# Tag Info

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.

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 ...

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.

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 $$x_n(k)= \begin{cases} ... 3 You are right: it is not totally bounded. Riesz's lemma directly leads to an infinite uniformly separated subset of unit ball, as the Wikipedia article shows. 3 I will give you a hint: Let us start with the definition, i.e. take sequences (x_n)_n in M and (y_n)_n in N with x_n + y_n \to z for some z \in X. We want to show z \in M+N. Now, let us write \Gamma : X \to X/M \oplus X/N, x \mapsto (x+M) \oplus (x+N). We have$$ \Gamma(x_n) = (0, x_n +N) = (0, x_n+y_n+N) \to (0, z + N). $$Why/how does ... 2 Reflexive spaces interest mathematicians because they have a lot of nice properties: Unit ball is weakly compact, so you can exploit compactness to prove exitence of fixed points, convergent subsequences and etc. Reflexive spaces are characterized by the property that weak and weak* topology coincide. You can forget about weak topology and work with much ... 2 Let e_m^* be the m-th unit vector in l_1 (the dual of c_0). Let \|A\| be the operator norm of A. I have omitted a few details, but the general idea is much the same as that involved in showing that the induced \infty matrix norm is the maximum row sum. Let x_N = \sum_{n=0}^N {\overline{e_m^*(A e_n)} \over | \overline{e_m^*(A e_n)}|}e_n, ... 2 I assume you mean C[0,1] with sup-norm. Hint: Try to show that for each n and arbitrary \varepsilon>0 there exists g\notin E_n such that \|g\|_\infty<\varepsilon. Try to show that f\in E_n and g\notin E_{2n} implies f+g\notin E_n. Using these two facts you should be able that if f\in E_n then in any ball B(f,r) there is a ... 2 You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional x^{\star} to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed ... 2 If t\gt0, then we want a\ge2|A| and$$ \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 $$In each case, \gamma_r is counterclockwise. When t\gt0, the finite part of \gamma_r is [a-i\infty,a+i\infty]. When t\lt0, the finite ... 2 Let f(s) = s^p for some p>1, then for every s\geq 0m we have$$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 ...

2

See the MathOverflow question http://mathoverflow.net/questions/5303/basis-of-linfinity to learn why it is consistent with ZF (without the axiom of choice) that there is no such operator. Discontinuous everywhere defined operators can be defined using Hamel bases. In particular, as Daniel Fischer points out, one can use a Hamel basis for $X$ to define ...

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.

2

This is basically continuation of the solution suggested in Daniel Fischer's comment. I will use notation $(w^{(n)})$ for the sequence of elements of $\ell_1$. They are sequences, I will use $w^{(n)}_k$ for the $k$-th term of the $n$-th sequence. We want to show $$\limsup_{n\to\infty} \|w^{(n)}-w\|_1 = \limsup_{n\to\infty} \|w^{(n)}\|_1 + \|w\|_1.$$ ...

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

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

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.

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 ...

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 $M$.

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

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

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 ...

1

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 $C^1([0,1])$.

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

If I'm reading Wojtaszczyk's book "Banach spaces for analysts" correctly, any absolutely summing operator maps weakly convergent sequences to norm convergent sequences. The Paley projection does not: if you denote by $(f_k)$ the trigonometric system and by $(e_n)$ the canonical basis of $\ell_2$, then $Pf_{2^n}=e_n$ for all $n\geq 0$, but $f_{2^n}\to 0$ ...

1

Let $P$ be an $L$-projection. From $$\lVert x\rVert = \lVert Px\rVert + \lVert x-Px\rVert$$ for all $x\in X$ it follows that $\lVert P\rVert \leqslant 1$ and $\lVert I-P\rVert \leqslant 1$, and hence $$\lVert P^\ast f\rVert = \lVert f\circ P\rVert \leqslant \lVert f\rVert\cdot \lVert P\rVert \leqslant \lVert f\rVert,$$ and analogously $\lVert (I-P)^\ast ... 1 Putting measures on infinite-dimensional spaces is hard and an interesting problem of current research. There is, for the same reasons you point out, no translation-invariant measure -- no analogue of Lebesgue measure -- on an infinite dimensional space. Probability is a good source of examples for such measures: Wiener measure is the prototypical example of ... 1 Let$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 ... 1 Your 'counterexample': With$x_n := (-1)^ne_n$we have$Ax_n = \frac 1n\cdot(\frac 1m)_m$. For$n \in \mathbb N$let$y_N = \sum_{n=1}^N x_n$, then$y_N \in c_0$with$\|y_n\| = 1$. Extending your$A$linearly to the span of$\{e_n\mid n \in \def\N{\mathbb N}\N\}\$, we have $$Ay_n = \sum_{n=1}^N Ax_n = \sum_{n=1}^N \frac 1n \cdot \left(\frac 1m\right)_m$$ ...

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