Strong convergence of operators

Solution 1:

Of course, $L(X,Y)$ means bounded linear operators.

That's a standard $\frac{\epsilon}{3}$ argument which does not require the linearity of $M$. Just that it be dense.

1- Assume (i) and (ii) are fulfilled.

Now take any $x\in X$. Let us prove that the sequence $\{T_nx\}$ is Cauchy. So let us take $\epsilon>0$. By density of $M$, there exists $y\in M$ such that $\|x-y\|\leq \frac{\epsilon}{3C}$. Then $$ \|T_nx-T_mx\|\leq \|T_nx-T_ny\|+ \|T_ny-T_my\|+\|T_my-T_mx\| $$ $$ \leq \|T_n\|\|x-y\|+\|T_ny-T_my\|+\|T_m\|\|x-y\| $$ $$ \leq C\frac{\epsilon}{3C}+\|T_ny-T_my\|+C\frac{\epsilon}{3C} $$ $$ =\frac{2\epsilon}{3}+\|T_ny-T_my\|. $$ Since $\{T_ny\}$ converges, we can find $N$ such that $\|T_ny-T_my\|\leq \frac{\epsilon}{3}$ for all $n,m\geq N$. Whence $\|T_nx-T_mx\|\leq \epsilon$.

Since $Y$ is complete, it follows that $\{T_nx\}$ converges to some $Tx$ for every $x\in X$. Since the pointwise limit of a sequence of linear operators is linear, $T$ is linear. And since $\|T_nx \|\leq C\|x\|$ for every $x$, we get $\|Tx\|\leq C\|x\|$ as well at the limit. Whence $T$ is bounded. This direction requires that $Y$ be complete, but not $X$.

2- Conversely, it $\{T_n\}$ converges strongly, then it must be uniformly bounded by the uniform boundedness principle (and its limit must be bounded). So (i) and (ii) are satisfied by $M=X$. This direction requires that $X$ be complete. But not $Y$.

Solution 2:

No, it need not be assumed that $M$ is a linear subspace. Without making that assumption, one can reduce to the case where $M$ is a linear subspace by noting that $\{x\in X:(T_nx)\text{ converges}\}$ is a linear subspace of $X$ containing $M$, and therefore it is a dense linear subspace of $X$. I think that making this reduction makes the proof clearer, but it is not necessary to put it in the statement of the lemma.