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2

Completeness is needed. To see this, for any inner product space $X$, let $\overline X$ be its completion. Then let $z \in \overline X \setminus X$ and define a bounded linear functional on $X$ by $$f (x) = \langle x, z\rangle$$ for all $x\in X$. This functional cannot be written as $f(x) = \langle x, y\rangle$ for some $y \in X$. Assume that it can be ...


1

If $g$ is not one-to-one, then $\Phi_g$ is not onto (since $\Phi_g(f)$ satisfies $\Phi_g(f)(s) = \Phi_g(f)(t)$ whenever $g(s) = g(t)$), so $0 \in \sigma$. $-1$ is an eigenvalue if $g$ is an involution that is not the identity. Indeed, take any $s \in [0,1]$ such that $g(s) \ne s$, let $h$ be a continuous function on $[0,1]$ with $h(s) \ne h(g(s))$, and ...


1

That's not true in general. The general theory says that if $A \subset B$ is dense in the Banach space $B$, then a linear functional $\phi$ on $A$ can be extended to $B$ if and only if there is $C$ so that $$ |\phi (f)| \le C ||f||_B$$ for all $f\in A$. For example, fix $x\in \Omega$. Then the functional $\phi_x (f) = f(x)$ defined on ...


1

First of all, the operator is unbounded if $p\geq 3$. From Minkowski's integral inequality, we see that $$||A(x)(t)||_{L^p(dt)} = \lvert\lvert\int_{-1}^1 \dfrac{x(s)}{(s-t)^{1/3}} ds \rvert\rvert_{L^p(dt)} \leq \int_{-1}^1 \lvert\lvert \dfrac{x(s)}{(s-t)^{1/3}} \rvert\rvert_{L^p(dt)} ds = \int_{-1}^1 \lvert x(s)\rvert \lvert\lvert (s-t)^{-1/3} ...


1

With a little use of Functional Analysis. First observe that $T(\,f)$ is a closed subspace of $C(\mathbb R)$, equipped with the topology of locally uniform convergence, as it is finite-dimensional. Let $j(x)$ be a non-negative $C^\infty$ function supported in $\big[-\frac{1}{2},\frac{1}{2}\big]$, with $\int_{-1/2}^{1/2}j(x)\,dx=1$. Then the convolution ...


0

Let $$ f_a(x)=\frac{1}{\sqrt{|x|}} \chi_{[-a,a]} $$ Observe that $f_a \in L^1$ but not in $L^2$. Now just compute the fourier transform and play with $a$. \begin{align} \hat{f_a}(y) &= \int_{-a}^a \frac{1}{\sqrt{|x|}}e^{-iyx} \, dx \\ &=\int_0^a\frac{1}{\sqrt{x}}e^{-iyx}+\frac{1}{\sqrt{x}}e^{iyx} \,dx \\ & =2 \int_0^{\sqrt{a}}e^{-iys^2} + ...


5

The only thing you can say that $fg$ has the same smoothness as $g$. That is, if $g$ is in $C^k$, then $fg$ is in $C^k$ as well. If $f$ in addition has compact support, then $g\in C^k$ implies that $fg$ is in $C^k$ with uniformly continuous derivatives up to order $k$. Similarly, if $g$ is in some Sobolev space $W^{m,p}$, then $fg\in W^{m,p}$ as well.


5

I don't think you can say anything without further assumption on $g$. The constant function $f=1$ is $\mathcal{C}^\infty$, and $fg=g$ can be arbitrary.


-1

My favorite one is Functional Analysis, Sobolev Spaces and Partial Differential Equations by Haim Brezis. Here is a link to a PDF.


0

Introductory Functional Analysis by Erwin Kreyszig This is the best book (in my opinion) for functional analysis, as it covers all the good stuff from metric, Banach and Hilbert spaces, to a good thorough introduction of spectral theory and operators (linear and self-adjoint). ...


0

The function's amplitude on $(a,b)$ is bounded by $m:= f(b)-f(a) \lt \infty$. Look at the right-side limit at a point of discontinuity $c$ (existence of the left-side limit follows in the same way). Fix $m \gt \epsilon \gt 0$. Pick a point $x_1$ in the interval to the right of $c$. If $y_1 := f(x_1) - f(c) \lt \epsilon$, you are done (monotonicity). If ...


1

Note that $f(x) \le f(c)$ for $a < x < c$. Therefore the following exists $$ \sup_{x < c} f(x) $$ and you can show that the following limit exists, too: $$ \lim_{x\uparrow c}f(x) = \sup_{x < c} f(x). $$


1

After fixing the definition of $\gamma$ (presumably a typo) to $\gamma(t)= -3e^{2i\pi t}$, there is one slip-up in the computation of the residue at $-i$ (which could also be just a typo): $$\operatorname{Res}_{-i} f = \frac{-ie^{-i\pi}}{-2i} = \frac{1}{2} e^{-i\pi}.$$ As it so happens, we have $e^{i\pi} = e^{-i\pi}$, but still the correct thing is to ...


1

As PhoemueX said: the statement follows from the Uniform Boundedness Principle applied to linear maps $\phi_n:X^*\to \mathbb{K}$ that are defined by $\phi_n(f)=f(x_n)$. ($\mathbb{K}$ stands for $\mathbb{R}$ or $\mathbb{C}$.) The principle applies here because $X^*$ is complete even when $X$ isn't.


0

(See my comment under Jochen's answer, which led to my writing this answer as a clarification. The first part is intended to be viewed in the context of Folland's presentation.) Folland's version of the Open Mapping Theorem is: Let $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces. If $T \in L(\mathcal{X}, \mathcal{Y})$ is surjective, then $T$ is ...


1

You don't need Zorn's lemma. The function $f$ can be defined explicitly as $$ f_1\chi_{\{|x|\le 1\}}+\sum_{n=2}^\infty f_{n} \chi_{\{n-1<|x|\le n\}} $$ which is evidently in $L^1_{\rm loc}$ (on every bounded interval, only finitely many terms are nonzero). The convergence of $f_n$ to $f$ on every bounded interval follows from the fact that the restriction ...


3

Suppose that $T$ is bounded. Then $$ \|x\|^{2}+\|Tx\|^{2}=((I+T^{\star}T)x,x) \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\|^{2} \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\| \le \|(I+T^{\star}T)x\|. $$ The last inequality can be use to show that the range of $I+T^{\star}T$ is closed, and $I+T^{\star}T$ has a bounded inverse on ...


1

Since $T^*T$ is Hermitian, the eigenvalues $\lambda_i$ are real. Suppose $v_i$ is an eigenvector corresponding to $\lambda_i$. Then \begin{eqnarray} \langle \lambda_i v_i, v_i\rangle =\langle T^*Tv_i, v_i\rangle= \|Tv_i\|^2\geq 0 \end{eqnarray} The eigenvalues $\lambda_i$ of $T^*T$ are real and nonnegative, and the eigenvalues of $I+T^*T$ are $1+\lambda_i$, ...


2

Let $f_1, \cdots f_n$ be basis of $T(f)$, where $f_i := f_{a_i}$ for simplicity. For any $h \in \mathbb R$, we have $$(*)\ \ f_i(x+ h) = \sum_j A(h)_{ij} f_j(x) $$ Note that we have $A(h+ g) = A(h) A(g)$ (matrix multiplication). Thus $A(h)$ is invertible and $A :\mathbb R \to GL(n, \mathbb R)$, $h\mapsto A(h)$ is a one parameter group in $GL(n, \mathbb ...


1

The claim is false in the general infinite dimensional setting. For example, let $X$ be the Banach space $\ell^p$, and define $A,B$ by$$A(x_1,x_2,\ldots)=(x_2,x_3,\ldots),\qquad B(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots).$$Then $AB$ is the identity, but $A$ and $B$ are not invertible.


0

For the question in the title, the answer is No. Take $B : \ell^2 \to \ell^2$ be the right-shift operator $$ (x_n) \mapsto (0,x_1,x_2,\ldots) $$ and $A$ to be the left shift $$ (x_n) \mapsto (x_2,x_3,\ldots) $$ Then $AB$ is the identity operator, and so invertible, but neither $A$ nor $B$ are invertible. But later on, you ask about $B^{-1} A^{-1}$ implying ...


3

Not necessarily. For example, define on $\ell^p$ $$ L(x_1,x_2,\dots) = (x_2,x_3,\dots)\\ R(x_1,x_2,\dots) = (0,x_1,x_2,\dots) $$ Then $LR = \text{id}_{\ell^p}$ is invertible, but neither $L$ nor $R$ are bijective. You might be able to say something in the case that both $AB$ and $BA$ are invertible.


0

Notice that the set $Y$ in $\mathbb R^2$ is actually the set of lines $y= \frac{1}{n}$ ($n \in \mathbb N$). Now, as $n$ becomes bigger and bigger these lines come closer can closer to the $x$-axis. Thus the answers are: Interior: Given any point $y$ in $Y$, you can't find a square around $y$ that is completely contained in $Y$. Thus, the interior is empty. ...


0

First note, that on $\Bbb R^2$ all the norms are equivalent. So instead of using this given norm, one can also consider the natural euclidean norm and get to the same results. This is just a detail, you can solve this problem without knowing this and using the given norm. Now, assume that $\left( x, \frac 1 n \right) \in Y$ is an interior point of $Y$. Then ...


0

Note that since $$ \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} < \infty $$ by Cauchy-Schwartz, the linear functional $$ \Lambda: (x_n) \mapsto \sum_{n=0}^{\infty} \frac{x_n}{n+1} $$ is a continuous map from $\ell^2 \to \mathbb{C}$ and so $\ker(\Lambda)$ is closed in $\ell^2$, whence $\ker(\Lambda)\cap \ell_0$ is closed in $\ell_0$


2

Consider $\mathbf{x} = (1,0,0,0,\ldots)$ and $\mathbf{y} = (0,1,0,0,\ldots)$. Then $\mathbf{x+y} = (1,1,0,0,\ldots)$ and $\mathbf{x-y} = (1,-1,0,0,\ldots)$. $$\|\mathbf{x}+\mathbf{y}\|_p^2 + \|\mathbf{x}-\mathbf{y}\|_p^2 = 2\times(1+1)^\frac{2}{p} = 2^{1+\frac{2}{p}}$$ Meanwhile, $$2(\|\mathbf{x}\|_p^2+\|\mathbf{y}\|_p^2) = 2\times(1+1) = 4$$ If the ...


0

For number one, $<0x,y>=<0,y>=0=<x,0>=<x,0y>$. So $0^*=0$.


0

The fact that there is an $\epsilon>0$ such that $B_{\epsilon}(x)\subset \overline{B_r(a)}^c$ does not imply $||x-a||\geq r+\epsilon>r$. This implication is not justified by the open-ness of $\overline{B_r(a)}^c$. You want this to say that $x$ is at least $\epsilon$ distance away from $\partial \overline{B_r(a)}$. This is the intuitive idea from ...


1

If $f_n$ converges converges uniformly on each of finitely many sets, then it converges uniformly on the union of these sets. The proof is of the "let $N=\max (N_1, \dots, N_k)$" variety. So yes to your first question. (This doesn't really have much to do with continuity.) No, on the second question: Let $D = [0,1/2] \cup \{2/3, 3/4, 4/5, \dots \}.$ Define ...


2

It's not a lucky guess. It's at the heart of Stein's method: you need a characterizing equation for your distribution. There isn't a unique equation, but in fact many and depending on the situation, one might use other characterizing equations. In the case of the normal distribution, this is actually simple integration by parts. Below, all expectations are ...


0

Let $(a_n) \in M$. The interior is empty. For instance, let $\epsilon > 0$ and take $b_n = a_n$ for all $n > 1$, and $b_1 = a_1 + \epsilon$. Now, $\lVert (a_n) - (b_n) \rVert = \epsilon$, and yet $(b_n) \notin M$. The closure is $\{(a_n) \mid 0 \leq a_n \leq 1, a_1 = 0\}$. To show this, once again let $0 < \epsilon < 1$ and take $(b_n) \in ...


0

Define projection $P_n$ on $\ell^2(\Bbb N)$ such that $$P_n(x_1,x_2,...)=(0,...,0,x_n,x_{n+1},...)$$ for $x=(x_1,x_2,...)\in\ell^2(\Bbb N)$. We show that $P_n\to 0$ (sot). For each $x\in \ell^2(\Bbb N)$, $\epsilon>0 $ there is $n_0>0$ such that $\sum_{n>n_0}|x_n|^2 < \epsilon$. This shows that $\|P_nx\|<\epsilon$ for $n>n_0$.


0

A linear map $f$ is continuous, and so is the function $y\mapsto\|y\|$. Therefore the real-valued function $\phi(x):=\|f(x)\|$ is bounded on the compact set $S^{d-1}\subset{\mathbb R}^d$ (the unit sphere): There is an $a>0$ such that $\bigl|\phi(x)\bigr|\leq a$ for all $x\in S^{d-1}$. From the linearity of $f$ it then follows that $\|f(x)\|\leq a\|x\|$ ...


1

Your guess is correct. The inequality can be proved for any Gateaux-differentiable $F$. This is clear, since you are proving a 1D fact, namely an estimate on the line joining $x$ and $x+h$. The proof is easy in the special case $F \colon X \to \mathbb{R}$: just consider $t \mapsto F(tx+(1-t)(x+h))$ and apply the MVT. If $F$ is vector-valued, then you need a ...


0

When working with linear maps, one frequently omits brackets, for brevity. This shouldn't surprise you, after all we also write $\log x$ or $\sin x$ instead of $\log(x)$ and $\sin(x)$.


1

When $\Gamma$ is uncountable, $c_0(\Gamma)$ is not separable. It consists of all functions $g : \Gamma \to \mathbb R$ (with countable support) such that for every $\epsilon>0$, the set $\{\gamma \in \Gamma : |g(\gamma)|>\epsilon\}$ is finite. But (you may ask) even if this particular pre-dual is nonseparable, might there not be some other pre-dual ...


1

We have $$ c_0(\Gamma) := \{x \in \mathbf K^\Gamma \mid \forall \epsilon > 0 \,\exists \Gamma' \subseteq \Gamma, |\Gamma'| < \infty \land \sup_{\gamma \in \Gamma - \Gamma'} |x(\gamma)| < \epsilon \} $$ which is a Banach space with respect to the norm $$ \def\norm#1{\left\|#1\right\|}\norm{x}_\infty := \sup_{\gamma \in \Gamma} ...


3

The operator is continuous from $W^{2,2}(\mathbb R)$ to $L^2(\mathbb R)$, hence closed. It is closed as operator from $D(A)\subset L^2(\mathbb R)$ to $L^2(\mathbb R)$: The key to proceed with your proof is the inequality $$ \|u'\|_{L^2(\mathbb R)}^2\le \|u\|_{L^2(\mathbb R)} \|u''\|_{L^2(\mathbb R)} . $$ This inequality can be proven for smooth functions ...


5

In some sense the norm you are using is the graph norm of the Laplacian, so the operator is bounded, since $$||-\Delta u||_{L^2} \le ||u||_{2,2} $$ and since continuity implies closedness, you are done.


1

No, because already for $n=1$ we have that the Sobolev space $H_0^2$ is a subset of your space.


1

If $M$ is not total in $H$, then $M^\perp\neq\left\{0\right\}$. Let $v\in M^\perp$, $v\neq 0$. Then $v$ and $-v$ are distinct elements of $H$ but for all $x\in M$, $\langle v,x\rangle=0=\langle-v,x\rangle$. This solves (the contrapositive of) the exercise.


3

Let $x = \{x_k\}$ denote an arbitrary sequence in $\ell^p$. For $j \in \Bbb N$, let $x^{(j)} = \{x^{(j)}_k\}$ denote the sequence given by $$ x^{(j)}_k = \begin{cases} x_k & k \leq j\\ 0 & k > j \end{cases} $$ Note in particular that $x^{(j)} \in V$ for all $j$. Claim: In the space $\ell^p(\Bbb N)$, $x^{(j)} \to x$ as $j \to \infty$. Proof: We ...


0

Which kind of convergence are we talking about. If $y=(y_1,\ldots,y_d)$ then it doesn't matter, but if $y=(y_1,y_2,y_3,\ldots)$ with infinitely many components, then it does. What does $x_n \to x$ mean? One characterization is $\|x_n-x\|\to 0$, i.e. convergence of a sequence of vectors is defined by convergence of a sequence of numbers. And this is ...


0

For arbitrarily $\epsilon > 0$ there exists an $N$ such that $n > N$ implies $\left|\langle y, x\rangle\right| = \left|\langle y, x_n - x \rangle\right| \leq \| y \| \| x_n - x \| < \epsilon$ . Hence ...


2

If $X$ is any Banach space, with $A, B \in L(X)$ bounded linear operators and $A$ invertible, we have, in the operator norm, using the hypothesis presented in the text of the question, $\Vert I - A^{-1}B \Vert = \Vert A^{-1}(A - B)\Vert \le \Vert A^{-1} \Vert \Vert A - B \Vert < 1. \tag{1}$ Since (1) shows that $\Vert I - A^{-1}B \Vert < 1, ...


1

Through the functional equation, you may also check that: $$ \sum_{n\geq 0}\frac{(-1)^n}{n!(x+n)}=\Gamma(x)-\Gamma(x,1)\tag{1}$$ where: $$ \Gamma(x,1) = \int_{1}^{+\infty} u^{x-1}e^{-u}\,du \tag{2}$$ is the incomplete $\Gamma$ function, regular (i.e. $C^{\infty}$) on the whole real line. Hence our function has the same regularity of the $\Gamma$ function, ...


0

If $x$ is a real number for which $x \not \in (\mathbb{Z} \cup \{0\})$, then there is a closed neighborhood of $x$, $D_x$ for which $D_x \cap (\mathbb{Z} \cup \{0\}) = \emptyset$. Note that each of the functions: $f_M(y) = \sum_{n=0}^M \frac{(-1)^n}{n!(y+n)}$ are continuous, and that this sequence converges uniformly to $f$ on $D_x$. You can show this by ...


1

Given a continuous mapping $F:X\rightarrow Y$ it induces continuous map $F^{**}:X^{**}\rightarrow Y^{**}$ given by the formula $$[F^{**}(x^{**})](y^*):=x^{**}(F^*(y^*)).$$ This assigment is functorial, i.e. it has two properties: $(id_X)^{**}=id_{X^{**}}$ and $(G\circ F)^{**}=G^{**}\circ F^{**}.$ Additionaly for every normed space $X$ we have continuos ...


2

Given a bijective linear isometry $T:X\rightarrow Y$, the dual map $T^*:Y^*\rightarrow X^*$ is also a bijective linear isometry. (This follows from the fact that $T^*$ has inverse $(T^{-1})^*$ and $\|T^*\|=\|T\|$.) From this, we have that $T^{**}:X^{**}\rightarrow Y^{**}$ is a bijective linear isometry as well. Let $J_X:X\rightarrow X^{**}$ is the canonical ...


1

The following version of the open mapping theorem holds: If $T:Y\to X$ is a bounded operator between Banach spaces with $B_X \subseteq c \,\overline{T(B_Y)}$ for some constant $c>0$ (and the unit balls $B_X$ and $B_Y$) then $T$ is surjective (and open). Sometimes such a $T$ is called almost open. (The proof is in every book on functional analysis, the ...



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