Dtyjlui

Let $X$ and $Y$ be Banach spaces.


(a) Let $T in mathcal{L}(X, Y )$. For each sequence $(x_n)_{n geq 1}$ in $X$ and each $x in X$, show that
$x_n →x$ weakly, as $n rightarrow infty$ ,implies that $Tx_n rightarrow Tx$ weakly, as $nrightarrow infty$.


(b) Let $T in mathcal{K}(X, Y )$. For each sequence $(x_n)_{n geq 1}$ in $X$ and each $x in X$, show that
$x_n →x$ weakly, as $n rightarrow infty$ ,implies that $||Tx_n -Tx|| rightarrow 0$ , as $nrightarrow infty$..


(c) Conversely, if $X$ is reflexive and separable, and $T in mathcal{L}(X,Y)$ satisfies that $∥Tx_n − Tx∥ rightarrow 0$, as $n rightarrow infty$, whenever $(x_n)_{ngeq 1}$ is a sequence in $X$ converging weakly to $x in X$, then $T in mathcal{K}(X, Y )$.



(d) Show that each $T in mathcal{L}(X,l_1(mathbb{N}))$ is compact, whenever $X$ is reflexive and separable.


(e) Let $Y$ be infinite dimensional. Show that no $T in mathcal{K}(X, Y )$ is open.


(f) Show that there is no reflexive separable Banach space $X$ such that $l_1(mathbb{N}) = T(X)$, for some $T in mathcal{L}(X,l_1(mathbb{N}))$.

My attempt:

(a) We have that $x_n rightarrow x$ weakly if and only if $f(x_n) rightarrow f(x)$ weakly for every $ f in X^*$. Now $Tx_n rightarrow Tx$ weakly if and only if $g(Tx_n) rightarrow g(Tx)$ weakly for every $g in Y^*$. But for every $g in Y^*$ we have $gT in X^*$. Therefore $Tx_n rightarrow Tx Leftrightarrow g(Tx_n) rightarrow g(Tx) Leftrightarrow (gT)x_n rightarrow (gT)x Leftrightarrow x_n rightarrow x$ weakly.

b) Since $T$ is compact I know that every sequence is sent to a sequence that has a convergent subsequence, but then I don't know how to proceed.

c) I have a hint for this problem:

Suppose that $T$ is not compact. Show that there exists $delta > 0$ and a sequence $(x_n)_{ngeq 1}$ in the unit ball of $X$ such that $∥Tx_n −Tx_m∥ geq delta$, for all $n neq m$.. Show next that $(x_n)_{ngeq 1}$ has a weakly convergent subsequence.

(d) Now let $(x_n)_{n geq 1}$ be a sequence in $ X$ and $x in X$ such that $ x_n rightarrow x$ weakly, as $n rightarrow infty$. By part (a) since $T in mathcal{L}(X,l_1(mathbb{N}))$ we get $Tx_n rightarrow Tx$ weakly, as $n rightarrow infty$. But weak convergence is the same as norm convergence in $l_1(mathbb{N})$. Therefore $||Tx_n-Tx|| rightarrow 0$, as $n rightarrow infty$. So we can use part $c)$ to deduce that $T in mathcal{K}(X,l_1(mathbb{N}))$, as requested.

(e) I know that if $Y$ is infinite dimensional the unit ball in $Y$ is not compact.

(f) Since $X$ is reflexive and separable, by part (d) we get that $T$ is compact. Suppose now that $l_1(mathbb{N})=T(X)$, i.e. $T$ is surjective. By the Open Mapping Theorem $T$ is open and this contradicts part (e).

(a) Looks good!

(b) Using this fact and (a), you should be set. If $Tx_n$ fails to converge to $Tx$ in norm, then there must be a subsequence $(Tx_{n_k})_k$ that maintains at least some $varepsilon > 0$ distance from $Tx$. Take a norm-convergent subsequence of this subsequence, and it will be a subsequence that fails to weakly converge to $Tx$ (as it norm-converges to another point). This contradicts (a).

(c) Not sure where separability comes in here, but if $T$ is not compact, then $T(B_X)$ is not totally bounded. That is, there exists a $delta > 0$ such that no finite number of balls radius $delta$ that cover $T(B_X)$. Hence, we may choose some $Tx_1 in T(B_X)$, and for each $n$,
$$Tx_{n+1} in T(B_X) setminus bigcup_{k=1}^n B[x_k; delta],$$
which is never empty by definition of $delta$. Thus, we have a sequence $(x_n) in B_X$ such that
$$|Tx_n - Tx_m| > delta$$
for all $n, m$. Using reflexivity, as well as Eberlein-Smulian, the sequence $(x_n)$ must have a weakly convergent subsequence. Note that this weakly convergent subsequence must also have its terms separated by at least $delta$, which means it cannot be mapped to a norm-convergent sequence (as such a sequence would have to be Cauchy). This shows the contrapositive of (c).

(d) Good.

(e) If $T$ is compact and open, then $T(B[0; 1])$ contains $T(B(0; 1))$, which is open, since $T$ is an open map. Thus, $T(B[0; 1])$ contains some ball $B[y; r]$. Since $T$ is compact, the closure of $T(B[0; 1])$ is compact, hence so is the closed subset $B[y; r]$. A scaling and translation argument yields that $Y$ has a compact unit ball, and hence is finite-dimensional.

(f) Also good.