Equivalence in strong convergence of operators

Question

I am trying to see if I have that $(T_n)\in L(X)$ bounded and that $T_nx$ converges to $Tx$ for every $x$ in a dense subset of $X$ a Banach space, then $T_n$ converges strongly to $T$.

Let suppose that $D$ is the dense subset, I was able to see that if $x\in cl D$ then $T_nx$ is a Cauchy sequence so we know that it converges , but I can't see that it converges to $T_x$. Also I tried seeing by this '$$||T_nx-T_x||=||T_nx-T_nd_n+T_nd_n-T_x||\leq ||T_n||||x-d_n||+||T_nd_n-T_x||,$$ where $d_k \rightarrow x,$ but I can't seem to prove what I want. Any advice is appreciated.

Answer 1

Case 1: $\sup_n\|T_n\|=\infty.$ Then your assertion need not be true. For example Let $X=c_0, D=c_{00}$ and $T_n(x)=(x_1,2x_2,3x_3,\ldots,nx_n,0,0,\ldots)$ for $x \in X.$ Then $\|T_n\|=n \to \infty$ which means $T_n$ cannot converge strongly in $X.$ However $T_n x \to 0$ for every $x \in D.$

Case 2: Suppose $M:=\sup_n\|T_n\|<\infty.$ Since $T \in L(D,X),$ so by Lemma below it has a unique norm preserving extension $T \in L(X).$ Then $\|T\|\leq M.$

Let $x \in X.$ Since $D$ is dense in $X,$ there exists $(x_k)$ in D such that $x_k \to x.$ Let $\epsilon >0,$ then there exists $k_0 \in \mathbb{N}$ such that $$\|x_{k_0}-x\|<\frac{\epsilon}{3M}.$$ By given condition $T_n x_{k_0} \to T x_{k_0}$. Therefore there exists $n_0\in \mathbb{N}$ such that $$\|T_nx_{k_0}-Tx_{k_0}\|<\frac{\epsilon}{3}$$ for all $n \geq n_0.$ Combining

$$\begin{align*}\|T_n x-Tx\|&\leq \|T_n x-T_n x_{k_0}\|+\|T_nx_{k_0}-Tx_{k_0}\|+\|Tx_{k_0}-Tx\|\\&<M \frac{\epsilon}{3M}+\frac{\epsilon}{3}+M\frac{\epsilon}{3M}\\&=\epsilon\end{align*}$$ for all $n \geq n_0.$

Lemma: Let $X:$ normed space, $Y:$ Banach space, $D \subseteq X:$ dense and $T_0 \in L(D,Y).$ Then there exists a unique $T \in B(X,Y)$ such that $T\mid_D=T_0$ and $\|T\|=\|T_0\|.$

Answer 2

This is not true. Let $X=\ell^{2}$ and $D=\ell^{0}$ (the space of sequences that have only finite number of non-zero terms). Then $D$ is dense in $X$. Let $T=0$ and $T_n(x=(0,0,...,0,nx_n,(n+1)x_{n+1},...,2nx_{2n},0,0,...)$. Then $T_n x\ \to Tx$ for all $x \in D$ but $T_nx$ does not tend to $Tx$ when $x=(1,\frac 1 2, \frac 1 3 ,...)$.

Of course, you argument works if $\|T_n\|$ is bounded.