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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

Given a linear map $T:\mathbb{C}^n\to \mathbb{C}^n$, for all $v\in\mathbb{C}^n$ we have $$\langle TT^*(v),v\rangle= \langle T^*(v),T^*(v)\rangle\geq 0,$$ because this is just the inner product of a vector with itself. Thus, $TT^*$ is positive. Furthermore, if $T$ is invertible, $T^*$ is invertible, so equality holds in the above if and only if $v=0$. ...


3

The function $t\cdot \chi_{[-1,1]}(t)$ is orthogonal to each $t^{2k}$ in $L^2[-1,3],$ hence is orthogonal to the the linear span of $\{t^{2k} : k=0,1,\dots \}$ in $L^2[-1,3],$ hence is orthogonal to the closure of this linear span in $L^2[-1,3].$ Therfore this closure cannot be all of $L^2[-1,3].$


3

The closure of the domain in $L^2$ is simply $L^2$: Obviously it holds $C_0^\infty(0,1)\subset D(A_0)$. The set of smooth function is dense in $L^2(0,1)$, hence its closure is $L^2(0,1)$. This implies that the closure of $D(A_0)$ is $L^2(0,1)$ as well.


2

I think the intended question was something like: Given $p(x)\in\mathbb{C}[x]$ with degree $n$, such that $$\int_{0}^{1}x^k p(x)\,dx=0 $$ for every $k\in\{1,2,\ldots,n-1\}$, show that the roots of $p(x)$ lie in the circle $|x|\leq 1$. We may notice that $p(x)$ has $n+1$ coefficients and we have $n-1$ linear constraints on them, so the space of ...


2

Let $A$ denote the linear span of $\{t^{2k}\}_{k \in \mathbb{N}}$. Then, if instead of $[-1,3]$ the domain was $[1,3]$ instead, we could use the Stone-Weierstrass theorem to conclude that $A$ is dense ins $C([1,3])$ and thus in $L^2([1,3])$ since $t^2$ separates the points of $[1,3]$ and $1 \in A$. However, for $[-1,3]$, we have $t^{2k}(-1) = t^{2k}(1)$ ...


2

In any Hilbert space (say $H=\ell^2(\mathbb N)$), let $T=I$ and $S=I-P$ where $P$ is a finite-rank projection. Then, as $ST=I-P$ and $\ker ST=\text{ran}\,P$, $$ 0=\ker T\subsetneq PH=\ker ST. $$ Both $S$ and $T$ are selfadjoint and Fredholm.


2

EDIT: This answer no longer applies because the OP changed the question. It is not true. Any linear map between finite dimensional vector spaces is Fredholm, so there are finite dimensional counterexamples. If you want an infinite dimensional counterexample consider $S,T:\ell^2\to\ell^2$, where $S(a_0,a_1,a_2,\ldots)=(0,a_1,a_2,\ldots)$ and $T$ is the ...


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 $L^1[0,\infty)$. ...


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

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-...


1

\begin{eqnarray} \langle T_z^*(\alpha),x \rangle_{H}&=&\langle \alpha , T_z(x)\rangle_\mathbb{K}\\ &=& \alpha \overline{T_z(x)}\\ &=& \alpha\langle z,x\rangle_H\\ &=& \langle \alpha z , x\rangle_H\\ \end{eqnarray} So $$T_z^*(\alpha)=\alpha z $$ It's clear now that $\|T_z\|=\|T_z^*\|=\|z\|$



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