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Show that $T$ and $S$ are closed. Then apply the closed graph theorem to conclude that $T$ and $S$ are continuous. To show that $T$ is closed, suppose $\{ x_{n} \}$ converges to $x$ and suppose $\{ Tx_{n} \}$ converges to $y$, and show that $Tx=y$; to do this observe that $$(Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in ... 4 For your first question, I think you are misunderstanding what the theorems say. When you restrict your \sigma-weakly continuous functional to the unit ball, you don't get a wot functional on the whole space: so 4.6.4 does not apply. For your second question, here is an example: fix an orthonormal basis \{e_n\} and let$$ ...

3

Suppose that $A$ is a bounded selfadjoint operator on a Hilbert space $\mathcal{H}$. Then, $$\|Ax\|^{2}=(Ax,Ax)=(A^{2}x,x)\le \|A^{2}x\|\|x\|.$$ Let $\{ x_{n} \}$ be a bounded sequence in $\mathcal{H}$ which is bounded by $M$. Assuming that $A^{2}$ is compact gives a subsequence $\{ x_{n_{k}}\}$ such that $\{A^{2}x_{n_{k}}\}$ is convergent ...

3

Let us take a closer look at $\lVert\,\cdot\,\rVert_0$. Since for $x\in X$ we have $\hat{x} = \{ x-y : y \in Y\}$, we can write $$\lVert \hat{x}\rVert_0 = \inf_{z\in\hat{x}} \lVert z\rVert = \inf \{ \lVert x-y\rVert : y \in Y\}.\tag{1}$$ So we have $\lVert \hat{x}\rVert_0 = 0$ if and only if for every $\varepsilon > 0$ there exists a $y_\varepsilon \in ... 3 The solution of your problem is an easy application of the proposition below: Theorem: Let$A$be a nonempty subset of$\mathbb{R}$and$A$is bounded below. Then there is a$m\in\mathbb{R}$such that$m$is a lower bound for$A$and given any$\epsilon>0$there exists$a\in A$such that$m\leq a<m+\epsilon$If$d=\inf_{y\in M} \|x-y\|^2$then for ... 3 I can show the claim when the interval is the same for each integral, let's say it is$[-1,1]$, and if there is a restriction on the roots of$f(t)$. I thought about this for a while and figured I might as well post it. I'm slightly hopeful with some more thinking a full solution will come. First, suppose$f(t)=\prod (t-\alpha_i)$with$|\alpha_i| > 3$. ... 3 Yes, the fact that$h(x) = e^{\delta\lvert x\rvert}f(x)$is Lebesgue integrable implies that$f$is also Lebesgue integrable. Since$x\mapsto e^{\delta \lvert x\rvert}$is continuous and everywhere$\geqslant 1$,$f$is (Lebesgue) measurable if and only if$h$is measurable, and since$\lvert f(x)\rvert \leqslant \lvert h(x)\rvert$, the finiteness of$\int ...

2

You can directly compute the derivative of $g$ by dominated convergece $\lim_{h\to 0}\dfrac{g(z+h) - g(z)}{h}=\lim_{h\to 0}\int_{\mathbb{R}}f(x)\dfrac{e^{-ix(z+h)}- e^{-ixz}}{h}d\mu_x$ And we have $$\dfrac{e^{-ix(z+h)}- e^{-ixz}}{h} = e^{-ixz} \dfrac{e^{-ixh}- 1}{h}$$ Let $z = \lambda + i\nu$ with $|\nu| < \delta$ $$\left|e^{-ixz} \dfrac{e^{-ixh}- ... 2 If you know the uniform boundedness principle (Banach-Steinhaus theorem), you could first show that the set C=\{Tx:\|x\|\le 1\} is weakly bounded. Then, if you apply Banach-Steinhaus theorem properly (in a subtle way), you get that C is bounded (which tells you that T is continuous). 2 Hint: e_n e_k = \delta_{nk}, where \delta is a Kronecker delta, i.e.$$\delta_{nk}=\begin{cases}1,&n=k,\\0,&n\ne k.\end{cases}$$2 Volumes 3 and 4 of Kadison-Ringrose consist of full solutions to the exercises in volumes 1 and 2. I don't know other examples first hand. That said, looking at full solutions is likely a bad way to go about functional analysis. You will certainly struggle with some exercises--we all do--but in my view that struggle is part of getting an understanding of ... 2 Assume the contrary, that for every k there exists a non zero polynomial p_k of degree n such that$$ \int_{-1}^1 |p_k(t)|dt \leq \frac{1}{k} \left( \int_0^2 |p_k(t)|^2dt \right)^{1/2}$$We can assume that \int_0^2|p_k|^2=1, because the inequality is scale invariant. This means that \int_{-1}^1 |p_k(t)|dt \leq \frac{1}{k} so p_k \to 0 in ... 2 You are right. For an infinite set S, we have$$\ell^2(S) = \bigcup_{\substack{T\subset S\\\operatorname{card} T = \aleph_0}} \ell^2(T),$$viewing \ell^2(T) as a subset of \ell^2(S) via the canonical embedding, since a summable family of real (or complex) numbers can only contain countably many nonzero terms. Since \operatorname{card} \ell^2(T) = ... 2 Suppose f is as stated with exponential bound Ce^{-\delta|x|}, and suppose that \overline{g} \in L^{2}(\mathbb{R}) is orthogonal to the functions \{ x^{n}f(x) \}_{n=0}^{\infty} in the L^{2} inner-product. Define$$ H(\lambda) = \int_{-\infty}^{\infty}g(x)f(x)e^{-i\lambda x}\,dx. $$Suppose that |\Im\lambda| \le \delta' < ... 2 For a complex Hilbert space, T is normal iff \|Tx\|=\|T^{\star}x\| for all x. This is due to the polarization identity, which is a useful algebraic tool:$$ (x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2}. $$If T is normal, then so is T-\lambda I for any complex \lambda, with adjoint T^{\star}-\overline{\lambda}I. ... 2 Notation: I'll probably be a little messy, and absorb a lot of constants as is standard with this sort of thing. If f is in the Schwartz space, then f is infinitely differentiable (in particular continuous), and f (as well as its derivatives) decay faster than any polynomial. Now, we wish to split the integral$$ u(x,t) = \int_{\mathbb{R}^n} ...

2

Take $$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$ for $n = 1, 2, \dots$ and for some $a_n$. You'll get $$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$ $$\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$ Then choose the sequance $a_n$ so that the first sum is convergent, while the second one ...

2

Use that a function is lower semi-continuous if and only if all of its lower level sets $f^{-1}(-\infty,t]=\{x \in X : f(x) \le t\}$ are closed. Let $a=\inf f(K)$ (either a real number or $-\infty$). Look at the family $F=\{K\cap f^{-1}(-\infty,t]: t>a\}$. This is a family (ordered by set-inclusion) of non-empty closed and hence compact subsets of $K$, ...

2

You can use the same idea that works for $+$: $$\|(\alpha,x)\|_{K\times X}=|\alpha|+\|x\|.$$ In any case, you can endow $K\times X$ with the product topology and no explicit norm is required. EDIT: \eqalign{\|\alpha x - \alpha_0 x_0\|_X & = \|\alpha x - \alpha_0 x + \alpha_0 x - \alpha_0 x_0\|_X\cr &\le\|\alpha x - \alpha_0 x\|_X + \|\alpha_0 ... 2 Let p(x)=(x-1)(x-2)\cdots(x-n-1). Then the polinom p has degree n and \|p\|=0 while p\neq 0 so the function in 1 is not a norm. Let p=c, c is constant. Degree of p is 0 and p\neq 0 but \|p\|=\sup_{x\in [0,1]}|p'(x)|=0 so 4 is not a norm. 2 Assume q<\infty. Note that you can express 2 as a convex linear combination of p and q, that is 2=sp+tq,\quad s=\frac{q-2}{q-p}, \;t=\frac{2-p}{q-p} $$And we have s+t=1, 0<s,t<1. If we set r=s^{-1} and r'=t^{-1}, then r and r' are conjugate exponents such that |f|^{sp}\in L^r(\mathbb{R}) and |f|^{tq}\in L^{r'}(\mathbb{R}) ... 2 The \ln function is concave. Therefore, for x,y > 0 and 0 \le t \le 1, you have (1-t)\ln x + t \ln y \le \ln((1-t)x+ty). Hence,$$ x^{1-t}y^{t} \le (1-t)x+ty. $$By taking a limit, the above continues to hold for x \ge 0 and y \ge 0. Now replace x by |f|^{p} and y by |f|^{q} to obtain$$ |f|^{(1-t)p+tq} ...

2

The Fourier transform is unitary on $L^{2}(\mathbb{R})$ if you choose the correct normalization: $$Ff = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ist}f(t)\,dt.$$ Some people instead absorb the constant into the exponent, but this is what I'm used to seeing because this $F$ is directly linked to the spectral resolution of the operator ...

2

For $\epsilon>0$, you can find $N$ s.t. for $x>N$, $f(x+1)-f(x)\in (l-\epsilon, l+\epsilon)$; then $f(x+k)-f(x)\in (lk-k\epsilon, lk+k\epsilon)$ for all $k\in\mathbb{N}$. We have $$f(x+k)=f(x)+(f(x+k)-f(x))\in (f(x)+ lk - k\epsilon, f(x) + lk + k\epsilon)$$ and $$\frac{f(x+k)}{x+k}\in \big(\frac{f(x)}{k}+l-\epsilon, \frac{f(x)}{k}+l+\epsilon\big). ... 2 Hint: rewrite$$ \langle F,g\rangle = \int_{\Bbb R} 1_{x>0} \sqrt{x} g(x) dx $$and do not forget the border terms. the integral$$ -\int_{\Bbb R} 1_{x>0} \frac{g'(x)}{2\sqrt{x}} dx $$does not converge around 0, but$$ -\int_{\Bbb R} 1_{x>0} \frac{g'(x) - g'(0)}{2\sqrt{x}} dx does, when g is smooth. Then: \begin{align} \langle F'', ... 2 Okay last statement is easy to prove "Finally, it is clear that no number less than α or greater than β can be a subsequential limit of the partial sums of (25)" Let if possible a<α is subsequential limit of the partial sums of (25). Then it would imply sequence partial sums of (25) has a subsequence converging to a i.e. after a certain stage, this ... 1 For the proof of the author's claim, note that c \geq 1 = c ^{-1} c if and only if 1 \geq c^{-1}. This example doesn't quite work since it doesn't meet the hypotheses set out by the author. Let's consider the C*-algebra A := C([2,10]) and the same function f, which you defined above. There are various notions of invertibility in the C*-algebra ... 1 A normed space comes with a natural metric, d(x,y) = ||x-y||. This natural metric gives us a topology \mathcal{T}, and a topology gives us a notion what a continuous function on that space is. Unless otherwise stated, this is the topology a normed space. Of course it's possible to consider other metrics d' on X, e.g. other ones that induce the same ... 1 For \lvert\lambda\rvert > \lVert T\rVert, we have a non-negative lower bound (strictly positive for x\neq 0) for \lVert (T-\lambda I)x\rVert, namely\lVert (T-\lambda I)x\rVert \geqslant (\lvert\lambda\rvert - \lVert T\rVert)\lVert x\rVert.$$Now set x = (T-\lambda I)^{-1} y to obtain the inequality$$\lVert y\rVert = \lVert(T-\lambda ...

1

I think you probably mean $\xi > \eta$? The overall inequality is $$|u(\xi) - u(\eta)|^2 \leq (b-a)\int_a^b (u')^2~dx.$$ Write this as $$u(\xi)^2 - 2 u(\xi)u(\eta) + u(\eta)^2 \leq (b-a)\int_a^b (u')^2~dx.$$ Integrate in $\xi$ from $a$ to $b$. $u(\eta)^2$ does not depend on $\xi$, so we obtain a factor of $(b-a)$. Similarly with the integral on the ...

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