紧算子商空间证明

Verifying well-definition

I’m checking the well-definition of $T$ on $\mathcal{X}/\mathcal{X}_0$ by ensuring $[Ax] = [Ay]$ if $x \equiv y$ in $\mathcal{X}/\mathcal{X}_0$.

Establishing compactness

I’m figuring out how $T$, the induced operator, achieves compactness on $\mathcal{X}/\mathcal{X}_0$ by showing that any bounded sequence in $\mathcal{X}/\mathcal{X}_0$ has a convergent subsequence mapping under $T$.

Connecting compactness

I’m thinking about how $T$, if it factors through a finite-dimensional space or is a composition of quotient maps and injections, would be compact, as these conditions often imply compactness.

Analyzing operator conditions

I'm exploring the theorem where if $A$ is a bounded operator on a Banach space $\mathcal{X}$ and $\mathcal{X}_0$ is invariant under $A$, then $T$ is compact.

Mapping compactness

I'm examining how the image of the unit ball in $\mathcal{X}/\mathcal{X}_0$ under $T$ is relatively compact, which aligns with the definition of a compact operator.

Contradicting the non-compact

I’m examining the scenario where $T$ is not compact by finding a bounded sequence in $\mathcal{X}$ whose $T$-image lacks a convergent subsequence in the quotient space, contradicting the compactness of $T$.

Clarifying assumption

I'm noting that assuming $A$ is a compact operator justifies that the induced operator $T$ on the quotient space is compact, as the theorem applies when $A$ is compact.

Concluding with final observations

I'm working through the idea that $T$ might have finite rank, especially if $\mathcal{X}_0$ is stable, or it could be zero, which is inherently compact.

Mapping the quotient

I'm curious about how $T$ can be represented as the composition of projection $\pi$, operator $A$, and map $s$, with $s$ potentially not being continuous.

Focusing on convergence

I’m confirming that for any bounded sequence $[x_n]$ in the quotient, a representative $x_n$ in $\mathcal{X}$ ensures $[A x_n]$ has a convergent subsequence in the quotient space.

Focusing on convergence

I'm curious about ensuring a sequence $(y_n)$ in $\mathcal{X}_0$ such that $\|A x_n - A x_m - y_n + y_m\|$ is small, leading to convergence to an element in $\mathcal{X}$.

Laying out the options

I’m exploring various results to see when $T$ is compact, like when the dimension of $\mathcal{X}/\mathcal{X}_0$ is finite or considering partial isometries and their quotients.

Confirming compactness

I’m confirming that if $T$ is an operator on $X$ with $Y$ as a closed $T$-invariant subspace and $T|_Y$ jointly compact, then the induced operator on $X/Y$ is compact.

Connecting the dots

I'm thinking through why the operator $T$ is compact by considering how a bounded sequence in $X/Y$ can be lifted to $X$, ensuring the difference is in $Y$, and then analyzing $T|_Y$.

Connecting abstract concepts

I'm gathering info on how $T$ being a compact operator relates to $\mathcal{X}_0$ being closed and $A(\mathcal{X}_0) \subseteq \mathcal{X}_0$, and the finite dimension or direct sum of the quotient.

Understanding compactness

I’m piecing together that $T$ is compact if the quotient space $\mathcal{X}/\mathcal{X}_0$ is finite-dimensional or the direct sum of the closure of the range, utilizing the properties of compact operators and bounded sequences.

Confirming assumptions

I’m thinking through the necessity of $A$ being a compact operator on $X$ to assert that $T$ is compact. This assumption is vital and might have been overlooked.

Clarifying assumptions

I'm digging into the crucial assumption that $A$ must be a compact operator on $X$ to conclude that $T$ is compact. This point appears to have been skipped over.

Filling in the gaps

I'm recognizing that $A$ needs to be a compact operator on $X$ to ensure $T$ is compact. This crucial condition might have been overlooked.

Determining compactness

I’m concluding that for $T$ to be a compact operator, $A$ must be compact on $X$. This matches standard functional analysis notation and ensures the compactness condition is satisfied.

Piecing together assumptions

It's becoming clear that $A$ should be a compact operator on $X$. This ensures $T$ is a compact operator, aligning with standard functional analysis notation.

Confirming compactness

I’m detailing how to show $T$ is a compact operator: take a bounded sequence $[x_n]$ in $\mathcal{X}/\mathcal{X}_0$, lift to a bounded sequence $x_n$ in $X$, and use $A$ being compact to find a convergent subsequence $[Ax_n]$ in the quotient.

Revisiting the assumption

I’m noting that the question might implicitly assume $\dim(X/X_0)$ is finite, ensuring $T$ is compact due to linear operator properties on a finite-dimensional space.

Analyzing compactness

I'm thinking about whether the dimension of $X/X_0$ has to be finite or countable for $T$ to be compact, and how the unit ball in the quotient space impacts compactness.

Reevaluating assumptions

I’m noting that $A^n(\mathcal{X}) \to \{0\}$ for all $x \in X$ seems overlooked, which could mean the closure is in $\mathcal{X}_0$, crucial for establishing $T$'s compactness.

Balancing bounded sequences

I’m piecing together how to lift a bounded sequence in the quotient to $X$, ensuring the differences lie in $\mathcal{X}_0$, to demonstrate $T$’s compactness.

Laying out the options

I’m considering showing that $T$ is "strictly singular," which involves analyzing how a bounded sequence in the quotient can be lifted to $X$ and ensuring $A(x_n)$ differs by elements in $\mathcal{X}_0$ to achieve convergence.

Examining operator behavior

I’m analyzing how $T$ might be compact if $A$ has a spectral radius of 0, using principles like quasinilpotency and the converging norm sequence $\lim_{n \to \infty} \|A^n\| = 0$.

Confirming operator traits

I concur with the notion that $T$ is compact when $A$ is compact on $X$, given the standard approach of analyzing bounded sequences and using the property that $A(\mathcal{X}_0) \subset \mathcal{X}_0$.

Confirming compactness

I’m verifying $T$ is compact by analyzing a bounded sequence $[x_n]$ in $\mathcal{X}/\mathcal{X}_0$, ensuring $A$ being compact leads to a convergent subsequence $[Ax_n]$.

Thought about compact operator in quotient space，持续 44 秒