumstances. You can also deduce that the domain ...



Top 50 recent answers are included