Q:

Norm vector spaces and Banach Space

Suppose that $\{T_j\}_{1}^{\infty}$ is a sequence of bounded linear transformations from a normed vector space $X$ into a Banach space $Y$, suppose $\lVert T_j\rVert\leq M < \infty$ for all $j$ and suppose there is a dense set $E\subset X$ such that $\{T_k(x)\}_{1}^{\infty}$ converges for every $x\in E$. Prove that $\{T_j(x)\}_{1}^{\infty}$ converges for $x\in X$.
Attempted proof: Let $x\in X$. We know $Y$ is complete, therefore if we show $\{T_j(x)\}_{1}^{\infty}$ is Cauchy then we are done. 
Let $\epsilon > 0$, choose an $N$ such that $$\lVert T_m(x) - T_n(x) \rVert < \epsilon \ \forall m,n\geq N$$ Since $E$ is dense in $X$ and $\{T_j(x')\}$ is convergent, we can choose an $x'\in E$ with $\lVert x - x'\rVert < \frac{\epsilon}{3M}$, and find an $N$ such that $$\lVert T_m(x') - T_n(x')\rVert < \frac{\epsilon}{3} \ \forall m,n\geq N$$ Then 
\begin{align*}
\lVert T_m(x) - T_n(x)\rVert &\leq \lVert T_m(x - x')\rVert + \lVert T_m(x') - T_n(x')\rVert + \lVert T_n(x' - x)\rVert\\
&< 2M\lVert x - x'\rVert + \frac{\epsilon}{3}\\
&< \frac{2\epsilon}{3} + \frac{\epsilon}{3}\\
&= \epsilon
\end{align*}

A:

Let $x\in X$. Then there is $y\in E$ so that $\|x-y\|_X$ is small.  Note
$$
\begin{split}
\| T_j x - T_k x\|_Y &= \| (T_j -T_k)x \| \\
&\le \|(T_j - T_k) (x-y) \|_Y + \|T_j y- T_ky \|_Y \\
&\le \| T_j - T_k\| \cdot \|x-y\|_X + \|T_j y- T_ky \|_Y
\end{split}
$$
Can you finish from here? 