# Tag Info

6

Make a change of variables $u = e^x$ $$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$ where $g(u) = \frac{f(\log(u))}{u}$ is a continuous function. You can now apply Weierstrass approximation theorem. Since the above holds for all $n$ we have that $$\int_{1}^e g(u) P(u)du = 0$$ for any polynomial $P(u)$. Now pick the polynomial to approximate ...

5

It seems that Nigel Kalton gave an answer to your question. He proved in [1], theorem 3.5, that for every compact metric space $K$, $C(K)$ is a $2$-absolute Lipschitz retract. [1]: Kalton, N. J. Extending Lipschitz maps into C(K)-spaces. Israel J. Math. 162 (2007), 275–315. Link to article Added by OP: For completeness, I'll outline the steps of ...

4

No, take the (incomplete) vector space $\Bbb Q$ which is a one-dimensional vector space over the rational numbers. Since it is not complete it is neither Banach nor Hilbert.

3

To avoid confusing yourself with sequences of sequences, I recommend thinking of the elements of $C_0$ as functions $f:\mathbb{N} \to\mathbb{R}$ such that $\lim_{n\to\infty}f(n)=0$. Then the claim becomes: if $(f_k)_{k=1}^\infty$ is a Cauchy sequence of functions with respect to the uniform norm, there is $f\in C_0$ such that $f_k\to f$ uniformly. The ...

2

Summary of comments: yes, such $D$ can be obtained as $$D=\bigcup_{n\in\mathbb Z}T^n E$$ where $E$ is any countable dense subset of $X$.

2

If $T: X \to Y$ is bounded and surjective, the Open Mapping Theorem says there is an isomorphism $S:\; X/\ker(T) \mapsto Y$ such that $T = S \circ \pi$, where $\pi: X/\ker(T) \to Y$ is the quotient map. The trouble is, a quotient of $X$ might not be isomorphic to a closed subspace of $X$, so there might not be a bounded right inverse. For example, every ...

2

The first inequality is the triangle (Minkowski) inequality for the $L^p$ norm, the second inequality is Minkowski inequality for the counting measure, http://en.wikipedia.org/wiki/Minkowski_inequality.

2

Let $g(x)=\frac12\sin(x)$. Then $f(x)=g(f(1-x))$ for all $x\in[0,1]$, so in particular each point $f(x)$ is a period-2 fixpoint of $g$, and $f(1/2)$ is a period-1 fixpoint of $g$. But note the following: $g(x)\le\frac x2$ for all $x>0$ $g(x)>0$ for all $x\le1$ $g(x)$ is odd. So for any $x>0$, if $g(x)>0$ then $g(g(x))\le x/4$, and if ...

2

It is not too hard to show that if $V$ is a complete $\Bbb{Q}$ vector space, one can extend the scalar multiplication uniquely continuously to $\Bbb{R}\times V\to V$, so that $V$ is also a $\Bbb{R}$ vector space. Hence, we assume this to begin with. Also, if the vector space is complete, it is natural to assume that the underlying field is complete too. ...

2

This question have been already answered on MO. For a more detailed discussion of the spectrum of arbitrary $L_\infty$-space see page 28 in Second duals of measure algebras. H. G. Dales, A. T.-M. Lau

2

What about the subspace $$S=\{f\in C(0,1); f(0)=f(1)\}$$ and the functional $T\colon f \mapsto f(0)$ defined on $S$. The you have two extensions $T_0\colon f\mapsto f(0)$ and $T_1\colon f\mapsto f(1)$. For these functionals we have $\|T\|=\|T_1\|=\|T_2\|=1$. In fact, for every $a\in[0,1]$ we have an extension $T_a\colon f\mapsto af(0)+(1-a)f(1)$ with ...

1

The following is true: If $X$ is a Banach space, then each Hahn-Banach (norm preserving) extension is unique iff the unit ball of $X^{\ast}$ is strictly convex. In particular, this would be true for a Hilbert space. Furthermore, there is an interesting theorem of Phelps (See this) which relates this property for a fixed subspace to a best approximation ...

1

We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$ Note that the Hölder norm is the "sum" of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder ...

1

Since your comment to my first answer is tricky to read, let me restate it. The follow-up question is: how does $$\frac{|v(x)-v(y)|}{|x-y|^\gamma} \le M_\alpha < \infty \qquad independent\ of\ x\neq y \qquad\qquad\qquad(*)$$ for every multi-index $\alpha$ of order $k$ imply $$u \in C^{k,\gamma}(\bar{U})?$$ The answer is as follows: in order to show $u ... 1 Since$T$is continuous, we have$|Tu|_Z\leq c|u|_X.$Hence $$|u|_X\leq|u|_Y\leq |u|_X(1+c),$$ i.e. the norms$|\cdot|_X,|\cdot|_Y$are equivalent. Thus the completion of$X$w.r.t.$|\cdot|_Y$is canonically isomorphic to$X.$1 The basic theorem in this regard is that the conclusion holds so long as the sequence of derivatives$df_n$converges locally uniformly. See 8.6.3 of Foundations of Modern Analysis by Dieudonné. 1 As mentioned in the comments, your first claim is a justly famous result of R. C. James. For the second question, you can use this result to show every continuous linear functional on the space attains its norm on the unit sphere of the space; then appeal to James' theorem. (note "attains its maximum on the unit ball" is the same as saying "attains its norm ... 1 The answer seems to be Yes if the space$Y$is reflexive. Extracting a subsequence if necessary, we may assume that the sequence$(\Vert Tx_n\Vert)$has a limit in$[0,\infty]$. If$\Vert Tx_n\Vert\to \infty$, the inequality is trivially satisfied. So assume that$\Vert Tx_n\Vert\to l<\infty$and let us show that$\Vert x\Vert\leq l$. Since the norm of ... 1 Yes. Namely if$Y = X \oplus U$and$Z = Y \oplus V$, then$Z = X \oplus (U \oplus V)$. 1 Let$e_n$denote the element of$\ell_1$whose$m$'th coordinate is$0$if$m\ne n$and$1$if$m=n$. To see more directly why the bounded linear map, referenced in my comment to the main post,$T:\ell_1\rightarrow\ell_1$defined by $$Te_1=e_1;\ Te_n=e_1+e_n, n>1$$ is not an adjoint operator, consider the sequence$(e_n)$in$\ell_1$. Note for any ... 1 Almost there... You have$|Sf(x)-Sg(x)| \le \int_0^x t^2 dt \|f-h\|_\infty ={ x^3 \over 3} \|f-h\|_\infty$, and so$\|Sf-Sg\|_\infty \le { 1\over 3} \|f-h\|_\infty$. 1 K. Davidson, "C*-Algebras by Example", Fields Institute Monographs, 1996. Update. Added a more complete list. I am also studying C * algebras. A more complete list can be this. O. Bratteli and D. W. Robinson, Operator algebras and quantum statistical mechanics. 1", Texts and Monographs in Physics, Springer-Verlag, 1987. K. R. Davidson,$C^*$-algebras by ... 1 Take the self-adjoint diagonal operator$Te_k=\frac1k e_k$on a Hilbert space with orthonormal base$\{e_k, k\in\mathbb N\}.$The kernel of$T$is$\{0\}$but$T$is not surjective, since e.g.$\sum_k \frac1k e_k$is not in the range. 1 It is obvious that$S= {~}^\perp(S^\perp)$implies that$S$is a closed subspace. For the oposite implication, assume that$S$is closed and$x\not\in S$. Let$T$be the linear subspace spanned by$S$and$x$. It is closed and each$y\in T$is of the form$y=s+\alpha x$for unique$s\in S$and number$\alpha$. Define$\xi_0$on$T$by$\xi_0(s+\alpha ...

1

The problem does not originate from the weak-* topology, it also is present within the weak topology on Hilbert spaces. On $\ell^2$, consider the set \begin{equation*} B := \{ \sqrt{n} \, e_n : n \in \mathbb N\}, \end{equation*} where $e_n$ are the canonical unit sequences. Now, it is possible to show that $0$ belongs to the weak closure of $B$, but no ...

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