Prove that there exists sequence $(u_n)$ in $E$ such that $\| u_n \|_E = 1$ and $\| Tu_n\|_F \rightarrow 0.$

Question

Let $E$ and $F$ be two Banach spaces, and let $T \in K(E,F)$. Assume $\dim E = \infty$. Prove that there exists sequence $(u_n)$ in $E$ such that $\| u_n \|_E = 1$ and $\| Tu_n\|_F \rightarrow 0.$

Comments: I'm trying to do it by contradiction. I'm trying to show that there exists $C> 0$ such that $\| Tu\|_F \geq C \| u \|_E $, for all $u \in E$. With this I can show that $R(T)$ is closed and I can use the Open Application Theorem.

Answer 1

Consider $\inf \{||Tu||:||u||=1\}$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||\geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||\geq a||u||$ for all $u$. If $\{x_n\}$ is any sequence in the unit ball of $E$ than $\{Tx_n\}$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||\geq a||x_n-x_m||$ (applied to the subsequence) shows that $\{x_n\}$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.

Answer 2

Let $G=\{Tx:x\in E\}.$ Suppose $\inf \{\|Tx\|_F: 1=\|x\|_E\}=1/k>0$. Then T is injective, and $T^{-1}:G\to E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$

Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^{-1}$ is Lipschitz-continuous on $G,$ the sequence $B=(T^{-1}g_n)_n$ is Cauchy in $E$. So $B$ has a limit $x\in E.$ Now $T$ is continuous, so the sequence $(T(T^{-1}g_n))_n= A$ converges to $Tx\in G.$ So $G$ is closed in $F.$

Let $C= \{Tx:\|x\|_E\leq 1\}.$ Then $cl_F (C)\subset G.$ But $cl_F(C)$ is compact and $T^{-1}$ is continuous on $G,$ so $T^{-1}(cl_F(C))$ is also compact.

If dim($E)=\infty$ this is absurd because it means the compact Hausdorff space $T^{-1}(cl_f(C))$ has a closed non-compact subset $\{x\in E: \|x\|_E\leq 1\}$.