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2

You can always define an unbounded operator on the whole space $X$, as long as $X$ is infinite dimensional. Simply take any unbounded linear functional $\varphi : X \to \Bbb{K}$ (with $\Bbb{K} \in \{\Bbb{R}, \Bbb{C}\}$) and some $x_0 \in X \setminus \{0\}$ and define $T : X \to X, x \mapsto \varphi(x) \cdot x_0$. For the existence of $\varphi$, see On ...


0

First, my version of your question: Take $f$ a function, say continuous on $\mathbb{R}$, analytic at $x=0$. You want to show that there exists a sequence $f_n$ of functions, say $\in C^3(\mathbb{R})$, such that for all fixed $x\in \mathbb{R}$, we have $f_n(x)\to f(x)$, and that there exists $M>0$, such that for all $n$ and $x\in \mathbb{R}$, we have ...


0

As an example, suppose that $F : [0,1]\rightarrow X$, where $X$ is a Hilbert space. Suppose that $\|F(t)\|$ is a Lebesgue integrable function. If $(F(t),x)$ is measurable for all $x$, then there is a unique vector--say $\int_{a}^{b}F\,dt$--such that $$ \int_{0}^{1}(F(t),x)\,dt = \left(\int_{0}^{1}F\,dt,x\right),\;\;\; x \in X. $$ This is a very ...


0

In QM a wave $\psi$ strangely gives the full information of particle and $|\psi|^2$ represents probability for finding that particle (just like $|I|^2$ is probability in EM-theory). When total probability is unity, $||\psi||_2=1$, we have the first connection to $L^2$ functions. Since QM is fundamentally probabilistic, we can only deal with expectations ...


0

Another case from physics is the set of Maxwell's equations in vacuum. There the energy \begin{equation*} \mathcal{E}(t)=\frac{1}{2}\int_{\mathbb{R}^{3}}d\mathbf{x}\{\mathbf{E(x},t% \mathbf{)}^{2}+\mathbf{B(x},t\mathbf{)}^{2}\} \end{equation*} is conserved, i.e. not $t$-dependent, which is easily obtained for smooth field functions $\mathbf{E(x},t\mathbf{)}$ ...


0

First, observe that it suffices to show $$ \sum_{i=1}^n w_ia_i \ge \left( \sum_{i=1}^n a_i^p\right)^{1/p} $$ for all sequences satisfying your conditions and $a_1=1$. Now let $(w_i)$ and $(a_i)$ be sequences satisfying the non-negativity and monotonicity properties. Let $n\ge 1$. Define $$ f_n (a_n) = \left(\sum_{i=1}^n w_ia_i\right)^p - \sum_{i=1}^n ...


2

(Unfortunately I don't remember much of the details, and I don't have time to look them up now, so this will be very sketchy, but perhaps it will be of some use until someone else gives a better answer...) There are indeed restrictions that $\mu$ must satisfy in order for the space of polynomials to be dense in $L^2(\mu)$. It has to do with limit ...


2

The definition of $L^p$ functions is motivated by their behavior under the operation of integration: Strictly speaking, the elements of $L^p$ are not functions, but equivalence classes of functions, where two functions are equivalent iff they disagree at most on a set of measure zero---in other words, the equivalence classes keep track only of the functions' ...


3

There are several reasons, you named some. You need to work with the space to get a feeling why these things are important. One reason you did not mention which makes them very popular is the fact that they are very well suited to study elliptic partial differential equations, esp of second order. By partial integration, assuming zero boundary conditions, ...


0

Strictly following the definition of $f$, one can note that $f$ is a function of one variable (which is $n$-dimensional vector), therefore you cannot write $f(x_1,\dots,x_n)$, since $f$ would be a function of $n$ variables (which are $1$-dimensional vectors, i. e. numbers) in that case. However, many would agree that you can write in both ways, especially ...


1

Start with a sequence in $c_0$, our goal is to show that its limit $\omega\in\ell^\infty$ is also in $c_0$. This limit exists by completeness of $\ell^\infty$ Incorrect. To prove that a set is closed, we start with a convergent sequence in that set, and show that the limit is contained in the set. The existence of limit is assumed, not obtained ...


1

Hints. We first observe that it suffices to consider step functions, as they are dense in $L^2[0,1]$. In particular, step functions $f$ for which there is an $N\in\mathbb N$, such that $f$ restricted in $[(j-1)/N,j/N]$ is constant, for every $j=1,\ldots,N$. Next, observe that, for every $j,N\in\mathbb N$, with $0<1\le N$, $$ ...


1

The convolution $\mu\ast\nu$ itself may be absolutely continuous... Here is an example. Let $(X_n)$ be i.i.d. with uniform distribution on $\{0,1\}$, thus, $P(X_n=0)P(X_n=1)=\frac12$ for every $n$. Consider $X=\sum\limits_{n\geqslant1}4^{-n}X_n$ and define $\mu$ as the distribution of $X$ and $\nu$ as the distribution of $2X$. Then $\mu$ is purely ...


1

The answer is yes and there is lots of literature on this issue (google "convolution of singular measures"). The earliest reference I could dig out is by Wiener and Wintner [1]. They gave a construction of a singular continuous measure $\mu$ such that $\mu*\mu$ is absolutely continuous. More precisely, for arbitrary $\epsilon>0$, they construct a ...


1

for (ii) $f(x) = \int \phi(x-y)\chi(y)dy$. Now if $x \in V$ then the integrand is non-zero if $\phi(x-y)$ is nonzero that is if $|x-y| < \delta/2$. Now the last condition holds if $y \in U$. Hence we can write $f(x) = \int_U \phi(x-y)\chi(y)dy = \int_U \phi(x-y)dy = \int \phi(x-y)dy = 1$. The second last inequality holds since the support of the ...


0

Substitution works for this example. For some $c>12$, suppose inductively that $T(n) \le \frac{cn}{\log n}$ for sufficiently large $n$. Then we have: \begin{align} T(n) &= T\left(\frac{n}{2}\right)+T\left(\frac{n}{4}\right)+T\left(\frac{n}{6}\right)+\frac{n}{\log n} & \\ & \le ...


1

Yes, the argument is correct. It rests on the fact that for an integrable function $g$, we have $$|\widehat g(\xi)|\leqslant \lVert g\rVert_1$$ due to the triangle inequality. Hence $$\sup_\xi|\widehat{f_n}(\xi)-\widehat f(\xi)|\leqslant \lVert f_n-f\rVert_1$$ and we conclude using convergence in $\mathbb L^1$.


0

No, $x^{(k)}$ is assumed to be a sequence of points of $c_0$. Here $k$ is the index. So $x^{(1)}$ is a sequence (namely a point in $c_0$), which has itself indices, which are denoted by $n$. So we write $x^{(1)} = (x^{(1)}_0, x^{(1)}_1,x^{(1)}_2,\ldots)$ .So every $x^{(k)}_n$ is indeed some real number: the $n$-th member of the sequence that is the point ...


3

Take an $x\in H\setminus \{0\}$. You need to show that $\langle x,Tx\rangle > 0$. Can you see that a certain projection with one-dimensional range could be helpful?


1

For (b). Since $\lambda_n \to +\infty$, we have $\frac{1}{\lambda_n} \to 0$. Hence, from (A.4) and due to the fact that $$ \frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n} \geq 0, $$ we get $$ 0 \leq \int_{\Omega}\left|\nabla w_n\right|^2 \leq \int_{\Omega}\left|\nabla ...


0

Since $e_1,\ldots,e_n$ are fixed, the vector $c^n$ is simply the vector of the components of $u_n$ in the basis $\{e_1,\ldots,e_n\}$. Thus, considering the vector $u_n$ or $c^n$ is completely equivalent.


0

In the first part, you showed that if $\{ f_{n} \}$ is a bounded sequence in $L^{1}_{\mu}$, then $\langle Kf_{n} | g \rangle$ has a convergent subsequence for a $g \in L^{\infty}_{\mu}$, where $\langle\cdot|\cdot\rangle$ denotes the duality pairing. You forgot that the subsequence $\langle Kf_{n} | k_{x}\rangle$ may be different for each $x$ in the second ...


1

First, if $x\in X$, then $$ \|\pi(x)\|_{X/Y}=\|x+Y\|_{X/Y}=\inf_{y\in Y}\|x+y\|_X\le \|x\|_X, $$ and hence $\pi$ is bounded. Next, as $$ \pi\big(B_X(0,1)\big)=\left\{x+Y:x\in B_X(0,1)\right\}, $$ let $\hat z\in B_{X/Y}(0,1)$. Then $\hat z=z+Y$, with $$ \|\hat z\|=\|\pi(z)\|_{X/Y}=\|z+Y\|_{X/Y}=\inf_{y\in Y}\|z+y\|_X<1, $$ and hence there exists a $y\in ...


1

Let $$ h(x)=\left\{\begin{array}{lll}\mathrm{e}^{-1/x^2} & \text{if} & x>0,\\ 0 & \text{if} & x\le 0.\end{array}\right. $$ Then $h\in C^\infty(\mathbb R)$. Then set $$ j(\boldsymbol{x})=c\,h\big(1-\|\boldsymbol{x}\|^2\big), $$ where $\boldsymbol{x}\in\mathbb R^n$, and $c>0$, so that $\int_{\mathbb ...


1

Rewrite $f_{\epsilon}(x)$ by substituting $y=y'-\frac{1}{\epsilon}x$: $$ \begin{align} f_{\epsilon}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y') e^{-(y'-\frac{1}{\epsilon}x)^{2}/2}\,dy' \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y')e^{-(\epsilon y'-x)^{2}/2\epsilon^{2}}\,dy' \end{align} ...


1

By the uniform boundedness principle, the domain of such a family cannot be a Banach space. A fruitful source for "counterexamples" to things that are true in Banach spaces are the $$\ell^p_{00} = \left\{ x\in \ell^p : \operatorname{card} \{k\in\mathbb{N} : x_k \neq 0 \} < \infty\right\}$$ spaces of sequences with finite support in the various ...


0

I would ordinarily leave a comment in this case, because I don't have any real answers. But it's too long to leave in a comment. For ODEs $-ay''+by'+cy=\lambda w y$, the common techniques are Multiply on the left by $\rho$ to get $\rho(ay''+by')= ((\rho a) y')'$ which (essentially) uniquely determines $\rho$. This requires minimal smoothness of ...


0

I think you are a bit confused by the notation. Proving that $c$ is a Banach space: Let $\{ x^n \}_n \subset c$ a Cauchy sequence. Every $x^n$ is a converging sequence in $\mathbb{R}$. We denote its elements by $\{ x^n_k \}_k$. This means that $$x^1 = x^1_1, x^1_2, x^1_3, x^1_4, x^1_5, \dots $$ $$x^2 = x^2_1, x^2_2, x^2_3, x^2_4, x^2_5, \dots $$ $$x^3 = ...


0

You apply that dimension is $\le 6$. That is not neccesarily the case, but thank you anyways. After some more work the proof goes as follows. Let $\Omega$ be bounded. The image $F(v)$ is the functional $$ F(v) : w \mapsto \int g(v) w $$ which is actually an $L^2$ functional. I have $$ |\int (g(v_1) - g(v_2)) w | \le \int |g'(\xi)| |v_1 - v_2| |w| \le ...


0

Here's a more elaborate example to give you a bounded group that is not $C_{0}$. To do this, I'll define a function $F : \mathbb{R}\rightarrow\mathbb{R}$ such that $F(x+y)=F(x)+F(y)$. Then $E(x)=e^{iF(x)}$ satisfies $E(x+y)=e^{iF(x+y)}=e^{iF(x)}e^{iF(y)}=E(x)E(y)$. Automatically $F(0)=0$ because $F(0)=F(0+0)=F(0)+F(0)$, which gives $E(0)=1$. I'll describe ...


2

If $G$ is bounded, then for $w\in H^1(\Omega)$, $$ |\langle g(v),w\rangle| \le \int_\Omega |g(v)|\cdot|w| dx \le M \|w\|_{L^1(\Omega)}, $$ and you are done. As you can see, there is a lot of room to relax the assumptions on $g$. In fact, one can allow a certain growth of $g(x)$ for $|x|\to\infty$. Then $g$ is well-defined. Since $g \in W^{2,\infty}$ it is ...


0

In order that $$\int_{1}^{+\infty}\frac{dx}{(\sqrt{x}\,(1+\log x))^p}$$ is finite we must have $p\geq 2$. On the other hand, $$\int_{0}^{1}\frac{dx}{(\sqrt{x}\,(1-\log x))^p}=\int_{0}^{+\infty}\frac{e^{(\frac{p}{2}-1)t}\,dt}{(1+t)^p}$$ is convergent only if $p\leq 2$.


1

Use polar coordinates: $x=z+\rho e^{i\theta}$, $$ \int_{B(z,r)}(x - z)^{n}dx = 2\pi \int_0^r \int_0^{2\pi}\rho^n e^{in\theta}\,\rho\,d\theta\,d\rho $$ Here $\int_0^{2\pi} e^{in\theta}\,d\theta=0$ when $n\ne 0$, because an antiderivative of $e^{in\theta}$ is $\frac{1}{in}e^{in\theta}$, which is $2\pi$-periodic.


1

Let $P_{M}$ be the orthogonal projection of $H$ onto $M$. Define $\tilde{\lambda}(x)=\lambda(P_{M}x)$. Then $\tilde{\lambda}$ is a linear functional satisfying $$ |\tilde{\lambda}(x)|= |\lambda(P_{M}x)|\le \|\lambda\|\|P_{M}x\|\le \|\lambda\|\|P_{M}\|\|x\| \le \|\lambda\|\|x\|. $$ So $\|\tilde{\lambda}\|\le\|\lambda\|$. The reverse inequality follows ...


1

Let $P_{k}$ be the orthogonal projection onto the closure of the range of $T_{k}$. Then $P_{k}P_{k'}=P_{k'}P_{k}=0$ for $k\ne k'$ because $(T_{k}x,T_{k'}y)=(T_{k'}^{\star}T_{k}x,y)=0$ for all $x,y \in H$. Similarly, if $Q_{k}$ is the orthogonal projection onto the closure of the range of $T_{k}^{\star}$, then $Q_{k}Q_{k'}=0$ for $k\ne k'$. Furthermore, $$ ...


1

First observation: For all $u,v\in\mathcal H$ and $j\ne k$ $$ \langle T_ju,T_k v\rangle=\langle T_j^*u,T_k^* v\rangle=0. $$ Thus the linear subspaces $T_j^*\mathcal H$, $j\in\mathbb Z$, are perpendicular to each other, and so are their closures. Set $$ Y=\bigoplus_{j\in\mathbb Z}\overline{T_j^*\mathcal H}. $$ This infinite direct sum contains elements of ...


1

A linear projection $P$ onto a subspace $\mathcal{M}$ has the properties that (a) the range of $P$ is $\mathcal{M}$, (b) projecting twice is the same as projecting once: $P^{2}=P$. Orthogonal projection is something peculiar to an inner product space, and it is the same as closest point projection for a subspace. There can be many projections onto a ...


0

For any vector $x$, we can write $$ x = x^{||} + x^\perp $$ Where $x^{||} \in M$ is given by $x^{||} = Px$ (as defined by the sum) and $x^{\perp} \in M^{\perp}$. (In order to verify that $x^{\perp} = x - Px$ is in $M^\perp$, it suffices to note that for each $e_n$, $\langle x - Px,e_n \rangle = 0$.) Note that for any $y \in M$, we have $$ \|x- y\|^2 = ...


0

My intuitive understanding of this is that we define a neighborhood as a set of functions that satisfy some property $P.$ Now if we are fortunate, all the functions in this set also satisfy a certain property $Q,$ which allows us to identify a particular function as a relative extremum. If we achieve this happy state, then if we place a stronger condition ...


3

Yes, let $X=L^{2}[0,1]$, and let $(Tx)(t)=tx(t)$, i.e., multiplication by $t$. This is the prototypical example in many regards. You don't have any eigenfunctions because $Tx=\lambda x$ would require $(t-\lambda)x(t)=0$ for a.e. $t\in[0,1]$ which means that $x=0$ a.e.. However, you have approximate eigenfunctions. For example, if $\lambda\in(0,1)$ and ...


1

You can see it directly using the orthogonal decomposition $$ X = \mathcal{N}(T)\oplus\mathcal{N}(T)^{\perp}. $$ If $Tx=y$, then $x=x'+x''$ for unique $x'\in\mathcal{N}(T)$, $x''\in\mathcal{N}(T)^{\perp}$. And $Tx=Tx''$. If $Tw=y$ also holds then $w''=x''$ because $w''-x''\in\mathcal{N}(T)^{\perp}\bigcap\mathcal{N}(T)$. So, with a small ...


0

Sorry, but I don't have the rights to add a comment. Can you explain the line $$ C_\pm(w)(A\pm i\mu I)=(A\pm i\mu I) +w B $$ I cannot see that this is true.


0

Think about $L^p(\Omega)$ first. The closure of $C_0^\infty(\Omega)$ in $L^p(\Omega)$ is all of $L^p(\Omega)$. That is, the vanishing at $\partial \Omega$ of functions in $C_0^\infty(\Omega)$ in no way affects the boundary behavior of functions in the closure. Now look at $W^{1,p}(\Omega)$. The closure of $C_0^\infty(\Omega)$ in $W^{1,p}(\Omega)$ is ...


3

Suppose that $x$ and $y$ are in $B(z,r)$. Then, for $\theta\in[0,1]$, $$ |\theta x+(1-\theta)y-z|=|\theta(x-z)+(1-\theta)(y-z)|\\ \leq\theta|x-z|+(1-\theta)|y-z|<\theta r+(1-\theta)r=r $$ so that $\theta x+(1-\theta)y$ is also in $B(z,r)$. The case for the closed ball is similar (the only difference is that the last inequality above will be weak.)


0

Let $H = K = \ell_2$. Define a function $\Phi$ by $\Phi( x,y ) = \tfrac{1}{2}\|x\|\|y\|I_{H\otimes K}$. If there were an extenstion of $\Phi$ to a bounded linear operator $\ell_2(\mathbb{N}\times \mathbb{N})\to B(H\otimes K)$ (still dented by $\Phi$), it would map the weakly null sequence $(e_n\otimes e_n)_{n=1}^\infty$ in the domain to a weakly null ...


2

Let $V$ be the span of $\phi_1,\dots,\phi_n$, and let $V^\perp$ be its orthogonal complement. Introduce new inner product on $V$ so that $\langle \phi_i,\phi_j \rangle_V = \delta_{ij}$ (this formula defines it for basis vectors; extend by linearity). Every vector $u$ in $H$ is the sum of vectors $u_1\in V$ and $u_2\in V^\perp$. Define $$\langle ...


0

An idea: maybe you can show that if $x_1M \subset M$ then $(x_1M \cap H^2) \subset (M\cap H^2)$ (+ remaining a CLOSED subspace, which I have no idea how to show). Once there, you could apply Beurling's Thm and you know that $M \cap H^2$ is of the form $\phi H^2$. By density of $H^2$ as a subspace of $H^1$, you can probably show that the space $M$ is also of ...


1

HINT: Recall that $\ell_\infty$ is a subset of $\Bbb{R^N}$, the set of all sequences. Now use cardinal arithmetics.


-1

HINT: Let $X$ an infinite dimensional real vector space and let $(e_i)_{i\in I}$ be a basis of $X$ as a vector space. Let $w\colon I \to (0, \infty)$, $i \mapsto w_i$ a function (the weights) such that both $(w_i)_{i\in I}$ and $(1/w_i)_{i \in I}$ are not bounded from above. This is possible since $I$ is infinite. Consider the norms $||\cdot ||_1$ , ...


0

Extend $f$ to the whole real real line, so that $f(x)=f(a)$, for $x<a$ and $f(x)=f(b)$ for $x>b$. Next set $$ f_\varepsilon(x)=\frac{1}{2\varepsilon}\int_{-\varepsilon}^\varepsilon f(x+t)\,dt. $$ Then $f_\varepsilon\to f$ uniformly in $[a,b]$, as $\varepsilon\to 0$, as $$ \lvert\, ...



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