Let $\;\mathcal B(H,F)\;$ be the set of all bounded linear operators $\;S:H\rightarrow F\;$ where $H,F\;$ are some Hilbert spaces.
To begin with, we consider $\;T_n \in \mathcal B(H,F)\;\forall n\;$ to be a Cauchy sequence. Then $\forall x \in H\;\;\vert\vert T_nx-T_mx \vert\vert \le \vert \vert T_n-T_m \vert \vert\cdot \vert \vert x \vert \vert \;$ which implies $\;T_nx\;$ is a Cauchy sequence on $F\;$. Since $\;F$ is Banach space, one can find $\;y \in F\;$ such that $T_nx \rightarrow y$ as $n \to \infty\;$
I want to show there is $\;x \in H\;$ such that $T_x=y\;$. This is why I define $T:H \rightarrow F\;$ as $Tx=\lim_{n\rightarrow \infty} T_nx\;\forall x \in H\;$.
My question is: How can I show the above is well defined?
I took $x,z \in H\;$ such that $\;T_x=T_z\; \Rightarrow \lim_{n \to \infty} T_nx=\lim_{n \to \infty} T_nz\;$
How do I proceed? I believe the answer is obvious but I 've been stuck. I would appreciate any help. Thanks in advance!!
What you're saying is you want to show $\lim_{n\rightarrow\infty} T_n x$ exists and is unique for each $x$. You found the limit using completeness, and it is unique because of uniqueness of limits in $F$ (since it's a metric space and has all of those nice properties that topological spaces can have).