I've to proof the following statement
Let $X,Y$ be to banach spaces and $(T_k)_{k \in \mathbb{N}} \subseteq L(X,Y)$ a bounded sequence of bounded linear operators. Further it exists a dense subset $D \subset X$ of $X$, such that for all $x \in D$ the sequence $(T_kx)_k \subset Y$ is a cauchy sequece in $Y$. Proof that there exists exactly one bounded linear operator $T \in L(X,Y)$, such that $(T_k)_k$ converges strong to $T$.
But I think I've a counter example. Can somebody verify this or if its not a counter example explain to me why not?
Choose for a bounded region $\Omega \subseteq \mathbb{R}^n$ with smooth boundary $X = H^1(\Omega), Y = L^2(\partial \Omega), D = C^{\infty}_0(\Omega)$ and define for $k \in \mathbb{N}\colon T_k f:= (-1)^k T_{\partial \Omega} f$ where $T_{\partial \Omega}\colon H^1(\omega) \to L^2(\partial \Omega)$ is a sobolev trace operator which is bounded and linear. Now $\|T_k\|$ is the same for each $k$ and by that the sequence is bounded. Further $D = C^{\infty}_0(\Omega)$ is dense in $X$ with $\forall f \in D \forall k \in \mathbb{N} \colon T_k f = 0$ in $Y$.
Now choose $f \in X$ with $ \|T_{\partial \Omega} f\|_Y > \epsilon$ and see for each $k=2n$ that $\|T_k f - T_{k+1}\|_y = 2 \|T_{\partial \Omega} f\|_Y > \epsilon$ and by that the sequence cannot converge strong.
Here is a counterexample. Let $X=Y=H$ be an infinite-dimensional Hilbert space with orthonormal basis $\{e_j\}$. Let $T_k$ be the projection onto the span of $\{e_1,\ldots,e_k\}$. Let $D$ be the span of $\{e_1,e_2,\ldots\}$.
For each $x\in D$, for $k$ big enough $T_kx=x$, so the sequence $\{T_kx\}$ is Cauchy. As we have $T_kx\to x$, we have $T_k\to I$ pointwise. But the converse is not uniform: $$ \|I-T_k\|=1 $$ for all $k$.
I cannot comment on your example because I have no idea what a Sobolev trace is. But you seem to imply that it is zero in a dense subset. If a bounded operator is zero on a dense subset, it is zero everywhere.