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The way the cartesian and tensor product combine can be seen in the isomorphism of complex valued functions $$L^{2}(R\times R)\cong L^{2}(R)\otimes L^{2}(R)$$ The inner product on the left hand side is a double integral while that on the right is your product of inner products (integrals in one variable) . This identifies functions on $R\times R$ with ...


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The construction for the locally convex inductive limit topology on $C^\infty_c$ is not technical at all: It is the locally convex tology generated by all semi-norms $p$ all whose restrictions to $C^\infty_K$ are continuous. However, the statement you claim is not true: There are many nets which converge to $0$ without having support in a fixed compact set. ...


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First you treat Hilbert spaces as algebraic objects, vector spaces. Thus you may take an algebraic tensor product $H_1 \underline{\otimes} H_2$, however an algebraic tensor product of Hilbert spaces is not a Hilbert space. Therefore, we have to fix this obstacle, we define the inner product (as you done it in your question) on their algebraic tensor product ...


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My favorite application of Baire is due to Daniel Fischer - thanks alot to Daniel!!! (See his Answer to Uncountable Lebesgue Null Set.) There exist uncountable Lebesgue null sets over the real line. (These are precisely the ones giving rise to those obscure singular continuous measures.)


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1) No, every such operator is bounded. For simplicity, I assume that all functions are real-valued. If $T$ is not bounded, there is a sequence $f_n$ with $\Vert f_n \Vert_p \leq 2^{-n}$ and $\Vert Tf_n\Vert_p \geq 2^n$ (why?). Then Fatou's Lemma (or monotone convergence) shows $f :=\sum_n |f_n|\in L^p$ with $\Vert f\Vert_p \leq 1$. But $|f_n|\leq ...


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For boundedness, I suggest you set $\epsilon=1$, that is $|\xi_j| < 1 + k_{N+1}$ for some $k_{N+1}\in \mathbb{R}$,so that you get a fixed upper bound for $|\xi_j|$. For convergence, Since $(x_n)$, where $x_n \colon= (\xi_j^{(n)})$ is a Cauchy sequence in $\ell^\infty$. Then, given $\epsilon > 0$, there exists an integer $N$ such that $\forall m,n ...


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This is not a proof of the general statement you ask. However, the answer is no if you're in an inner-product space and the sequence is orthonormal: If $u,v$ are orthonormal, then $$ |u-v|^2 = \langle u-v, u-v \rangle = |u|^2 + |v|^2 = 2 $$ as the cross terms cancel. Therefore, if $\{v_n\}_{n=1}^{\infty}$ is an orthonormal sequence, it will not be Cauchy. ...


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How to find these... By linearity of the integral one has: $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|^p\mathrm{d}\lambda=\int|f|^p\mathrm{d}\lambda+\int|g|^p\mathrm{d}\lambda$$ So take a block and shift it slightly: $$f:=\chi_{[0,1]}:\quad f_\varepsilon(x):=f(x-\varepsilon)$$ Then for an appropriate choice: ...


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I'll give you a hint. Let $$f(x)=\begin{cases}2,x\in[0,3/4],\\0,x>3/4,\end{cases}\quad g(x)=f(1-x).$$ Then the parallelogram law says that in hilbert spaces we have $$2\|f\|^2+2\|g\|^2=\|f-g\|^2+\|f+g\|^2.$$ Can you calculate the norms above?


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On the one hand, I feel that your question is almost too general for math.se, but on the other hand I also think it is a fairly interesting one! In the following, I will provide some basic ideas regarding your question (just enough so that you will be able to research further on your own!) and also provide a suggestion for your particular case. The question ...


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Let $S$ be any densely-defined selfadjint linear operator on a Hilbert Space with spectral measure $E$. The range of $E[-\lambda,\lambda]$ is in the domain of every positive power of $S$. In fact, $$ \mathscr{C}=\bigcap_{n=1}^{\infty}\mathcal{D}(S^{n}) $$ is a core for $S$. This is because the range of $E[-\lambda,\lambda]$ is in ...


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If you have to problems with products, first think about the case $I:=\{1,2\}$. Then an element of $\prod_{i\in I}\mathcal L(E,F_i)$ can be written as $(T_1,T_2)$, where $T_i\in \mathcal L(E,F_i)$. We can define for $x\in E$, $$(\varphi(T_1,T_2))(x):=(T_1(x),T_2(x))$$ This gives a isometric isomorphic map between $\prod_{i\in I}\mathcal L(E,F_i)$ and ...


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Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.


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I am afraid that it is rather straight-forward: We have that $$ Dw_k \cdot Dv_k -|Du|^2 =Dw_k\cdot Dv_k-Dw_k\cdot Du+Dw_k\cdot Du-Du\cdot Du= Dw_k\cdot (Dv_k- Du)+Du\cdot(Dw_k-Du), $$ and hence $$ Dw_k \cdot Dv_k -|Du|^2 \le \lvert Dw_k\cdot (Dv_k- Du)\rvert+\lvert Du\cdot(Dw_k-Du)\rvert \le \lvert Dw_k\rvert\lvert Dv_k- Du\rvert+\lvert Du\rvert \lvert ...


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For each $x\in E$ define $\kappa(x)\in E^{\prime \prime}$ by $\langle \kappa(x), f\rangle = \langle f, x\rangle$. Then $E_0^a = \bigcap_{x\in E} \ker \kappa(x)$ is closed as the intersection of a family of closed sets. (The kernels are closed by continuity of $\kappa(x)$.)


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By the Cauchy-Schwarz inequality, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \sqrt{\int_U |Dw_k|^2} \sqrt{\int_U |Dv_k - Du|^2}.$$ Since the Sobolev norm has $\|f\|_{H^1}^2 = \int_U (|f|^2 + |Df|^2) \ge \int_U |Df|^2$, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \|w_k\|_{H^1} \|v_k - u\|_{H^1}.$$ Since $w_k$ converges in $H^1$ norm, $\sup_k \|w_k\|_{H^1} ...


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See if you can figure it out from the following hint: If $E$ is finite dim, by defn. it has a finite basis. Say $x_1, x_2, \dots x_n$ is a basis of $E$. For any $x \in E$, x can be written uniquely as $x = a_1x_1 + a_2x_2 + \dots a_n x_n$. Try to prove the following: 1) For $j \in \{1, \dots n \}$, the maps $x_j^*: X \to \mathbb{R}$ defined by $$ ...


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Suppose that $E$ is finite dimensional. Let $x_1, \ldots, x_n$ be a basis for $E$. Consider the coordinate functionals $f_1, \ldots, f_n$ given by $$\langle f_i, \sum_{k=1}^n \lambda_k x_k \rangle = \lambda_i\quad (i\leqslant n)$$ Prove that $f_1, \ldots, f_n$ are linearly independent. Having this done, note that they span $E^\prime$. Indeed, if $f$ is any ...


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Hint: let $h=t-1$, if $x\neq 0$ $$\frac{\varphi(tx)-\varphi(x)}{t-1}=x\frac{\varphi(x+hx)-\varphi(x)}{hx}.$$ By mean value theorem $\varphi(x+hx)-\varphi(x) = \varphi'(\eta)hx $, where $\eta\in [\min\{x,x+hx\}, \max\{x,x+hx\}]$. Again by mean value theorem $$|x||\varphi'(\eta)-\varphi'(x)|\leq |x(\eta-x)| \sup_y |\varphi''(y)| \leq |x^2h| \sup_y ...


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A sublinear function (or functional) is a function f:V to F on a vector space over F which satisfies: f(ax)=af(x) for any positive a in F and any x in V. f(x+y)=


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There are many norms. This is but an example. Let $W=\{w_i\}_{i=1}^\infty$, $w_i>0$, be a weight-sequence. Then $$ \|x\|_{p,W}=\sum_{í=1}^\infty|x_i|^pw_i $$ defines a norm. If $W$ is bounded, then it is a norm on $\ell^p$. If moreover $\inf_{i} w_i>0$, then it is equivalent to the $\ell^p$ norm. But if $\inf_{i}w_i=0$, the norms are not equivalent.


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You could let $A = \mathbb C$ and define the following trivial multiplication: For all $z,z' \in \mathbb C$: $$zz' = z'z = 0$$ Now assume that $\tau \in \Omega (A)$ was a character. Then $$\tau(z)^2 = \tau(z^2) = \tau(0) = 0$$ But since characters are non-zero by definition it therefore follows that the character space $\Omega (A)$ is empty.


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Smoothness of a function and the decay of $a_n$ at infinity are connected. For smooth functions ( $C^{\infty}$), $a_n$ coverges to $0$ faster than any power of $1/n$, that is $$\lim_{n \to \infty} n^k \cdot a_n = 0$$ or $|a_n| = O(\frac{1}{n^k})\ \forall k>0$. For real analytic functions ( equivalent to : restriction to the circle of a holomorphic ...


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There is not. The closest you can come to explicitness, so far as I know, is to let $p$ be a free ultrafilter on $\Bbb N$ and extend $L$ to the $p$-limit: $p$-$\lim_nx_n=a$ iff for all $\epsilon>0$ $\{n\in\Bbb N:|x_n-a|<\epsilon\}\in p$.


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While $\Vert A\Vert$ as defined above does define a norm of $\mathbb{C}^{n\times m}$, to show that it is an operator norm this is not enough. You must demonstrate the following: There exist vector spaces $V$ and $W$ equipped with norms $\Vert\cdot\Vert_V$ and $\Vert\cdot\Vert_W$, such that the space of bounded linear operators $A:V\to W$ is precisely ...


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I'm very sorry but I realised there is a mistake in my "proof" above (also for some reason I was logged in only as a guest account when I posted my answer which is why I'm writing under a different profile now). The mistake is as follows: in estimating the first term in the right hand side of the final inequality, I'd estimated as follows $$|S_n^{**} ...


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Since your $f$ is continuous and $S$ is compact so is $f(S)$ but in $\mathbb{R}$. Hence $f(S)$ contains its infimum.


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First of all, I believe you meant $f(a)\leq\alpha$, for every $a\in A$, and $\alpha<f(b)$, for every $b\in B$. Let us prove that it is not always possible to find this linear functional. Consider the vector space of real sequences $\mathbb{R}^{\mathbb{N}}$. Let $A$ be the subspace of $\mathbb{R}^{\mathbb{N}}$ formed by sequences with finitely many non ...


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Note you've proved that your result is true for reflexive Banach spaces X. I think you can actually prove it's true in general, though it's been a few years since I've done any functional analysis so you might want to check this carefully. The basic idea is to follow your argument, noting we're fine if $x^{**}$ is in $J_X(X)$ (where $J_X$ is canonical ...


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You can define linear functionals $g_j:Im(T) \to \mathbb C$ by $$ g_j(\sum_{i=1}^n a_i y_i)=a_j. $$ Another way to phrase this is that $g_j$ is defined by setting $$ g_j(y_i)=\delta_{ij} $$ and then extending linearly. Using the fact that all norms on finite-dimensional spaces are equivalent, this can easily be shown to be continuous. Then $g_j \circ T$ is ...


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$|\frac{1-e^{-jk}}{jk+1}|^2 \leq \frac{1}{j^2k^2}$ for large $j,k$.


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We have $$\left|\frac{1-e^{-jk}}{jk+1}\right| < \frac{1}{jk}$$ so $$\sum_{j,k=1}^\infty \left|\frac{1-e^{-jk}}{jk+1}\right|^2 < \sum_{j,k=1}^\infty \frac{1}{j^2k^2} = \left(\sum_{j=1}^\infty \frac{1}{j^2}\right)^2 < \infty$$ since $\sum_{j=1}^\infty \frac{1}{j^2} = \frac{\pi^2}{6}$.


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A sequence space is a space of sequences. In this case I think you're talking about a vector space, each member of which is a sequence ${\bf x} = (x_1, x_2, x_3, \ldots)$ of real numbers (or complex, if you prefer complex scalars). A permutation $\pi$ of the natural numbers $\mathbb N$, i.e. a one-to-one function from $\mathbb N$ onto $\mathbb N$, acts on ...


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Wendland's Scattered Data Approximation covers this topic in chapter 6. Below, I have paraphrased relevant definitions. Definition 6.1 A continuous function $\Phi:\mathbb{R}^d\rightarrow \mathbb{C}$ is called positive definite on $\mathbb{R}^d$ if for all $N\in\mathbb{N}$, all pairwise distinct $X=\{x_1,\dots,x_N\}\subset\mathbb{R}^d$, and all ...


1

The polarization identity for a complex Hilbert space $H$ is $$ (x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2},\;\;\; x,y\in H. $$ If $T$ is a linear isometry, then $$ \begin{align} (T^{\star}Tx,y) & =(Tx,Ty) \\ & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|Tx+i^{n}Ty\|^{2} \\ & = ...


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Isometric means $$ \|Tx\|=\|x\|, $$ for all $x\in H$. Equivalently, for all $x,y\in H$ $$ \|T(x+y)\|^2=\|(x+y)\|^2. $$ But $$ \|(x+y)\|^2=\langle x+y,x+y\rangle=\|x\|^2+2\langle x,y\rangle+\|y\|^2, $$ while $$ \|T(x+y)\|^2=\langle T(x+y),T(x+y)\rangle=\|Tx\|^2+2\langle Tx,Ty\rangle+\|Ty\|^2. $$ Thus, for all $x,y\in H$ $$ \langle Tx,Ty\rangle=\langle ...


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Assuming $H$ is a Hiblert space: Let $\{e_i\}$ be an orthonormal Schauder basis for the Hilbert space. We note that $$ T^*T = I \iff \forall x,y: \langle x,T^*Ty\rangle = \langle x,y\rangle\\ \iff \forall x,y: \langle Tx, Ty \rangle = \langle x,y \rangle $$ Try to deduce this last statement using the fact that $$ \forall x,y: \langle Tx, Tx \rangle = ...


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Consider the formal differential operator (i.e., has no specific domain): $$ L= -\frac{1}{2r^{2}}\partial_{r}r^{2}\partial_{r} = -\frac{1}{2}\partial_{r}^{2}-\frac{1}{r}\partial_{r}. $$ The solutions of $L\psi=0$ are easily derived: $$ \psi = C\frac{1}{r}+D. $$ Neither of these solutions is square-integrable at $\infty$ with respect ...


2

Since $\mathcal{V}$ is convex and attains no negative values, for every $c\in [0,\infty)$ the set $S_c = \mathcal{V}^{-1}([0,c])$ is convex. Since $\mathcal{V}$ is lower semicontinuous, $S_c$ is closed. Since $$\mathcal{V}(\rho) = \int_0^1 V(\rho(x))\,dx \geqslant \int_{\{x : \lvert \rho(x)\rvert \geqslant 1\}} V(\rho(x))\,dx \geqslant \delta\int_{\{x : ...


0

Let $n(k)=\lfloor\mathrm e^{2k\pi}\rfloor$ for every positive integer $k$, then $$2k\pi+\log(1-\mathrm e^{-2k\pi})\leqslant\log n(k)\leqslant2k\pi,$$ hence $u_{n(k)}\geqslant\cos(\log(1-\mathrm e^{-2k\pi}))$. Since $\log(1-\mathrm e^{-2k\pi})\to0$, $u_{n(k)}\to+1$ when $k\to\infty$. Let $m(k)=\lfloor\mathrm e^{(2k+1)\pi}\rfloor$ for every positive integer ...


2

If you start with a unit vector $v \in \mathcal{H}$, then you can define a subspace $\mathcal{H}_{v}$ as the closure in $\mathcal{H}$ of all polynomials in $N$, $N^{\star}$ acting on $v$. $\mathcal{H}_{v}$ is invariant under $N$, $N^{\star}$. This defines a unitary map $$ \mathcal{F}_{v} : \mathcal{H}_{v}\rightarrow L^{2}_{\mu_{v}} $$ where ...


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I think I got it. There is no such counterexample. I welcome and appreciate any comments on the proof below, which hinges on ideas presented by Klee (1952). $\textbf{Claim:}\quad$If $(X,\|\cdot\|)$ is a completely metrizable normed vector space, then it is a Banach space. Proof:$\quad$Suppose that $(X,\|\cdot\|)$ is a normed vector space and $d$ is such a ...


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The answer is yes, any such space must be a Banach space. This result was proved by Victor Klee in 1952 and answered a question first asked by Banach in 1932. V. L. Klee, Invariant metrics in groups (Solution of a problem of Banach), Proc. Amer. Math. Soc., 3 (1952), 484–487.


0

To continue with your approach, you could observe that the map $x\mapsto \|x\|$ is differentiable with nonzero derivative on $\mathbb R^n\setminus 0$. Therefore, it maps open sets to open sets. So, the set $\{\|f(x)\|:f(x)\ne 0\}$ is open in $\mathbb R$, which gives the result. A slicker way (I think) is to introduce $\phi(x)=\|f(x)\|^2$, which is also ...


3

Numerically, we get some very interesting results for the matrix $M_{ij}=(i-1)^{j-1}$ if it is expressed in the Fourier basis with alternating row signs, $$A := \begin{bmatrix}1 \\ & -1 \\ && 1 \\&&& -1 \\ &&&& \ddots \end{bmatrix}\mathcal{F}^{-1} \left[\begin{matrix} 0^0 & 0^1 & 0^2 & 0^3 & \ldots\\ ...


0

The linear transformations of $\Bbb R^\infty$ Take the form of row-finite matrices (all rows only have finitely many nonzero entries) or column finite matrices (analogously) depending on if you're writing the vectors on the right or left of transformations. This matrix is neither row nor column finite, so it can't represent a linear transformation, ...


3

3. $D(t^n) = nt^{n-1}$. But $\|t^n\| = 1$ and $\|D(t^n)\| \to \infty$. So $D$ is not a bounded operator. You know, right, that for linear operator on normed spaces, continuity holds iff the operator is bounded?


1

When you are on the free space and your operators have constant coefficients, you can diagonalize them via the Fourier transform. (To "diagonalize" a linear operator on a Hilbert space here means finding a unitary transformation that turns it into a multiplication operator on $L^2$-space). This solves the self-adjointness issue, since it is easy to tell when ...


3

Sometimes is good to look at the simple examples first. Let $X=Y=\mathbb C^2$, and $$ f=\begin{bmatrix}1&1\\0&1\end{bmatrix}. $$ Take $X_0=\left\{\begin{bmatrix}x\\0\end{bmatrix}:\ x\in\mathbb C\right\}$, and $Y_0=f(X_0)=X_0$. Now you can check that $$ f'(Y_0)=\left\{\begin{bmatrix}x\\ x\end{bmatrix}:\ x\in\mathbb C\right\}\not\subset X_0. $$



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