$\epsilon$/3 - Proof for converging sequence of functions I tried to prove the following statement and would like to ask if it's okay like that. 
$\underline{Statement}$: 
Let [B,$||.||_B$] be a complete normed linear space, $T_n :B \rightarrow B$ a sequence of linear maps which are bounded uniformly in n. Let $D \subset B$ be a dense subset of B and $T_nx \rightarrow Tx$ if $x \in D$.
Then $T_nx \rightarrow Tx \ \ \forall x\in B $.
$\underline{Proof}:$ 
$(x_n)_n$ Cauchy sequence in B and B complete $\curvearrowright$ $x_n \rightarrow x \in B$,   so that with $j>j_0$: $$ ||T_nx_j - T_nx||_B \leq ||T_n||~||x_j - x||_B < \frac{\epsilon}{3} \ \ (1) $$
$(x_n)_n$ Cauchy sequence in B and $j,k > k_0$, then: $$ ||T_nx_j - T_nx_k||_B \leq ||T_n||~||x_j - x_k||_B < \frac{\epsilon}{3} \ \ (2) $$
D dense in B $\curvearrowright \forall x \in B \ \exists (y_j) \subset D:~ \lim_{j \to \infty} y_j = x $; 
with $y_j \in D$, $\epsilon$ fixed and $ n>n_0 \curvearrowright ||T_ny_j - Ty_j||_B < \frac{\epsilon}{3}$ so that $$ T_ny_j \rightarrow T_nx \ \ (3) $$
Pick any fixed $x \in B$, J = max{$j_0,k_0$}, then for $j,k > J$, $n > n_0$
$$ ||T_nx - Tx||_B \leq \underbrace{||T_nx-T_nx_j||_B}_{<\frac{\epsilon}{3},~(1)} + \underbrace{||T_nx_j - T_ny_j||_B}_{<\frac{\epsilon}{3},~(2)} + \underbrace{||T_ny_j - Tx||_B}_{<\frac{\epsilon}{3},~(3)} < \epsilon $$ 
 A: Your argument is a little hard to follow. Try to write in complete sentences. State for which domains do the inequalities hold. For instance, in $(1)$ and $(2)$, do they work for every $n$ or some particular $n$? Also, you don't seem to use the hypothesis that the $(T_n)$ are uniformly bounded. If not, then how are you getting $(1)$ and $(2)$? Near inequality $(3)$, does your $n_0$ depend on $j$? What is the purpose op picking $k$ at the end?
Here's a similar epsilonic argument.
Proof:
Fix $x\in B$ and let $\varepsilon>0$. Set $M:=\sup_n\|T_n\|<\infty$. Due to the density of $D$, there exists a sequence $(x_j)$ in $D$ such that $x_j\to x$. Then there is $j_0$ such that every $n\in\mathbb{N}$ and $j\geq j_0$ satisfies
$$ \|T_nx_j - T_nx\| \leq \|T_n\|\|x_j-x\|
   \leq M\|x_j - x\| < \varepsilon/3. \tag{1}$$
Since the sequence $(x_j)$ consists of elements in $D$ and $T_nd\to Td$ for all $d\in D$, we have that for each $j$ there exists $n_j$ such that if $n\geq n_j$, then
$$ \|T_nx_j-T_nx\| < \varepsilon/3. \tag{2}$$
Because $T$ is bounded (assuming this), then there is $j_1$ such that $j\geq j_1$ implies
$$ \|Tx_j-Tx\| < \varepsilon/3. \tag{3} $$
If $j\geq\max\{j_0,j_1\}$ and $n\geq n_j$, then we obtain from $(1)$, $(2)$, and $(3)$ that
\begin{align*}
\|T_nx-Tx\| &\leq \|T_nx-T_nx_j\|+\|T_nx_j-Tx_j\|+\|Tx_j-Tx\| \\
   &< \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\
   &= \varepsilon \tag*{$\square$}
\end{align*}

Here's almost the same argument, except this one avoids all $\varepsilon$'s.
Proof:
Suppose $x\in B$. By the fact that $D$ is dense there exists a sequence $(x_k)$ in $D$ such that $x_k\to x$. Set $M:=\sup_{n}\|T_n\|$, which exists by hypothesis. Then, assuming that $T$ is also bounded and linear, we obtain
\begin{align*}
\|T_n x - Tx\| &\leq \|T_nx - T_nx_k\|+\|T_nx_k-Tx_k\|+\|Tx_k-Tx\| \\
&\leq M\|x-x_k\| + \|T_nx_k-Tx_k\|+\|T\|\|x_k-x\|
\end{align*}
for every $k\in\mathbb{N}$.
Taking the the limit supremum of the above inequality as $n\to\infty$ yields
$$
\limsup_{n\to\infty}\|T_nx-Tx\| \leq M\|x-x_k\|+\|T\||x_k-x||
$$
for every $k\in\mathbb{N}$, because $\|T_nx_k-Tx_k\|\to0$ as $n\to\infty$ by hypothesis. Finally, taking $k\to\infty$ implies
$$
\limsup_{n\to\infty}\|T_nx-Tx\|\leq \lim_{k\to\infty}(M\|x-x_k\|+\|T\|\|x_k-x\|)=0.
$$
Therefore $T_nx\to Tx$ for every $x\in B$. $\square$

In both proofs, there was no use of the assumption that $B$ is complete. We did however assume that $T$ was bounded and linear. Under certain hypothesis, the fact that $B$ is complete could imply that $T$ is bounded and linear on $B$. If you can clear up what we are allowed to assume about $T$, then I'll happily edit my post and add this part.
