New answers tagged

1

This depends on the domain and range of the the operator $A$. For a linear operator $A : X \to Y$, where $X$ and $Y$ are normed spaces, one can show that if $A$ is bounded and of finite rank, i.e., $\dim A(X) < \infty$, then the operator $A$ is compact. So, if you consider normed spaces, then your reasoning is fine.


0

If $\lambda=q(s)$ for some $s\in[0,1]$, then $T_q 1_s=q 1_s=q(s) 1_s=\lambda 1_s$. Thus, $\lambda\in \sigma(T_q)$. As the spectrum of an operator is closed, it follows that $\overline{\{q(t)\mid t\in[0,1]\}}\subset \sigma(T_q)$. If $\lambda\notin\overline{\{q(t)\mid t\in[0,1]\}}$, let $\phi=\frac 1{q-\lambda}$. Since $\lambda\notin\overline{\{q(t)\mid ...


2

For $1$, start with any bounded sequence $\{ f_n \}$ in $C[0,1]$ that has no convergent subsequence. Then $F_n\int_{0}^{x}f_n(t)dt$ gives you a bounded sequence $\{ F_n \} \subset C^1[0,1]$ whose image under $T$ is $\{ f_n \}$. For $2$, the same technique holds. Start with a bounded sequence in $C^1[0,1]$ with no convergent subsequence, and integrate. For ...


0

Answer for (1): Using the Poincare-Wirtinger inequality, there is $C>0$ so that (write $u = u(f)$) $$\tag{1} \| u\|_2 = \| u - u_D\|_2 \le C\| \nabla u\|_2,$$ where $u_D := \int_D u $ is assumed to be zero here. Then we have $$\begin{split} \|\nabla u\|_2^2 &= \int_D |\nabla u|^2 dx \\ &= \sum_{i=1}^n \int_D u_i u_i dx \\ &= ...


1

If you you expand $(T-\lambda I)^n$ you get $R+(-\lambda)^nI$ with $R$ compact, and now you can apply the same argument as before.


2

Let $\{v_n\}$ be a dense countable subset of $V$. Then we can find: $$ \alpha_1,\ \ \ \text{ such that } \|T_{\alpha_1}(v_1)-v_1\|<1. $$ $$ \alpha_2,\ \ \ \text{ such that } \|T_{\alpha_2}(v_j)-v_j\|<\frac12,\ \ j=1,2. $$ $$ \alpha_3,\ \ \ \text{ such that } \|T_{\alpha_3}(v_j)-v_j\|<\frac13,\ \ j=1,2,3. $$ $$ \vdots $$ $$ \alpha_n,\ \ \ \text{ such ...


0

There is a theorem that so-called finite-rank operators between normed spaces are compact. It goes like this: Theorem 8.1-4 [Kreyszig: Introductory Functional Analysis with Applications] Let $X$ and $Y$ be normed spaces and $T:X\to Y$ a linear operator. Then if $T$ is bounded and dim$(T(X))<\infty$ the operator is $T$ is compact. (Proof given ...


0

This is an integral operator with kernel $K:C([0,\pi])^2\to\mathbb R$ defined by $K(x,t)=\sin x+\cos t$. Since $$\|K\|_2^2=\int_0^\pi\int_0^\pi(\sin x+\cos t)^2\ \mathsf dx\ \mathsf dt =\pi^2<\infty,$$ $K$ is a Hilbert-Schmidt kernel and therefore the operator $k$ is compact. (The proof of this result does use a sequence of finite-rank operators to ...


0

For your first question, $$Ku(x) = \int_0^\pi (\sin(x)+\cos(t)) u(t) dt = \sin(x) \underbrace{\int_0^\pi u(t) dt}_a + \underbrace{\int_0^\pi \cos(t)u(t) dt }_b$$ For your second question, if the range of an operator is a finite dimensional vector space, then it is compact, because the unit ball of a finite vector space is compact



Top 50 recent answers are included