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5

A Fredholm operator $T$ is an operator for which the solutions of the nonhomogeneous linear problem $Tx = y$ can be described using "finitely many pieces of data" just like in the finite dimensional case even though the operator acts on a possible infinite dimensional space. More explicitly, if $T$ is Fredholm then $\ker(T)$ is finite dimensional and so we ...


4

Define $\Lambda\in(\ell^1)^*$ by $$\Lambda x = \sum\frac{n}{n+1}x_n.$$Note that $||\Lambda||=1$. In fact it's clear that $$|\Lambda x|<||x||\quad(x\ne0).$$ If $\Lambda x=0$ then $x\ne e_1$, hence $$\frac12=\Lambda(e_1-x) <||e_1-x||_1.$$


3

In fact, $A$ can't be a compact operator (if $X$ is infinite dimensional). This condition is sometimes referred to as "$A$ is bounded away from $0$". To show that it can't be compact, it suffices to show that there is a bounded sequence whose image has no convergent subsequence. Note (by the usual application of Riesz's lemma) that there exists a bounded ...


2

Theorem: Let $H$ be a real or complex Hilbert space, and let $\{e_{\alpha}\}_{\alpha\in\Lambda}$ be an orthonormal subset of $H$. The following are equivalent: 1. $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is a complete orthonormal set, meaning that the only $x\in H$ that is orthogonal to every $e_{\alpha}$ is $x=0$. 2. Parseval's identity $\|x\|^2=\...


2

A typo slipped in; a $k$ became $j$ for no reason. Fixing that, you're almost there, re showing it's a Banach algebra: $$\begin{align}\dots=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{(k-j)}(t)}{(k-j)!}\dfrac{g^{(j)}(t)|}{j!}} &=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=j}^{n}\dfrac{|f^{(k-...


2

Let's recall the definition of accumulation point: Given a metric space $(X,d)$ and a subset $A\subseteq X$, an accumulation point of $A$ is a point $p$ of $X$ such that for all $\varepsilon>0$, $(B_{\varepsilon}(x)-\{x\})\cap A$ is non-empty, i.e., there is a point $a_{\varepsilon}\in A$ different from $x$ such that $d(x,a_{\varepsilon})<\...


2

The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$. ...


2

$ \| A B \| = \sum_r \sup_j |\sum_k a_{j,k} b_{k,j+r}| $ $ \leq \sum_r \sum_k |a_{j(r),k}| |b_{k,j(r)+r}| $ --- $j(r)$ is where the $\sup$ is attained (possibly up to $\epsilon$...) $ = \sum_r \sum_k |a_{j(r),k+j(r)}| |b_{k+j(r),j(r)+r}| $ $ = \sum_k \sum_r |a_{j(r),k+j(r)}| |b_{k+j(r),j(r)+r}| $ $ \leq \sum_k \sup_j |a_{j,k+j}| \sum_r |b_{k+j(r), j(r) +...


2

Regarding convergence and completeness: For $n\in N$ let $f_n(x)=0$ for $x\leq 1/2-1/(n+2)$ and $f_n(x)=1$ for $x\geq 1/2.$ Let $f_n(x)$ be linear for $x\in [1/2-1/(n+2),1/2].$ Then $(f_n)_{n\in N}$ is a Cauchy sequence with respect to the norm $\|f-g\|=[\int_0^1|f(x)-g(x)|^2\;dx]^{1/2}.$ Let $h(x)=0$ for $x\leq 1/2$ and $h(x)=1$ for $x>1/2.$ The ...


2

Hint: if $x$ is in the unit ball of $c_0$, there is some $i$ such that $|x_i| < 1$. What happens if you increase or decrease $x_i$ a little bit?


2

If isomorphic means isomorphic as rings then no, they're not isomorphic. Keen fact: Suppose $K_1$ and $K_2$ are compact Hausdorff spaces. Then $C(K_1)$ is isomorphic to $C(K_2)$ if and only if $K_1$ and $K_2$ are homeomorphic. (Sketch: $C(K)$ is a Banach algebra with maximal ideal space $K$.) This is a keen thing, because it shows that the ...


2

No. Consider $y(t) = t^m$ and $x(t) = t^n$ for some $n\neq m$. Then $$2 = \| x + y\| = \|x \| + \|y\| $$ but $x\neq \lambda y$.


2

They are tight, in the sense that we have $\text{trace}(AB) = \lambda_{max}(A)\; \text{trace}(B)$ if $A = I$. Similarly in the second one if $B=I$.


2

If I understand the question correctly, if we have any linear operator $A$ for which the exponential $\exp(A)$ is meaningful, then $A$ commutes with $-A$, and so $$ \exp(A) \exp(-A) = \exp(A + (-A)) = \exp(\mbox{the zero operator}) = I. $$ The inverse of $(I + B)^{-1}$ (again, if the latter is defined) is $(I+B)$.


2

Short answer: Bernstein. First note that since $\widehat{S_0u}$ has compact support $S_0u$ is smooth, in fact real-analytic. So we forget about $S_0u$, at least for now. Say that dyadic block $2^{j-1}\leq|\xi|\leq 2^{j+1}$ is $A_j$. There exists a $C^\infty_c$ function which equals $i\xi$ on $A_0$, hence there exists a Schwarz function $F$ with $$\hat F(\...


2

Let $h'(x) := \max(f'(x), g'(x))$. $h'$ is continuous. Get a primitive function $h$ with $h(0) = \max(f(0), g(0))$. This should do it.


2

Sometimes the map $(\cdot, \cdot) : X \times X^* \to \mathbb{R}$ defined via $$(x,y) := y(x)$$ is denoted as duality map. In my opinion, this notion is in particular used if one has a space $Y$ which is isometric to $X^*$, i.e., for the map $$(x,y) := (I\,y)(x),$$ where $I : Y \to X^*$ is an isometric isomorphism. However, typically one might use acute ...


2

It is often very difficult to calculate. A point $y\in X$ is in the weak closure if you can not enclose it in a weak neighborhood disjoint from $S$, i.e. if for every $\epsilon>0$ and linear functionals $\ell_1,\ldots,\ell_k\in X'$ the set $N=\{ x\in X : |\ell_i(y-x)|<\epsilon, i=1,\ldots,k \}$ intersects $S$. Note that $k$ must be finite. In finite ...


2

Since $\mathcal F$ is any $\sigma$-algebra on $[0,1]$, this is essentially the general case of: Halmos [1] showed that the range of a non-negative, finite measure is a closed subset of real numbers. [1] Halmos, Paul R. On the set of values of a finite measure. Bull. Amer. Math. Soc. 53 (1947), no. 2, 138--141. http://projecteuclid.org/euclid.bams/...


1

If one of the two sets is bounded, the statement is true. One even gets the stronger result you mentioned. For unbounded sets it is false. I do not know if there are really simple counterexamples (e.g. twodimensional?), but I would try something like: With $n=3$, set $A = \{x \colon x_2 = x_3 = 0\}$ and $$B= epi f = \{x \colon x_3 \geq f(x_1,x_2) \}.$$ If ...


1

If all you assume about your involution is that it's an involution. that is, $(x+y)^*=x^*+y^*$, $(xy)^*=x^*y^*$, $x^{**}=x$ and $(cx)^*=\overline cx^*$, then most of what you expect doesn't follow. In particular you assume above that $\phi(x^*)=\overline{\phi(x)}$, and that doesn't follow: Consider $C([-1,1])$. Define $$f^*(t)=\overline{f(-t).}$$ That's an ...


1

First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets. Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$...


1

Suppose that $T-\lambda I$ is invertible, then it has trivial kernel and is bounded. Particularly you can solve the equation $$(T-\lambda I)x = y$$ for any $y\in\ell^p$. Writing $x = (x_m)$ and $y = (y_m)$, we see that $$ (\alpha_m-\lambda)x_m = y_m,$$ i.e. $$x_m = \frac{1}{\alpha_m-\lambda}y_m.$$ Note that $\lambda=\alpha_m$ is a serious problem here. ...


1

Hint: A compact space is separable. Write $X=U_nB(0,n)$ $K(B(0,n)$ is separable since it is relatively compact, $\bigcup_nK(B(0,n))=K(X)$ is separable since it is the union of separable spaces. For your second question let $(v_n\neq 0)$ be a dense family in $Y$, write $w_n={{v_n}\over{\|v_n\|}}$. Consider $K:l^1\rightarrow Y$ defined by $K(e_i)={w_i\over i}$...


1

Let $$(T_n(\underline x))_k:=\begin{cases}\alpha_kx_k&\text{if }k\le n\\ 0&\text{if }k>n\end{cases}$$ You can prove that $$\left\lVert T-T_n\right\rVert_{\mathfrak L(\ell^p,\ell^p)}\le \sup\{\lvert \alpha_k\rvert\,:\,k>n\}\stackrel{n\to\infty}\longrightarrow 0$$ So $T$ is limit in $\mathfrak L(\ell^p,\ell^p)$ of finite-rank operators. Hence, ...


1

Here's a proof why $l^p(\mathbb N)$ is not locally convex, this is just for simplicity, it can be easily generalized. If it would be locally convex, then the unit ball $B_1(0)$ would contain a convex neighborhood U of $0$. Then there must be $\delta>0$ with $B_{2\delta}(0)\subset U$, hence also $\mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0)$. Let ...


1

Hint: $g_n$ is weak-$*$ convergent.


1

This is not true. Take for example $p=1$ and the function $f_\epsilon(x)=\frac{1}{x}$ for $x>\epsilon$ and 0 else. Then $f_\epsilon$ is in $L^1(\epsilon,T)$ with norm equal to $ln(\epsilon)-ln(T)$. Bug the norm diverges as $\epsilon$ goes to 0. Hence $\frac{1}{x}$ is not in $L^1(0,T)$.


1

If $e_n$ are an orthonormal basis of $H$, consider the sequence $a_n = (1+1/n) e_n$. The set $A = \{a_n : \; n = 1,2,3,\ldots\}$ is a closed (and thus complete) set, because the distance between any two members of $A$ is greater than $1$. There is no best approximation of $0$: $\|a_n - 0\| = 1 + 1/n \to 1$ as $n \to \infty$, but the infimum is not attained....


1

For a simple example to demonstrate the importance of the basis being orthonormal, consider $\mathbb{R}^2$ with the standard inner product and the basis $h_1=(1,0)$ and $h_2=(1,1)$. If $h=(0,1)$ then $$ h=-h_1+h_2 $$ but $\langle h,h_1\rangle=0$. By the way, a Hilbert space basis of $H$ is different from a basis of $H$ as an abstract vector space (called a ...



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