Let $T_n,T$ be operators on a Banach space $X$ such that $x_n\to x\implies T_n(x_n)\to T(x)$. Does $T_n\to T$ in norm?

Question

Fix some Banach space $X$, and let $T_n,T\in\mathscr{L}(X)$ be bounded linear operators on $X$, such that for any convergent sequence $x_n\to x$ in $X$, we have that $T_n(x_n)\to T(x)$. Then does $T_n\to T$ in norm?

Clearly, this property implies strong convergence, and is implied by convergence in norm. But is it equivalent to either of these properties?

Answer 1

Consider $C([0,1])$ the space of continuous functions defined on $[0,1]$ and $f_n$ defined by $f(x)=0, x\geq {1\over n}$ or $x\leq {1\over{n+1}}$. the restriction of $f$ to $[{1\over {n+1}}, {{2n+1}\over{2n(n+1)}}]$ is affine and $f({{2n+1}\over{2n(n+1)}})=1$, the restriction of $f$ to $[{{2n+1}\over{2n(n+1)}},{1\over n}]$ is affine. Then $lim_nf_n(x)=0$ and $\|f_n-0\|=sup_{x\in [0,1]}|f_n(x)-0|=1$.

Answer 2

Let $X = \ell^2$ and consider $T_nx = (0,\dots,0,x_n, x_{n+1}, \dots)$ for $x = (x_1, x_2, \dots) \in \ell^2$, together with $T = 0$. Your assumptions are true, but $T_n$ does not converge to $T$ in norm.