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5

Since $A$ and $B$ are closed subspaces of the Banach space $X$, they are themselves Banach spaces, and hence so is $A\times B$, endowed with the norm $\lVert (a,b)\rVert = \lVert a\rVert_X + \lVert b\rVert_X$. Now consider the map $$T \colon A\times B \to X; \quad T(a,b) = a+b.$$ We have $\lVert T(a,b)\rVert_X \leqslant \lVert (a,b)\rVert$, so $T$ is ...


3

Let denote $T$ the two linear functional. Let the sequence $p_n(x)=nx^n$ then $$||p_n||_1=n\int_0^1 t^ndt=\frac n{n+1}<1$$ hence the sequence is bounded but $T(p_n)=n^2p_n$ and we have $$||T(p_n)||_1=\frac{n^2}{n+1}\xrightarrow{n\to\infty}\infty$$ hence $(T(p_n))$ isn't bounded and then $T$ isn't continuous. We have ...


3

For any $f \in C([0,1])$ you have $Tf(x) = f(1-x)$ so that $$\|f\| = \int_0^1 x^2 f(x) \, dx \quad \text{and} \quad \|Tf\| = \int_0^1 x^2 |f(1-x)|\, dx.$$ If $f$ is concentrated near $0$, the first norm will be small and the second norm will be large. One way to look for counterexamples is to temporarily disregard continuity: for instance you could take ...


3

We do not need to consider $\Omega(A)\cup \{0\}$, but it makes things easier, since that is a compact (Hausdorff, if that isn't part of your definition of compact) space. Every open subspace of a compact (Hausdorff) space is locally compact, and since the weak* topology is Hausdorff, singletons are closed, whence $$\Omega(A) = (\Omega(A)\cup \{0\}) ...


3

Let $\Omega$ be the disk of radius $r<1$ centered at the origin. Define $u(x,y) = (-\log(x^2+y^2))^{1/3}$. Clearly, this is not a continuous function. But on every line not passing through the origin, this function is Lipschitz, therefore absolutely continuous. It remains to check that its partial derivatives are in $L^2$. By the chain rule, ...


2

One can significantly simplify your proof. Assume $B(X,Y)$ is Banach. The normed space $Y$ can be isometrically be embedded into the $B(X,Y)$ via the map $$ I:Y\mapsto B(X,Y):y\mapsto f(\cdot) y $$ for some $f\in X^*$ of norm 1. Since embedding is isometric, then $\operatorname{Im} I$ is a closed subspace of Banach space $B(X,Y)$. Hence $\operatorname{Im} ...


2

A few things that come to mind: The topology of uniform convergence is too fine on e.g. $C(X)$ where $X$ is an open subset of some $\mathbb{R}^n$. In many situations, you don't have uniform convergence, but locally uniform convergence, and the locally uniform convergence is sufficient for many theorems (the limit of a locally uniformly convergent sequence ...


2

One can prove strong convergence following this article: The product of projection operators I. Halperin http://acta.fyx.hu/acta/showCustomerArticle.action?id=7164&dataObjectType=article The original result is due to Neumann. Halperin's note generalizes this to $m$ projections: There it is proven that $(P_1P_2\dots P_m)^n x \to z$ with ...


2

First, let me say that since the family $\{g_n\}$ is not only equicontinuous but equilipschitz (that is, there is a Lipschitz constant that works for all these functions at once), we could prove the uniform boundedness of the family in one short statement. I encourage you to look for that proof. Let us prove something more general: Proposition: Let $K$ a ...


2

We can easily construct an inverse to $\phi^*$, it is given by $(\phi^*)^{-1}=(\phi^{-1})^*$, indeed we have: $$\left<(\phi^{-1})^*\phi^*v,x\right>=\left<\phi^*v,\phi^{-1}(x)\right>=\left<v,\phi\phi^{-1}(x)\right>=\left<v,x\right>$$ for all $x\in X$ and $v\in X^*$. Thus $(\phi^{-1})^*\phi^*=1$, and similarly $\phi^*(\phi^{-1})^*=1$.


2

First we want to show that $$c'=\inf\{\|a-b\|:a\in A,b\in B, \|a\|,\|b\|\ge 1\}>0$$ as suggested by Jochen. Suppose $a_n-b_n\to 0$ but $\|a_n\|,\|b_n\|\ge 1$. Then $$0\le|\|a_n\|-\|b_n\||\le \|a_n-b_n\|\to 0$$ so $\|a_n\|-\|b_n\|\to 0$. Let $c=\inf\{\|a-b\|:a\in A,b\in B, \|a\|=\|b\|= 1\}$. Then $$\begin{align} \|a_n-b_n\| &\ge ...


2

The claim is false, and there are actually counterexamples in any vector space of dimension $\ge 1$. Choose any isomorphism $T$ which is not the identity; then $\mathcal{N}(T) = 0$ and $\mathcal{R}(T) = V = \mathcal{D}(T)$. Since $T$ is an isomorphism, $$T^2 = T \iff T^{-1} T^2 = T^{-1} T \iff T = I$$


2

Apparently you forgot about the negative powers. If for example we consider $a \in \ell^1(\mathbb{Z})$ given by $a_n = 2^{-\lvert n\rvert}$, the corresponding Laurent series $$f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n = \sum_{n=1}^\infty a_{-n}z^{-n} + \sum_{n=0}^\infty a_nz^n = \sum_{n=1}^\infty \frac{1}{(2z)^n} + \sum_{n=0}^\infty ...


2

Obviously, $y \in C$ implies $f(y) \leq S_C(f)$ for any $f \in X^{\ast}$. Now we prove that whenever $y \notin C$ we can define $g \in X^{\ast}$ such that $g(y)> S_C(g)$. Without loss of generality, we may assume that $0 \in C$. As $C$ is closed, we know that $d:=d(y,C)>0$. Now we define the open blow-up of $C$ by $$C_d := C+B(0,d/2) := \{x; x=c+b, c ...


2

Assuming you have experience with cardinal arithmetic, the following show that its size is the cardinality of the continuum: First, $\mathbb{C}\cong \mathbb{R}\times \mathbb{R}$. Thus, $$|\mathbb{C}|=|\mathbb{R}\times \mathbb{R}|=|\mathbb{R}|\cdot |\mathbb{R}|=\max\{|\mathbb{R}|,|\mathbb{R}|\}=|\mathbb{R}|$$ Next, we know that $|\mathbb{R}|=2^{\aleph_0}$ ...


1

It is possible to expand $$ \|U(f+g)-U(f)-U(g)\|^{2} $$ into 9 inner-product terms involving $U$ applied to one of $f+g$, $f$, $g$ in the first coordinate with another such term in the second coordinate. Asumming $(Uf,Ug)=(f,g)$ for all $f, g \in X$, then $U$ can be removed from both coordinates of all 9 terms, resulting in the trivial ...


1

I think the author is being a little sloppy in this paragraph, which raised your question: Lower semicontinuity was earlier defined by the openness of upper level sets. The weak topology being non-metrizable, one should not casually insert "in other words" followed by the definition of sequential weak lower semicontinuity. For example, Ciarlet ...


1

In the weak* topology, the dual is a (topological) subspace of $$\mathbb{C}^X = \prod_{x\in X} \mathbb{C}.$$ A product of complete spaces is always complete, so $\mathbb{C}^X$ is complete ($\mathbb{C}$ is complete), and thus $X'$ is complete in the weak* topology if and only if it is a closed subspace of $\mathbb{C}^X$. The algebraic dual $X^\ast$ (the ...


1

Yes since if you consider that fact that if $ F\in C_c(\mathbb {R^n},\mathbb {R^m})$ then $F=( f_1,...,f_m)$ where $ f_i\in C_c(\mathbb {R^n},\mathbb {R}),\forall i=1,...,m$. Then since $C_c(\mathbb {R^n},\mathbb {R})$ is dense in $L^1(\mathbb {R})$, $\forall g\in L^1(\mathbb {R}),\forall \epsilon\gt 0$ there exists $ f\in C_c(\mathbb {R^n},\mathbb {R})$ ...


1

As noticed in the comments there are some inconsistency of notations in the question but I will try nevertheless to give an answer to your question. To do so I had to make some assumption on what is the precise context. I choose what seems to me to make sense. I hope it fits with what you really need. For the first question: If the vectors $\xi$ and $\eta$ ...


1

Directly, $$ P_{0} = 1.\\ $$ Notice that $\|P_{0}\|=1$. Then $$ Q_{1}=x-(x,P_{0})P_{0}=x-\frac{1}{2}1 $$ is orthogonal to $P_{0}$, and leads to $P_{1}=\|Q_{1}\|^{-1}Q_{1}$, where $$ \|Q_{1}\|^{2}=\int_{0}^{1}(x-1/2)^{2}\,dx=\left.\frac{1}{3}(x-1/2)^{3}\right|_{0}^{1}=\frac{1}{3\cdot 2^{3}}. $$ So $$ P_{1}=2\sqrt{6}(x-1/2). $$ Next, $$ ...


1

It's actually trivial if you write everything down. For the direct implication: Let $\epsilon > 0$ Let $N \in \mathbb N$ such that $n \geq N \Rightarrow \forall x \in S, |f_n(x)-f(x)|\leq \epsilon$ Then (by the very definition of $\sup$), for any $n \geq N$ we have that $\sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$ Equivalently said, ...


1

As closed subspaces of the Banach space $B$, both $P := \operatorname{Im} f$ and $Q := \operatorname{Ker} f$ are themselves Banach spaces. The map $a \colon P\times Q \to B;\; (p,q) \mapsto p+q$ is a bijection. (Why?) $a$ is continuous. (Why?) The inverse of $a$ is $$b \mapsto (f(b), b-f(b)),$$ therefore $f$ and $\operatorname{id}_B - f$ are continuous. ...


1

Not really an answer but a rather long comment. I think it is enough to consider $f_i$ (that is $f_i$ restrict to $X_i$). In this case you get the fixed points are the vectors that solve $(I-A_i)^{-1}b_i$. If $I-A_i$ doesn't have an inverse, the use the pseudo-inverse. Also $f_i$ may not have a fixed point or as you pointed $(I-A_i)^{-1}b_i\not\in X_i$. In ...


1

Projection: $$P:\mathcal{l}^2\to\mathcal{l}^2:P(b_1)=b_1,P(b_2)=b_2,\ldots,P(b)=0\text{ with }b_n:=\sum_{k=1}^n\frac{1}{k}e_k$$ Discontinuity: $$x_n\to(\frac{1}{1},\frac{1}{2},\ldots),P(x_n)\nrightarrow P(\frac{1}{1},\frac{1}{2},\ldots)\text{ with }x_n:=b_n$$ Remark: This can be established even nicer when noticing that any continuous operator must have ...


1

Not sure if you can use this, but a basic property of convex functions is that nonnegative combinations of them are convex. This extends to nonnegative integrals as well. If $h(x,t)$ is convex in $x$ and $g$ is nonnegative over the domain $\Omega$, and $$ f(x) = \int_\Omega h(x,t)g(t)dt $$ then $f$ is convex. This is a consequence of the definition of ...


1

Let me make my question more general and clear, maybe giving an answer, thanks to @p.s, of course! Let $f\colon K\subseteq\mathbb{R}^n\times\mathbb{R}\to\mathbb{R}$, given by $$ f(\mathbf{a},b)=\int_{\Omega}\! h(\mathbf{a},b,\mathbf{x})g(\mathbf{x}) \,\mathrm{d}\mathbf{x}, $$ where ...


1

If $(a_n)_{n\geqslant 1}$ is a sequence of complex numbers converges to some $a$, then so does the sequence $(n^{-1}\sum_{j=1}^na_j)_{n\geqslant 1}$. For a fixed linear continuous functional, use this result with $a_k:=\langle f,x_k\rangle_{X',X}$. Not completely related, but if $X$ was a Hilbert space, then we could extract a subsequence $(x_{n_k})_k$ of ...


1

The eigenvalues of $AB$ and $BA$ are always the same. As for the eigenvalues of $A \tilde{A}$ being the squares of the absolute values of the eigenvalues of $A$, have you tried some examples? Try e.g. $A = \pmatrix{1 & 2\cr 1 & 1\cr}$ and $D = \pmatrix{2 & 0\cr 0 & 1\cr}$.


1

A very simple innner product space which is not Hilbert is the spacee $\phi$ of sequences with only finitely many non-zero terms endowed with the scalar product of $\ell^2$. Let $L$ be the kernel of the continuous linear functional $x\mapsto \sum\limits_{n=1}^\infty x_n/n$ (the scalar product with $(1/n)_{n\in\mathbb N} \in\ell^2\setminus \phi$). Then ...



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