New answers tagged

0

$$ - \frac{1}{2} \leq \frac{uv}{u^2 + v^2} \leq \frac{1}{2} $$ because $$ (u-v)^2 \geq 0 $$ and $$ (u+v)^2 \geq 0 $$ The stuff above is just, well, true. If you then take $$ u = x, \; \; v = y^2 $$ you get something useful.


1

Distributions are correctly thought of as continuous maps $F:\mathcal D\to\mathbb C$ where $\mathcal D$ is "test functions," or smooth functions with compact support in some region $\Omega$. In light of this abstraction, they do not enjoy all of the same operations that you have on traditional functions, although many extend by self-adjointness. For example, ...


1

Notice that for all $t \in \Bbb R$, $$\left| \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt) \right| \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| \underbrace {|\cos (2nt)|} _{\le 1} \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| = \frac 1 {4n^2-1}$$ and $\sum \frac 1 {4n^2-1}$ is convergent (use the limit comparison test and compare it with $\sum \frac 1 ...


0

HINT: For the left to right direction it’s probably easier to show the contrapositive; suppose that $\inf_{n\in\Bbb N}a_n=0$, and show that $S^{(a)}$ is not open. For this you want to find an $x\in S^{(a)}$ such that no open ball centred at $x$ lies entirely inside $S^{(a)}$. In fact it turns out that you don’t have to search hard for such an $x$: any $x\in ...


2

If $x=0$ you're done, since $p(y)\le p(y)$. Similarly if $y=0$. So assume $x\ne0$ and $y\ne0$. Let $C=\{x:p(x)\le 1\}$. Assume $C$ is convex. Now $$\frac x{p(x)},\frac y{p(y)}\in C.$$ Let $$t=\frac{p(x)}{p(x)+p(y)}\in[0,1].$$Then $$C\ni t\frac{x}{p(x)}+(1-t)\frac{y}{p(y)}=\frac{x+y}{p(x)+p(y)}.$$So $$p\left(\frac{x+y}{p(x)+p(y)}\right)\le1,$$hence ...


1

The equality $$ \int_{\Omega \setminus F} (\dots) \, dx = \int_{\Omega^{(i)}} \int_{ \{x_i \in \mathbb{R} \, : \, (x_1, \dots, x_i, \dots, x_N) \in \Omega \setminus F \}} (\dots) \, dx_i \, \dots \, dx_{i-1} \, d{x_{i+1}} $$ is Fubini's theorem. It has nothing to do with the functions being integrated, or with the assumption on $F$. It's just doing ...


1

Idea: Consider $C[0,1]$ with the supremum metric $d(x, y) = \sup_{t\in [0,1]} |x(t) - y(t)|$ and $F : C[0,1]\to C[0,1]$ be given by $$(Fx) (t) = \begin{cases} \frac{1}{2} x(3t) + \frac{1}{2},& 0 \leq t \leq \frac{1}{3}\\\ f(t), & \frac{1}{3} < t\leq \frac{2}{3}\\\ \frac{1}{2}x(3t - 2) - \frac{1}{2},&\frac{2}{3} < t \leq 1 \end{cases} $$ ...


2

Since the properties involved are about boundedness and convergence, they survive $*$-homomorphisms, which means that they pass to quotients. And any separable C$^*$-algebra is a quotient of $C^*(\mathbb F_\infty)$. In particular, $C[0,1]$ is, so $C^*( \mathbb F_\infty)$ does not have a boundedly complete Schauder basis by the article you linked to.


1

To answer your question, yes. For a proof, first require a few prerequisites. From F. G. Friedlander and M. Joshi's Introduction to the Theory of Distributions, we have Theorem 1.2.1. Let $f \in C_c^k(\mathbb R)$, let $\rho \in C_c^\infty(\mathbb R)$ be such that $\rho \geq 0$ and $\int \rho = 1$, let $\rho_\varepsilon = \varepsilon^{-1} \rho( x / ...


0

Implicit in your identity $f(x-y)=f(x)/f(y)$ is $f(y)\ne 0$. Letting $u=x-y$, $v=y$ gives the standard exponential property $$ f(u)f(v)=f(u+v). \tag{*} $$ And $f(0)f(0)=f(0)$ implies $f(0)=1$. Differentiating with respect to $u$ and then setting $u=0$: $$ f'(u)f(v)=f'(u+v) \\ f'(0)f(v)=f'(v) ...


1

Fourier's Treatise on heat conduction was first submitted in 1807. In that work he introduced the method of separation of variables, and various expansions in orthogonal functions. Fourier died in 1830. Gauss looked at Potential Theory at roughly the same time. By 1830, Gauss was heavily involved in the study of Electromagnetism. He gave a minimization ...


5

Assuming merely that $f$ is differentiable, $f'$ is measurable because $\{x\mid f'(x)>L\}$ is Borel for every $L$: Since we know $f$ to be differentiable everywhere, the $x$ for which $f'(x)>L$ are those where $\dfrac{f(x+1/n)-f(x)}{1/n}> L$ for all sufficiently large $n$. And once we choose an $n$, the set $$ A_n = \left\{ x \;\Big|\; ...


3

You have $$C_0(\mathbb R)=\{f\in C(\beta\mathbb R):\ f|_{\beta\mathbb R\setminus\mathbb R}=0\}.$$


3

Suppose each $f_n$ is bounded, i.e. for each $n$, we have $\|f_n\|=\sup_{\|x\|=1}|f_n(x)|<\infty$. Define $T_n:X\to \ell^1(\mathbb N)$ by $x\mapsto \sum_{i=1}^n f_i(x)e_i$, where $\{e_i\}$ is the canonical basis for $\ell^1(\mathbb N)$. If $\|x\|=1$ then $$\sup_n\|T_n x\|=\sum_{i=1}^\infty|f_i(x)|<\infty, $$ so by Banach-Steinhaus we have ...


2

For a subspaces $A'$ of $X'$ the topology on $X$ given by the seminorms $x\mapsto |a'(x)|$ with $a'\in A'$ is somtimes denoted $\sigma(X,A')$. A standard result is that the dual of $(X,\sigma(X,A'))$ is equal to $A'$. In particular, $\sigma(X,A')$ is strictly coarser than $\sigma(X,X')$ if $A'\neq X'$. The situation becomes more symmetric for general dual ...


0

It is well known in narrow circles that $$ d(Z,\ell_2^n)\leq\sqrt{n}\tag{1} $$ Furthermore it is easy to see straight from the definition that $$ d(U, V)\leq d(U,W)d(W,V)\tag{2} $$ Now set $U=X$, $V=\ell_\infty^n$, $W=\ell_2^n$ and apply $(1)$.


5

When $x=y=0$ we have $f(0)=f(0)/f(0)$, which implies $f(0)=1$. Let $x-y=a$: then $f(a)f(y)=f(a+y)$. For fixed $a$ we have $f(a)f'(y)=f'(a+y)$. When $y=0$, we get $f(a)f'(0)=f'(a+0)$, or $pf(a)=f'(a)$, or $$f(a)=\frac{f'(a)}{p} \tag{1}$$ When $y=5$, we get $f(a)f'(5)=f'(a+5)$, or $f(a) q=f'(a+5)$, or $$f(a)=\frac{f'(a+5)}{q} \tag{2}$$ From (1) and ...


2

You already know $f(0)=1.$ Also, $f(-y)=f(0-y)=1/f(y)$ is the inverse of $f(y).$ Hence, $f(x+y)=f(x-(-y))=f(x)f(y).$ This shows that $f$ is a homomorphism from $(\mathbb R,+,0)$ to $(\mathbb R^*,\times,1).$ Then we shall show that $f(q)=f(1)^q,\forall q\in\mathbb Q.$ First, for $q=0,$ we already know that $f(0)=1=f(1)^0.$ Then, for any $\alpha\in\mathbb ...


0

For sure, $f(0) = 1$. Indeed $$f(x-x) = \frac{f(x)}{f(x)} = 1.$$ Moreover: $$f(-x) = f(0-x) = \frac{f(0)}{f(x)} = \frac{1}{f(x)} \Rightarrow f(x)f(-x) = 1.$$ Using logarithms, we get: $$\log(f(x)f(-x)) = \log(f(x)) + \log(f(-x)) = \log(1) = 0.$$ This suggests that $f(x) = g(x)^{x}$, where $g(x)$ is a even function (i.e. $g(-x) = g(x)$). Indeed: ...


0

The key is that linear functions on vector spaces are one-to-one iff they only map $0$ to $0$, or said another way, iff the kernel contains exactly 0. For the forward direction, suppose $L:X\to Y$ is not one-to-one. Then $\exists$ $x \ne y$ such that $L(x) = L(y)$. By linearity, this implies $L(x) - L(y) = L(x - y) = 0.$ Therefore the kernel of $L$ ...


4

Dominated convergence and related theorems don't apply here anyway. The norm here is almost certainly the supremum norm since this is the norm that makes $C([0,1])$ a complete space. Also, you'll need to to know something about $K$ in order to do this. For example, if $K$ is continuous the following argument works: what you need to do is find a $C > 0$ ...


1

Any one-to-one quasinilpotent operator will do (quasinilpotent are the operators $T$ with $\sigma(T)=\{0\}$); the Volterra operator from user3808066's answer is one example. Any finite set $K\subset\mathbb C$ can be realized this way: if $K=\{k_1,\ldots,k_n\}$ and $T$ is the Volterra operator, then $$ \bigoplus_{j=1}^nT+k_jI $$ has spectrum ...


1

The key fact you want to use is that a Hilbert space is strictly convex. This means that if $x,y\in X$ are distinct unit vectors and $0<t<1$, then $\|tx+(1-t)y\|<1$ (note that this inequality holds with $\leq$ for any normed space, so the key thing here is that the inequality is strict). To apply this to the problem, suppose $T$ is an isometry and ...


0

For continuity to make sense here, it's not necessary to have a norm on $E \times E$; having a topology on $E \times E$ would suffice. The natural choice of topology on $E \times E$ would be the product topology. In this case, showing that $+: E\times E \to E$ is continuous at a point $(x,y)\in E\times E$ means to show that, for every open neighborhood $W$ ...


2

A metric space is said to be "complete" if every Cauchy sequence converges. For example: Let $(X, \mu)$ be a measure space. Then $L^P(X)$ is complete under the $L^P$ norm, for $p \in [1,\infty]$. [It is a Banach space.] Every finite dimensional normed vector space is also complete. (This this can be explained by the Lipschitz equivalence to the euclidian ...


0

First, I prove this where $f$ is continuous. If $f$ is continuous on $[0,1]$ then it is uniformly continuous on $[0,1]$ and is bounded. Let $\varepsilon>0$ be given. Then there is some $\delta_1>0$ such that for all $x_0\in [0,1]$, if $0<|x_0-x|<\delta_1$ then $|f(x_0)-f(x)|<\sqrt[p]{\varepsilon^p/2}$. Since $f$ is continuous on a closed ...


2

If $p<\infty$, I'd use the density of $C_c(0,1)$ in $L^p$, together with the fact that a continuous function with compact support is uniformly continuous.


0

Here is a simple solution (albeit I am partial to sledgehammer approaches): It is not too hard to show that the extreme points of the closed unit ball are $\pm e_k$. Since $\overline{B}(0,1)$ is convex and closed it follows that $\operatorname{\overline{co}} \{ \pm e_k \}_k \subset \overline{B}(0,1)$. Suppose $x \in \overline{B}(0,1)$, and let $x_n ...


0

Consider the space $\ell_1$ with the weak star topology. It is a locally convex topological wector space, and by Banach - Alaoglu theorem the unit ball is compact in this space. Hence by Krein - Milman theorem it is a closed convex envelope of its extreme points.


0

Since $A\subseteq A^*$ (because $A$ is symmetric), it's enough to prove that $D(A^*)\subseteq D(A)$. Take $y\in D(A^*)=D(A^*\pm\mathrm i)$. As $(A^*\pm\mathrm i)y\in\mathcal R(A^*\pm\mathrm i)\subseteq \mathcal{H}=\mathcal R(A\pm \mathrm i)$, there exists $x\in D(A\pm\mathrm i)$ such that $$(A^*\pm\mathrm i)y=(A\pm\mathrm i)x=(A^*\pm\mathrm i)x$$ and thus ...


1

Here is a solution, sorry that it is rather long, I'm sure a lot can be cut. Let $x_1,x_2 \in [y_1,y_2]$, then since $[x_1,x_2]$ is the convex span of $x_1$ and $x_2$ you have $[x_1,x_2] \subset [y_1,y_2]$. Similarly $y_1,y_2 \in [x_1,x_2]$ implies $[y_1,y_2] \subset [x_1,x_2]$. In the following $\alpha, \tilde \alpha, \beta, \gamma \in [0,1]$. Assume ...


1

To make Davide Giraudo's comment more clear, here are the steps: $$ \int |f(x)|^p = \int \left((1+x^2)|f(x)|\right)^p\frac{1}{(1+x^2)^p} $$ From here, since $(1+x^2)f(x)$ is bounded (because $f\in \mathcal{S}$), and $\frac{1}{(1+x^2)^p}\leq \frac{1}{1+x^2}\in L^1$ (for $p\geq 1$), one gets $$ \int |f(x)|^p \leq \|(1+x^2)f(x)\|^p_\infty \int ...


2

denote $X=||x||$ and $Y=||y||$, then your inequality becomes $$2^{2-\frac{2}{p}}(X^p+Y^p)^{\frac{2}{p}}\le 2(X^2+Y^2)$$ or equivalently $$\sqrt[p]{\frac{X^p+Y^p}{2}}\le\sqrt[2]{\frac{X^2+Y^2}{2}}$$ what is just an inequality between power means https://www.artofproblemsolving.com/wiki/index.php?title=Power_Mean_Inequality


1

For hermitian $T$ you have $\ker(T) \cap \rm{im}(T) = \{0\}$, as $\langle x,x \rangle = 0 \iff x=0$, so if $x = Tz$ and $x \in \ker(T)$ you have $\langle x, x\rangle=\langle x, Tz\rangle=\langle Tx,z\rangle = 0$. Now for a positive bounded operator $T$ on a Hilbert space you have a root $T^{1/2}$ that is also positive. Then $\langle x, Tx\rangle=\langle ...


2

What you did for the first part is correct. For the converse, however, there is a problem. You have to prove that the sequence $\left(M_n\right)_{n\geqslant 1}$ is bounded. Your $K$ depends on $X$, and the task is to prove that $K$ is finite. You can solve the question using the uniform boundedness principle: since $Tx$ belongs to $\ell^\infty$ for each ...


2

Hint: Say $b_1,\dots,b_n$ is a basis for $T(W)$. Then for every $w\in W$ there is a unique expansion $$Tw=\sum_{j=1}^n a_jw_j.$$Say $$a_j=\Lambda_jw\quad(w\in W).$$Since every linear functional on a finite-dimensional space is bounded, there exist $c$ so that $$|\Lambda_j w|\le c||Tw||.$$


1

This is functional in general. Let $X$ be a set an let $\mathcal V\subseteq\wp(X)$. Then the collection of finite intersections of sets in $\mathcal V$ automatically has the basic properties of a base of a topology. This in the understanding that the empty intersection equals $X$. Denoting the collection by $\mathcal{V}^{\stackrel{\cap}{f}}$ we have ...


2

We have: $$\left(\bigcup_{i \in I} A_i\right) \cap \left(\bigcup_{j \in J} B_j\right) = \bigcup_{(i,j) \in I \times J} (A_i \cap B_j)$$ so that the intersection of two unioned families is again a unioned family, over a bigger index set, and if the $A_i,B_j$ come from a family which is closed under finite intersections, the latter is also a union from that. ...


1

I think that a simpler proof comes from $$\left|f(x)\right|\leq\frac{K}{1+x^2}\quad \Longrightarrow\quad \int_{\mathbb{R}}\left|f(x)\right|^p\,dx \leq K^p\cdot\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(p-\frac{1}{2}\right)}{\Gamma(p)}.$$


1

Use the definition of weak convergence you have with $\chi:=f\mathbf 1\{|f|\lt R\}$ where $R$ is fixed. Then we have, by Cauchy-Schwarz inequality, $$\left(\mathbb E\left[f\cdot f\mathbf 1\left\{|f|\lt R\right\}\right]\right)^2=\lim_{T\to +\infty}\left(\mathbb E\left[f_T\cdot f\mathbf 1\left\{|f|\lt R\right\}\right]\right)^2\leqslant \liminf_{T\to ...


0

$\forall a \in H, \forall x,y \in H, \forall \lambda \in \mathbb{K}$, we have : $$<T(x+\lambda y)|T(a)>=<x+\lambda y|a>=<x|a>+\lambda <y|a>=<T(x)|T(a)>+\lambda <T(y)|T(a)>=<T(x)+\lambda T(y) | T(a)>$$ It is true for all $a$, so we have $T(x+\lambda y)=T(x)+\lambda T(y)$, so $T$ is linear. For the isometry : ...


0

Fix $x\in H $. Then $ y\longmapsto (x|y)_1$ is a linear functional. It is bounded, because $$ |\, (x|y)_1\,|\leq (x|x)_1^{1/2}\, (y|y)_1^{1/2}\leq (x|x)_1^{1/2}\, (y|y)^{1/2}. $$ By the Riesz representation theorem, there exists $x'\in H $ such that $$ (x|y)_1=(x'|y). $$ Now one shows that $x'$ is unique, so $x\longmapsto Tx=x'$ is well defined. The ...


0

The distributional derivative of the Dirac delta distribution is the distribution $\delta′$ defined on compactly supported smooth test functions $\varphi$ by $$ \delta'[\varphi] = -\delta[\varphi']=-\varphi'(0). $$ The first equality here is a kind of integration by parts, for if $\delta$ were a true function then $$ \int_{-\infty}^\infty ...


0

As one might aspect, this works by applying Hölder's inequality twice: Let $s=p$, $t=\frac{qr}{q+r}$. Then $\frac 1 s+\frac 1 t=1$, and Hölder's inequality yields $$ \|fgh\|_1\leq \|f\|_s\|gh\|_t=\|f\|_p\| |gh|^t \|_1^{\frac 1 t}. $$ Next let $\sigma=\frac q t=\frac{q+r}{r}$ and $\tau=\frac r t=\frac{q+r}{q}$. Then $\frac 1\sigma+\frac 1\tau=1$, and a ...


1

A counter example is the Volterra operator, it is a bounded linear operator between Hilbert spaces, with no eigenvalues while the spectrum is $\{0\}$. It is defined as $$ V:L^2(0,1)\to L^2(0,1), f\mapsto \left(t\mapsto \int_0^t f(x)\mathrm{d}x\right). $$


0

The dual unit ball is weak$^*$ compact by Alaoglu. If $S$ is a countable dense subset of $X$ then the topology on $X^*$ describing pointwise convergence on $S$ is metrizable (in particular, Hausdorff) and coarser that the weak$^*$ topology. But a compact space does not have strictly coarser Hausdorff topologies so that the dual unit ball with the weak$^*$ ...


0

For any Banach space $X$, if there is an exhaustion $X=\bigcup_n K_n$ by compact sets, then one of the $K_n$ has an interior point by Baire's theorem. This implies that $X$ is finite dimensional.


1

I think the answer is no. Each $K_i$ is separable, hence $\cup K_i$ is separable. But $C_b(\mathbb R^n)$ is nonseparable. To see the latter, think of a countable family of closed pairwise disjoint balls $B_m$ and continuous functions $f_m$ such that for every $m$: i)the support of $f_m$ is contained in $B_m;$ ii) $0\le f_m\le 1;$ iii) $f_m = 1$ somewhere in ...


1

First a few comments about your statement: 1) I assume you are thinking in a certain $f$ which is a continuous linear functional (otherwise the argument does not say anything). 2) What do you mean by "a basis element of the the vector space". If you mean any element $e$ such that $\{e\}$ can be extended to a basis of the space, then $e$ can be any non $0$ ...



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