sequence of continuous functions on a Banach space X 
Let $X$ be a Banach space and $(T_n)_{n\in \mathbb{N}}$ be a sequence in $L(X)$. Then the following assertions are equivalent:
(a)  $(T_nx)_{n\in \mathbb{N}}$ converges for all $x\in X$.
(b)$(T_nx)_{n\in \mathbb{N}} \subset L(X)$ is bounded and $(T_nx)_{n\in \mathbb{N}}$ converges for all $x$ in some dense subset $D$ of $X$.
(c) $(T_nx)_{n\in \mathbb{N}}$ converges uniformly for all $x\in C$ and every compact set $C$ in $X$.

$L(X)$ is the space of bounded linear functions that map $X$ into $X$.
I want to prove $(a) \Rightarrow (c)\Rightarrow (b)\Rightarrow (a)$. in first step, from (a), let  $Tx=lim_{n\rightarrow \infty}T_nx$ for each $x\in X$ then i conclude $T\in L(X)$.But i don't know what to do then. Can someone suggest me?
 A: $(a) \implies (c)$
Let $(T_n)$ be the given sequence and $Tx := \lim_n T_n x$. Banach-Steinhaus gives you some constant $D > 0$ such that $\|T_n\| \leq B$ and $\|T\| \leq D$.
Furthermore let $C \subseteq X$ be a compact set and $\varepsilon > 0$. From $(a)$ you know that for each $x \in C$ you find $N_x \in \mathbb{N}$ such that for all $n \geq N_x$ you have
$$
\|T_n x - T_x \| \leq \frac{\varepsilon}{3}.
$$
Now let $y \in B_{\tfrac{\varepsilon}{3D}}(x)$. Then you get
$$
\|T_n y - T_x\| \leq \|T_n (y - x) \| + \|T_n x - Tx\| + \|T (x - y)\| \\
\leq \|T_n\| \|y  - x\| + \frac{\varepsilon}{3} + \|T\| \|x - y\|
< \varepsilon.
$$
Note that $C \subseteq \bigcup_{x \in C} B_{\tfrac{\varepsilon}{3D}}(x)$ and the fact that $C$ is compact imply that there exist $x_1,\dots,x_k$ such that 
$$
C \subseteq \bigcup_{r = 1}^k B_{\tfrac{\varepsilon}{3D}}(x_r).
$$
This implies that for all $n \geq \max\{N_{x_1},\dots,N_{x_k}\}$ and all $y \in C$ you have
$$
\|T_n y - T y \| \leq \varepsilon
$$
that is uniform convergence on the compact set $C$.
Obviously $(c) \implies (a)$ since one-point sets $\{x\}$ are compact.
$(a) \implies (b)$ is also obvious, since $X$ is a dense set and the boundedness of $\|T_n\|$ follows once again from Banach-Steinhaus.
$(b) \implies (a)$ is a straightforward $\frac{\varepsilon}{3}$-argument. You should try this on your own first.
