# Questions tagged [fredholm-operators]

The fredholm-operators tag has no usage guidance.

12
questions

**4**

votes

**1**answer

124 views

### Closure of the space of Fredholm operators

Let $X,Y$ be two Banach spaces.
A bounded operator $A$ is Fredholm if $\ker A$ and $\mathrm{coker} A$ are finite dimensional. Denote by $Fred(X,Y) \subset \mathcal{L}(X,Y)$ the space of Fredholm ...

**2**

votes

**1**answer

103 views

### Characterization of absolutely-continuous spectrum

The essential spectrum of a bounded linear operator $A$ on a separable Hilbert space $\mathcal{H}$ is defined as $$ \sigma_{\mathrm{ess}}(A) \equiv\left\{z\in\mathbb{C}\left.\right|A-z\mathbb{1}\text{ ...

**6**

votes

**3**answers

348 views

### Essential spectrum of multiplication operator

Let $a\in \mathcal{L}(L^2([0, 1], \mathbb{R}))$ be a multiplication operator. I wonder whether there is any work on calculating its essential spectrum. Is there any way to explicitly compute its ...

**0**

votes

**1**answer

136 views

### Proof of the analytic Fredholm theorem in Borthwick

I've stumbled across a proof of the analytic Fredholm theorem given in Theorem 6.1 in Spectral Theory of Infinite-Area Hyperbolic Surfaces by David Borthwick (see below).
Given the notion of being &...

**4**

votes

**1**answer

151 views

### Equivalence of families indexes of Fredholm operators

Let $F=F(H,H)$ be the space of bounded Fredholm operators in a Hilbert space $H$ with topology inherited from the norm operator topology, and let $X$ be a compact topological space.
For a continuous ...

**1**

vote

**0**answers

59 views

### Empty Weyl/Fredholm spectrum of an operator on an infinite dimensional Banach space

Let $X$ be a complex infinite dimensional Banach space, and let $T \in B(X)$ be nonscalar. The Fredholm spectrum of $T$ is defined by:
$$ \sigma_{\Phi} (T) := \lbrace \lambda \in \mathbb{C} : T- \...

**1**

vote

**0**answers

24 views

### Analytic solution of a compact operator eigenvector (integral equation)

In Sewall Wright's Evolution of Mendalian Population, the equation for the nonrecurrent mutation is
$$\frac{\phi(x)}n = \binom n {nx}\int_0^1 q^{nx}(1-q)^{n(1-x)}\phi(q)\,dq,\quad \forall x\in[0,1],$$
...

**1**

vote

**0**answers

46 views

### Operational quantities characterizing upper semi-Fredholm operators

An operator $T:X\rightarrow Y$ is said to be upper semi-Fredholm if its range is closed and its kernel is finite-dimensional. M. Schechter (1972) introduced a quantity $$\nu(T):=\sup_{\operatorname{...

**4**

votes

**0**answers

126 views

### Generalizations of Sard-Smale Theorem

Sard-Smale theorem holds for Fredholm maps $f:M\rightarrow B$ between separable Banach manifolds $M,N$. There are some constrains relating the Fredholm index $\operatorname{ind}(f)$ of $f$ to its ...

**0**

votes

**1**answer

131 views

### Eigenvalues of an integral operator

Let $K\in L^2((0,1)\times(0,1))$ and consider the operator defined in $L^2(0,1)$ by
$$Lu(x):=u(x)-\int_0^1K(s,x)u(s)ds.$$
What kind of assumption might I impose on $K$ such that $\lambda=1$ will be ...

**5**

votes

**0**answers

169 views

### Index of the Fredholm operator

I have two vector bundles $E_1$, $E_2$ over $M$ and an embedding of the smooth sections $\lambda : \Gamma(M, E_1) \rightarrow \Gamma(M, E_1 \oplus E_2)$. I consider a Fredholm differential operator $...

**5**

votes

**1**answer

98 views

### Fredholm elements of a Lie algebra

An element $a$ of a Lie algebra $L$ is called a Fredholm element if the adjoint operator $\mathrm{ad}_a:L \to L$ is a Fredholm linear map. That is: its kernel is a finite-dimensional space and its ...