Tag Info

New answers tagged

0

HINT: You have $\|\frac{1}{2}x + \frac{1}{2}y\| = 1$. Let's show for instance that $\|\frac{1}{3}x + \frac{2}{3}y\| = 1$. Note that $$\left\|\frac{1}{3}x + \frac{2}{3}y\right\| \le \frac{1}{3}\|x\| + \frac{2}{3}\|y\|=1$$ If we had $\|\frac{1}{3}x + \frac{2}{3}y\| <1$ then $$\left\|\frac{1}{2}x + \frac{1}{2}y\,\right\| = \left\|\frac{1}{4} x + ...


0

The spaces with Fréchet differentiable norm are usually called Fréchet smooth spaces... which of course is just a name. Šmulian gave a useful characterization of this property: it holds if and only if for every unit vector $x$ and every sequence of unit functionals $f_n$ such that $f_n(x)\to 1$, the sequence $\{f_n\}$ is norm-convergent. The stronger ...


3

I think the authors are somewhat inconsistent in what they mean by the domain of an operator. Let's ignore the boundary conditions for now and focus on the order of smoothness. The domain of $(-\Delta)^s$ as an operator into $L^2$ is a Sobolev space of order $2s$. Indeed, $s$ fraction of the Laplacian is like $2s$ derivatives. The domain of ...


3

You're thinking too Euclidean! It is entirely possible in an arbitrary normed linear space to have points other than antipodal points that satisfy this property. As an example, consider the points $(1, 0)$ and $(0, 1)$ in the unit sphere under the $1$-norm on $\mathbb{R}^2$. Good question though!


1

So, you want to differentiate the map $T(X) = x/\|x\| $. For any vector $h$, the derivative of $T(x+th)$ with respect to $t$ at $t=0$ can be computed using the ordinary quotient rule: $$\frac{d}{dt}_{| t=0} \frac{x+th}{\|x+th\|} = \frac{h\|x\| - x\langle x,h\rangle/\|x\| }{\|x \|^2 }$$ So, the derivative at $x$ is the linear operator $$ h\mapsto ...


0

Independence of trace, is a result of $trace(AB)=trace(BA)$ & when you change basis, there exist a unitary matrix like $P$ such that $A=PAP^{-1}$.


0

Here is a counterexample on $(0, \infty)$ (it can be symmetrically extended to all of $\mathbb{R}$. Let $\omega(x)$ be a function s.t. for all integers $n \geq 0$ $$\omega(x) = 1 \,\,\,\,\,\, \forall x \in [n, n + 1/2)$$ and $$\omega(x) = 1 + 2n\left(x - (n + 1/2)\right) \,\,\,\,\,\, \forall x \in [n + 1/2, n + 1)$$ Then on each segment $[n, n+1)$, the ...


1

$$\sum_{j=1}^d |n_j|^{2k} \geq \max_{j=1,\ldots,d} |n_j|^{2k}$$ implies, using the mentioned inequality, $$(2\pi)^{2k} \sum_{j=1}^d n_j^{2k} \geq (2\pi)^{2k} \max_{j=1,\ldots,d} |n_j|^{2k} \geq \left( \frac{2\pi}{\sqrt{d}} \right)^{2k} \|n\|_2^{2k}$$


0

I think you have a problem of notation. I mean $AN$ is in effect a $X/M$ subspace while $ { n+M| n\in N}=N+M$ is a subspace of $X$. So $AN={ n+M| n\in N}$ it's formally uncorrect and you should write $A^{-1}AN={ n+M| n\in N}$.


0

There is no separable commutative Banach algebra that contains isometric copies of all separable commutative Banach algebras. Indeed, if $p$ and $q$ are commuting projections in a Banach algebra then $\|p-q\|\geqslant 1$ so each set of commuting projections is discrete. Now, projections in a Banach algebra can have arbitrarily large norms. Consequently, a ...


0

For $\lambda\in [0,1) $, you are taking the positive (rank-one) operator $\lambda |\psi\rangle\langle\psi|$ and you are adding a positive multiple of the identity, so the resulting operator is invertible. Thus your dimension (the range) will be $2^N $.


0

By Riesz representation theorem I can say ... No you can't, because in the Riesz representation theorem, the functional is assumed to be bounded. This theorem doesn't help you prove that it's bounded. Instead, follow the approach pointed out by A.G.: by the [trace theorem], the trace of $v$ on $\Gamma$ has $L^2$ norm bounded by $C\|v\|_{H^1(\Omega)}$ ...


1

Let's fix some notations. Denote the Haar measure on $\Bbb{T}^d$ by $\lambda$ instead of $m$. The measures suggested should probably be normalized by $$ d\mu_{v,N}(y)=\frac{1}{(2N+1)^d}\left\|\sum_{m\in\mathbb{Z}^d, \lvert m_i \rvert \leq N}\chi_{-m}(y) U(m)v\right\|^2 d\lambda(y).$$ The characters of $\Bbb{T}^d$ are the functions $$\chi_m: \Bbb{T}^d \to ...


1

For $x\in (0,\infty),x\ne 1,$ define $f(x) = \frac{1}{\sqrt x \ln x}\frac{x-1}{x+1}.$


0

I don't know if this is what you were looking for, but it is possible to create a function which is in $L^2$ but not in $L^p$ for any $p\ne2$. Note that $$ f_n(x)=\left(x^{\frac{n-1}{n}}+x^{\frac{n+1}{n}}\right)^{-1/2} $$ is in $L^p$ only for $\frac{2n}{n+1}\lt p\lt\frac{2n}{n-1}$. For $p$ outside that range, $f_n(x)\not\in L^p$. Any $p\ne2$ is outside that ...


0

PROBLEM Answer Explanation Demonstration Given the real line $\mathbb{R}$. Consider the metrics: $$d(x,y):=|y-x|$$ $$d'(x,y):=\arctan|y-x|$$ Then one obtains: $$\mathcal{N}=\mathcal{N}'\quad\mathcal{U}\neq\mathcal{U}$$ Concluding problem. PROOF Identification Given normed spaces $\Omega$ and $\Omega'$. Regard the category: ...


0

You may be interested in the policy iteration algorithm. To clear something up the value function is not time-dependent and solves $$ v(w)=\max_{w'\in [0,w]}\{ u(w-w')+\beta v(w')\}. $$ This should be solved on the interval $[0,w_0]$ for $w_0$ the initial allotment of bread. For policy iteration, fix a policy $c:[0,w_0]\rightarrow \mathbb{R}$ with ...


1

You can start with "the norms induce the same topology". Then use the fact that a linear transformation is continuous if and only if it is bounded. And this is one of your inequalities. For the other direction, use the inverse of that linear transformation.


1

Maybe it will be useful to consider an example of two norms $F$ and $G$ of a vector space $X$ not being equivalent to each other. What it means is that at least one of the quantities $\sup\limits_{x \in X}\frac{F(x)}{G(x)}$ or $\sup\limits_{x \in X}\frac{G(x)}{F(x)}$ is unbounded, i.e. there is a sequence $(x_n)_{n \geq 0}$ of vectors in the space such that ...


1

Suppose you have a sequence which converges in $G $. The lower bound implies it converges in $F $ to the same limit. Suppose you have a sequence which ddoes not converge in $G $. The upper bound implies it does not converge in $F $. That's all you need, since metric spaces are sequential spaces.


1

Are you sure about your definition of $\mathcal H_0^{2,c}$? Shouldn't this be the class of $L^2$-bounded continuous martingales? This is a indeed a Hilbert space with the scalar product $\langle L,N\rangle =E[L_\infty N_\infty]$. This makes sense because $L$ and $N$ are uniformly integrable and therefore $L_\infty=\lim L_t$ exists by the martingale ...


1

The Fréchet derivative $\nabla f(x)$ of a function $f:X\to\mathbb{R}$ at $x$ is an element of $X^*$. And therefore, $\nabla f$ is a map from $X$ to $X^*$. The Lipschitz condition makes sense whenever there is a map between metric spaces; and both $X$ and $X^*$ have metrics induced by their norms. Explicitly, $$\| \nabla f(x) - \nabla f(y)\|_{X^*} ...


0

In what you wrote first, $v_n$ should be an element of the dual space and then it means convergence in the dual norm (norm of the functionals over $W_0^{1,p}$). The second thing, that you wrote, i.e $\langle x^*,v_n\rangle\rightarrow 0\quad \forall x^*\in (W_0^{1,p})^*$ means weak convergence of the sequence $\{v_n\}_{n=1}^\infty\subset W_0^{1,p}$ to $0\in ...


0

As Martin pointed out, $\langle f,v,\rangle$ is just notation for the action of the functional $f\in X^*$ on the element $v\in X$; it does not mean (in general) an inner product. To be less ambiguous, one can write the pairing as $$ _{X^*}\langle f,v\rangle_X $$ to emphasize the fact that $f$ and $v$ live in two different spaces, so the action is not an ...


1

The symbols $\langle x^*,v_n\rangle$ just express $x^*(v_n)$, the functional $x^*$ evaluated at $v_n$. It is a common notation, inspired in the Hilbert space case, where the dual is the same original space.


0

It is a general fact that the tensor product of bounded operators gives a bounded operator. Since your operator is of the form $S\otimes 1$, it is enough to show that $S$ is bounded. As it is defined in terms of a series, we need to check that such series converges. Write $q_k=\sqrt{1-q^{2k}}$. Note that $0\leq q_k\leq 1$ (assuming $|q|\leq1$). Given any ...


6

First note that the sequence is bounded: $$ |\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty. $$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it ...


3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...


0

By linearity one has: $$f=g:\quad\left(\int|f+g|\mathrm{d}\mu\right)^2+\left(\int|f-g|\mathrm{d}\mu\right)^2=2\left(\int|f|\mathrm{d}\mu\right)^2+2\left(\int|g|\mathrm{d}\mu\right)^2$$ $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|\mathrm{d}\lambda=\int|f|\mathrm{d}\lambda+\int|g|\mathrm{d}\lambda$$ So shift a block: ...


2

There is nothing wrong about your argumentation. Here is a "real" counterexample: Set $$f(x) := 1_{[0,1/2]}(x) \qquad \text{and} \qquad g(x) := 1_{[1/2,1]}(x).$$ Then $$\|f\|_1 = \|g\|_1 = \frac{1}{2}$$ and therefore $$2 (\|f\|_1^2+ \|g\|_1^2) = 1.$$ On the other hand, it is not difficult to see that $$\|f+g\|_1 = \|f-g\|_1 = 1.$$ Hence, ...


0

For $p> 1$ it is true that $\int{f_ng}\rightarrow\int{fg}$. This follows from Theorem 3. Because the cited theorem has a different notation, I will write the body of it with your notation (in terms of $p$, $L_p$ ) and then apply it. Theorem 3. Let $M$ be a positive constant and $(X,\mu)$ be a complete measure space. Suppose that $0<p\leq \infty$ and ...


0

Let us assume that $K$ is bounded and invertible (hence, boundedly invertible). Then, you have \begin{align*} (F \circ K)^* (x^*) &= \sup_x \{ (x^*, x) - (F \circ K)(x) \} \\ &= \sup_y \{ (x^*, K^{-1} y) - F(y) \} \\ &= \sup_y \{ (K^{-*} x^*, y) - F(y) \} \\ &= F^*(K^{-*}(x^*)) = (F^* \circ K^{-*})(x^*). \end{align*}


1

Let $(A_n)_{n\in\mathbb{N}}$ be the set of disjoint measurable subsets of $L_p(\Omega,\mu)$ that all of positive measure. Then consider bounded liner operator $$ P:L_p(\Omega,\mu)\to L_p(\Omega,\mu):f\mapsto \sum_{n=1}^\infty\left(\mu(A_n)^{-1}\int_{A_n}f(\omega)d\mu(\omega)\right)\chi_{A_n} $$ One can show that ...


12

Note that $C_c \subset C_0$, but $C_c \neq C_0$. For example, $f(x) = \dfrac{1}{x^2+1}$ belongs to $C_0$ but not $C_c$. What you seem to be assuming is that $\lim_{|x|\to\infty}f(x) = 0$ implies that there is some $N > 0$ with $f(x) = 0$ for all $|x| > N$. This is not true, as the above example demonstrates. That is, a function can limit to zero at ...


0

The first one contains functions which are nonzero over the whole real line. The second one does not.


2

For convenience, scale $\phi$ so that $\|M_{\phi}\|_{\mathcal{L}(X)}=1$. For $\delta >0$, the $\chi_{\delta}$ be the characteristic function of the set where $|\phi| \ge 1+\delta$. Then, for any $f \in L^{1}$, $$ (1+\delta)\int |f\chi_{\delta}|d\mu \le \int |f\chi_{\delta}| |\phi|d\mu \le \int |f\chi_{\delta}|d\mu. $$ Thus $\|f\chi_{\delta}\|=0$ ...


3

Let $Af = \int_{0}^{x}f(t)dt$ in $L^{2}[0,1]$. Then $A : L^{2}\rightarrow L^{2}$ is bounded. Let $W$ consist of all continuously differentiable $g \in L^{2}[0,1]$ for which $g(0)=g(1)=0$. $W$ is dense in $L^{2}[0,1]$ because $\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$ is an orthogonal basis of $L^{2}[0,1]$. However, $A^{-1}W$ is not dense because $f \in ...


3

This is false. Consider $$\mu = \sum_{n=-\infty}^\infty \infty \delta_n + \lambda$$ where $\infty \delta_n$ is the measure that has infinite point mass at $n$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$. Let $\phi(n) = n,$ for $n\in \mathbb{Z}$ and $\phi(x)=1$ for $x$ not an integer. Then $f \in L^1(\mu)$ implies $f(n) = 0$ for all $n\in ...


3

Edit Here is a proof that works if $\mu$ is semifinite. Suppose $\phi \notin L_\infty(X, \mu).$ Then given any $R>0$ there is a set of positive measure $E \subset X$ such that $\vert \phi \vert > R$ on $E.$ Let $f \in L_1(X,\mu)$ such that $f$ vanishes outside of $E$. Taking absolute value, we can assume $f$ is real and non-negative. Then $$\Vert Mf ...


0

Yeaaahh, I got it! :D Great thanks to David C. Ullrich!!! Counterexample Given the Hilbert space $\mathbb{C}^4$. Regard the matrices: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Then they are ...


1

Because $W_0^{1,p}$ is compactly embedded in $L_p$, the embedding operator is obviously compact so it maps weakly convergent sequences in strongly convergent, i.e a weakly convergent subsequence in $W_0^{1,p}$ is strongly convergent in $L_p$. Finally, you can extract a further subsequence, which is pointwise a.e convergent: see Does convergence in $L^{p}$ ...


0

The absolute value of $|a|$ is $a $ or $-a $, depending on whether $a $ is nonegative or not. In your case, the condition $n\geq m $ implies $f_m (t)\geq f_n (t) $, so $$|f_n (t)-f_m (t)|=-(f_n (t)-f_m (t))=f_m(t)-f_n (t). $$ For your second question, when you calculate your integrals you are forgetting the antiderivatives.


2

What you need to show is that $\phi $ is one-to-one, and that it is a $*$-homomorphism. That lets you define the norm on $M_n (A) $ and the two properties you want are inherited automatically from the codomain. Then you need to show that $M_n (A) $ is closed. This is done by showing that norm convergence is equivalent to entrywise norm convergence, and as ...


3

Uniform closure, of course not. Closure in the topology of pointwise convergence yes, as has been pointed out, but this is trivial (immediate from the fact that a polynomial can take any values you want on any finite set) and I don't see how it's good for anything. There's a closure for which the answer is yes, and which also actually allows one to prove ...


0

1) Yes. 2) CP on a Hilbert space does not make sense. You need $X\subset B (H) $ a C$^*$-algebra, and $Y\subset B (K) $ on the codomain. There, you have to convince yourself that entry wise sot convergence in $M_n (Y) $ implies sot convergence in $M_n (Y) $. Note that $\langle T_jx,x\rangle \to\langle Tx,x\rangle$ is wot convergence, not sot.


1

From $a^*ab^*b=0$ we get $$0=a^*ab^*ba^*a=(ba^*a)^*ba^*a, $$ so $ba^*a =0$. But then $ba^*ab^*=0$, which implies $ab^*=ba^*=0$ The others are similar.


1

The topology you use on $C $ is the product topology, so basic open sets are products of an open set in $E $ and an open set in $\mathbb R $. We can write $$ C=\bigcup_{t>0}\{x:\ p (x)<t\}\times (-\infty,t). $$


1

All polynomials are dense in borel functions in pointwise convergence topology because polynomials are dense in continous functions in uniform convergence topology, and those are dense in borel in pointwise convergence topology. However if you restrict to polynomials vanishing at zero you will only get borel functions vanishing at zero.


1

The integral (also known as tha $L^1$ metric) and the uniform metric cannot be defined on any space. The uniform metric is defined on sets of bounded, real (or complex) valued functions defined on a set $X$. If $f,g\colon X\to\mathbb{R}$ are bounded, then $$ d_\text{u}(f,g)=\max_{x\in X}|f(x)-g(x)|. $$ It measures the greates difference between the values of ...


2

It may be of some value to write an explicit answer (which is pretty similar to the comment by user251257, only containing more details). Let $e:U\to\mathbb{R}$ be differentiable, and let $e':U\to(\mathbb{R}^n)^*$ denote the differential of $e$. Let $u\in U$, and let $h\in\mathbb{R}^n$. By differentiability, we ...



Top 50 recent answers are included