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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 ...

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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

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