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First observation: by the uniform boundedness principle, the sequence $(\lVert T^n\rVert)_{n\geqslant 1}$ is bounded. Let $x\in H$. Using compactness of $T$ and the fact that $T^nx\to 0$ weakly, we can find $n_k\uparrow \infty$ such that $\lVert T^{n_k}x\rVert\to 0$. Since $$\sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^nx\rVert\leqslant \sup_{n_k\leqslant ... 0 (I am not sure if this is satisfactory.) A trace class operator can be represented in the form \sum (\cdot, u_k)v_k with \sum \|u_k\|\,\|v_k\|<\infty. If you take a trace class integral operator of this form and write down its kernel, it will have the formula \sum \overline{u_k(y)}\, v_k(x). For the diagonal you get \sum \overline{u_k(x)}\, ... 3 No. The identity operator is not compact in any infinite dimensional Banach space. 1 There is no problem for boundedness. However, for equi-continuity, you have to be more careful. First of all, we should show that \lim_{i\to\infty}\sup_n\dots\to 0 (there won't be any problem because the estimates are uniform in n, but we have to take the supremum). Second, this proves the equi-continuity at t and we want a uniform equi-continuity. ... 5 In the definition of Z, you probably want |v|>N instead of v>N. Also, in item 1, the definition of B, you have the Sobolev norm of x, so it's better to use norm notation for that. Let's also not use subscripts in superscripts... say, p<q and the embedding is H^q\to H^p. The H^q norm is given by$$\|f\|_{H^q}^2 = ...
No, a bounded set of continuous functions is different from a set of bounded continuous functions. In your case of $C[0,1]$, note that every $f \in C[0,1]$ is a bounded continuous function, due to the compactness of $[0,1]$. A subset $B \subseteq C[0,1]$ is a bounded set if there is one bound for all the functions in $B$, that is the number $$\sup_{f \in ... 0 In the context of your question a set B is bounded means that it is a collection of vectors that are all bounded in norm by some value M. However the norm you are probably using on C[0,1] is \| x \| = \sup_{t \in [0,1]} |x(t)|. This should show you the connection between your two phrases. 0 F_n(t) is a trigonometric polynomial of "degree" n-1, as exhibited by your second formula. Therefore the functions$$g_t(\theta):= F_n(t-\theta)={1\over2}+\sum_{k=1}^{n-1}\left(1-{k\over n}\right)\bigl(\cos(kt)\cos(k\theta)+\sin(kt)\sin(k\theta)\bigr)$$are trigonometric polynomials in \theta for each fixed t. It follows that the function ... 0 This is what I think is happening: Assume that w\in L^\infty and \mathrm{supp}\,w\in B for some ball B\subset\mathbb{R}^n. Take g\in L^2. Then we have$$ \|wg\|_{L^2}\leq\|w\|_{L^\infty}\|g\|_{L^2}. $$Hence by the basic property of the resolvent, we have R_\tau wg\in H^2 and$$ \|R_\tau wg\|_{H^2}\leq C\|w\|_{L^\infty}\|g\|_{L^2}. $$With the ... 3 If 1\le p<q<\infty, then the embedding$$L^q(0, 1)\subset L^p(0, 1) trivially is weakly compact. Indeed, an $L^q$-bounded sequence is $L^p$-bounded too, and by Banach-Alaoglu's theorem (which applies since $q\ne 1, \infty$) it has weakly convergent subsequences.