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1

The only problematic part of your proof is the implication $3)\implies 2)$. You should argue as follows Assume $S$ is bounded. Consider arbitrary $(y_n)_{n\in\mathbb{N}}\subset\overline{u(S)}$, then $(y_n)_{n\in\mathbb{N}}$ is also bounded. For each $n\in\mathbb{N}$ we can find $x_n\in S$ such that $$ \Vert y_n-u(x_n)\Vert\leq 2^{-n}.\tag{*} $$ Since ...


0

Since $$ \int_{a}^{b}dt\int_{a}^{b}ds|K(t,s)|^{2}<\infty $$ $K(t,s)$ is a Hilbert-Schmidt operator and such operators are compact.


2

Okay: $\|Tf\|_2^2=\int_a^b|\sum_{j=1}^n\int_a^b\phi_j(t)\psi_j(s)f(s)ds|^2dt\le\int_a^b|\sum_{j=1}^n|\phi_j(t)|\int_a^b|\psi_j(s)f(s)|ds|^2dt$ $\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\int_a^b|\sum_{j=1}^n \phi_j(t)|^2dt\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\|\sum_{j=1}^n \phi_j(t)\|_2^2$ Thus: $\|Tf\|\le\sup_j\|\psi_j\|_2\|\sum_{j=1}^n \phi_j(t)\|_2\|f\|_2$ I.e. $T$ ...


0

Notice that the range of the unit ball is contained in the compact set $$K:=\left\{\sum_{j=1}^na_j\phi_j, |a_j|\leqslant \lVert \psi_j\rVert_2\right\}.$$


-1

The definition of a compact operator $u \in B(X,Y)$ is that $u$ maps bounded sets of $X$ into relatively compact sets of Y. Therefore, (1) and (2) are equivalent by definition and no proof is required. This definition is equivalent to saying if $x_n$ is a bounded sequence in $X$, then $u(x_n)$ has a Cauchy subsequence (not necessarily convergent) in $Y$. To ...


1

General helpful little fact If $A$ is a bounded operator and there is an infinite dimensional subspace $M$ on which $A$ has a lower bound $c>0$, i.e., $$\|Ax\|\ge c\|x\|\quad \text{ for all }x\in M \tag{1}$$ then $A$ is not compact. (Proof: pick a bounded uniformly separated sequence in $M$; its image under $M$ has the same properties.) In your ...


2

It's a lemma of J.-L. Lions, see An application of J.-L. Lion's Lemma where a book reference is given (Brezis, Functional Analysis...). You don't get quantitative control of $C$ from the compactness argument by which the lemma is usually proved. Instead, we can use the Sobolev embedding with the Peter-Paul trick as here. Below I write $\|u\|_p$ ...



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