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2

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 = ...


0

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.



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