New answers tagged

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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.


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If isomorphic means isomorphic as rings then no, they're not isomorphic. Keen fact: Suppose $K_1$ and $K_2$ are compact Hausdorff spaces. Then $C(K_1)$ is isomorphic to $C(K_2)$ if and only if $K_1$ and $K_2$ are homeomorphic. (Sketch: $C(K)$ is a Banach algebra with maximal ideal space $K$.) This is a keen thing, because it shows that the ...


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Alternative answer:your operator induces a continuous bijection between quotient space and image subspace. By the open mapping theorem if it were compact then the unit ball in both spaces were compact, thus finite dimensional by Riesz


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First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets. Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$...


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Regarding convergence and completeness: For $n\in N$ let $f_n(x)=0$ for $x\leq 1/2-1/(n+2)$ and $f_n(x)=1$ for $x\geq 1/2.$ Let $f_n(x)$ be linear for $x\in [1/2-1/(n+2),1/2].$ Then $(f_n)_{n\in N}$ is a Cauchy sequence with respect to the norm $\|f-g\|=[\int_0^1|f(x)-g(x)|^2\;dx]^{1/2}.$ Let $h(x)=0$ for $x\leq 1/2$ and $h(x)=1$ for $x>1/2.$ The ...


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Here is an alternative proof, using as a starting point that $\ell^{\infty}$ is a Banach space. As $c$ is a subspace of $\ell^{\infty}$, it suffices to show that $c$ is closed in $\ell^{\infty}$. To see this let $\mathbf{x}=(x_1,x_2,\dotsc)\in\ell^{\infty}$, and suppose that $\{\mathbf{x}^n\}_{n=1}^{\infty}$ is a sequence in $c$ converging to $\ell^{\infty}$...


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The function $t\cdot \chi_{[-1,1]}(t)$ is orthogonal to each $t^{2k}$ in $L^2[-1,3],$ hence is orthogonal to the the linear span of $\{t^{2k} : k=0,1,\dots \}$ in $L^2[-1,3],$ hence is orthogonal to the closure of this linear span in $L^2[-1,3].$ Therfore this closure cannot be all of $L^2[-1,3].$


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Let $A$ denote the linear span of $\{t^{2k}\}_{k \in \mathbb{N}}$. Then, if instead of $[-1,3]$ the domain was $[1,3]$ instead, we could use the Stone-Weierstrass theorem to conclude that $A$ is dense ins $C([1,3])$ and thus in $L^2([1,3])$ since $t^2$ separates the points of $[1,3]$ and $1 \in A$. However, for $[-1,3]$, we have $t^{2k}(-1) = t^{2k}(1)$ ...


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Unless I am missing something, $Y$ is not even closed (so it cannot be complemented either). Indeed, $Y$ is the image of the compact operator that multiplies elements of $E$ by $(1/n)$. Compact operators cannot have closed range unless they are finite-rank operators.


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I will make the following assumption: (2'): $\exists (x_n)\subset E$ and $c>0$ such that $x_n$ converges weakly to $0$ and $\|x_n\|>c$ for all $n$. This follows from (2), since non-convergence in norm implies the existence of a subsequence bounded away from $0$. (As a side note, this also implies (2).) Let $(x_n)$, $c$ satisfy (2'), and let $x\in ...


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One such proof is via the so-called Kuratowski embedding.


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Your sequence of equalities should have been $$ \|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt= \int _0^\lambda f(t)^2=\int _0^1f(t)^2\chi_{[0,\lambda]}^2\leq \|f\|\|\chi_{[0,\lambda]}\|=\lambda \|f\|. $$ Now, I don't know what inequality you are using, but I cannot make sense of it. And you get $\|f\|$ instead of $\|f\|^2$, which is a bad sign. The ...


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If the basis is unconditional (with, as David Mitra points out, the unconditional basis constant also bounded by something independent of $n$) then yes, at least with some constant independent of $n$. Using $c$ to denote anything not depending on $n$: Of course $|a_k|\le c||\sum a_je_j||$ just because it's a normalized basis. On the other hand, if $-1\le ...


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That's right. Reflexive spaces are both $M$-embedded and $L$-embedded.


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In this context, $\cong$ is to be interpreted as “isometrically isomorphic.” This means that there exists a map $f:X\to J(X)$ such that $f$ is linear; $f$ is bijective; and $f$ preserves norms: $\|f(x)\|_{J(x)}=\|x\|_X$ for each $x\in X$. The natural candidate for such a function $f$ is the canonical mapping $J:X\to J(X)$ itself. By construction, it is ...


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For each $x\in I=[-1,1]$, define $T_{x}(y):=\frac{1}{2}\ln(1+y^{2})+\sin(\ln(x+2))$. Observe that $T_{x}'(y)=\frac{y^{2}}{1+y^{2}}\leq \frac{1}{2}$ on $I$, so we can show $|T_{x}(y')-T_{x}(y)|\leq \frac{1}{2}|y'-y|$ and $T_{x}$ has a unique fixed point for each $x$ by Banach fixed point theorem. Define $f(x)$ as fixed point of $T_{x}$ for each $x\in I$. To ...


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You should indeed use the Banach fixed-point theorem. The "obvious" map to consider is $T : C[-1,1] \to C[-1,1]$ $$(Tf)(x) = \frac{1}{2}\ln(1+f(x)^2) + \sin(\ln(x+2))$$ Check that $T$ is a contraction mapping. The inequality $$\vert \ln(1+x^2)-\ln(1+y^2) \vert \leq \vert x - y \vert$$ might prove useful. (The latter inequality follows e.g. from the mean-...


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Ben, the answer is no. Note that if $X$ is reflexive, then $B(X)$ is isometric to $B(X^*)$ via $T\mapsto T^*$. Note that this map is an anti-isomorphism of Banach algebras. As for less trivial examples, $B(\ell_p)$ is Banach-space isomorphic to $B(L_p)$ as well to $B(X)$ for any other separable, infinite-dimensional $\mathscr{L}_p$-space and $\ell_p$ is ...


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No. Not that that makes any sense to me, but no. In fact it turns out that $\kappa_{X^{**}}=\kappa_X^{**}$ if and only if $X$ is reflexive. In fact the truth behind my erroneous feeling that the two must be equal in general is this: $$\kappa_{X^{**}}\kappa_X=\kappa_X^{**}\kappa_X.$$ Proofs: Writing $x\in X$, $x^*\in X^*$, etc.: If you unpack two ...


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Two answers. First, you can easily derive the general case from the case where $Z$ is closed. Second, and I think more important, $Z$ being closed is really irrelevant in the first place! (i) You can derive the general case from the case where $Z$ is closed: If $d=0$ you can just set $\Lambda=0$ and you're done. Suppose $d>0$. Then $y\notin\overline Z$...


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Let $P(x)=\inf_{z\in Z}\|x-z\|$. It is easy to show that $P(ax)=aP(x),a>0$ and $P(x+y)\leq P(x)+P(y)$. Meanwhile, $P(z)=0, \forall z\in Z$ and $P(y)=d$. We can show that $P(x)\leq \|x\|$ by choosing $z=0$ in the definition. Next, we define another function $L(x)$. $L(z)=0, \forall z\in Z$ and $L(y)=d$. Let $L$ be defined on the subspace $Y=\{by+z|z\in Z,...


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The quantity $\|f\| = |f(a)| + \|f\|_{TV}$ is a natural norm on $BV[a,b].$ If $f\in AC,$ then this norm equals $|f(a)| + \int_a^b|f'|.$ Suppose $f_n$ is a sequence in $AC$ that is Cauchy in this norm. Then $f_n(a)$ is a Cauchy sequence in $\mathbb R,$ and $f_n'$ is Cauchy in $L^1[a,b].$ Thus $f_n(a) \to c$ in $ \mathbb R$ and $f_n'\to g$ in $L^1[a,b].$ We ...


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By definition, $f: [a,b] \to B$ is Riemann integrable on $[a,b]$ with integral $L \in B$ if for each $\epsilon > 0$ there is $\delta > 0$ such that for every tagged partition $x_0,\ldots,x_n$, $t_0, \ldots, t_{n-1}$ of $[a,b]$ with mesh $<\delta$, $$\left\| \sum_{i=0}^{n-1} (x_{i+1}-x_i) f(t_i) - L \right\| < \epsilon \tag{1}$$ I claim that $...


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We give the proof the OP wants, also an example showing that the result fails for functions on the line. The standard proof via the vector-valued Cauchy Integral Formula seems like the "right" proof, because that vector-valued CIF is going to be fundamentally important soon anyway, when we get to Banach algebras and operator theory and so on. But, if we ...


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Let $(h_n)$ a sequence in $V$ converging in $\mathbb{C}$. Let $u_n = \frac{f(z_0+h_n)-f(z_0)}{h_n}$ The bounded $\sup$ condition implies $||u_p - u_q|| \leq M |h_p-h_q|$ Hence $(u_n)$ is a Cauchy sequence in $X$ so it converges in $X$ by completeness. Therefore $\lim \frac{f(z_0+h_n)-f(z_0)}{h_n}$ exists in X.


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Yes, take a subsequence $(x_{n_k})_k$ such that $\|x_{n_k}\|\leq 2^{-k}$. Then $$\|\sum_{k=1}^{m_2}x_{n_k}-\sum_{k=1}^{m_1}x_{n_k}\|=\|\sum_{k=m_1+1}^{m_2}x_{n_k}\|\leq\sum_{k=m_1+1}^{m_2}\|x_{n_k}\|\leq\sum_{k=m_1+1}^{m_2}2^{-k}=2^{-m_1}-2^{-m_2}\to0$$ as $m_1,m_2\to0$. Thus the partial sums of $x_{n_k}$ are Cauchy.


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Let $d$ be a metric for $X.$ Let $ E$ be a dense subset of $X$. For any $x\in X$ and $r>0,$ there exists $q\in (0,r/3)\cap \mathbb Q $ and $e\in E$ such that $d(x,e)<q.$ $$ \text { This gives }\quad x\in B_d(e,2q)\subset B_d(x,r).$$ (Because $y\in B_d(e,2q)\implies d(x,y)\leq d(x,e)+d(e,y)< q+2q<r$.) Therefore $B=\{B_d(e,2q):q\in \...


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Let $D\subset X$ be countable and dense and $Z\subset X$. Assume there is no $z_0\in Z$ with a sequence $(z_n)_{n=1}^\infty$ of points in $Z$ converging to $z_0$. Then for every $z\in Z$ there exists $r=r(z)>0$ such that $B_r(z)\cap Z=\{z\}$. By decreasing $r(z)$ if necessary, we may assume that $r(z)=\frac1{n(z)}$ with $n(z)\in\Bbb N$. For $m\in \Bbb N$, ...


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Yes, and we do not even need to assume completeness. Suppose not, towards a contradiction. Then we can find open balls $U_z=B_{\epsilon_z}(z)$ around each $z\in Z$ such that $U_z\cap Z=\{z\}$. Let $q>0$ be such that uncountably many $z\in Z$ have $\epsilon_z>q$ (Exercise: why does such a $q$ exist?). Let $Z'=\{z\in Z: \epsilon_z>q\}$. Now consider ...


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My first thought was surely not. But actually the answer is yes. Say $f$ is a non-zero bounded linear functional on $X$. Let $N$ be the null space of $f$. Now $X\setminus N$ certainly spans $X$, so there exists a Hamel basis $B\subset X\setminus N$. So $f(b)\ne0$ for every $b\in B$; now modify $B$, replacing every element by an appropriate scalar multiple ...


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The following proof is from "On the convergence of sample probability distributions" of Varadarajan. As you may not have access to jstor, i write it down here. Remark that it is important that we lie in a probability space. So that it is a particuliar case of your question. Let $f:M \to X$ measurable, suppose $\mu(M)=1$, let $\mu_f$ the "law" of $f$, ie: $\...


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More generally, if $\{T, S_1, S_2, \ldots\}$ is a uniformly continuous family of maps from metric space $X$ to metric space $Y$, then $\{x \in X: \lim_{n \to \infty} S_n(x) = T(x)\}$ is closed. This does not require completeness of either metric space.


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Your argument is fine, besides the typo where it should be $(\|S_n\|+\|T\|)$ instead of $\|S_n+T\|$. And I don't think you need for $X$ to be complete.


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Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a) $ for $a $ positive; The fact that a $*$-homomorphism reduces the spectral ...



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