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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 ...

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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 $$\frac12=\Lambda(e_1-x) <||e_1-x||_1.$$

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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 ...

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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$became$jfor 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. Then2 = \| 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 Leth'(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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