A linear operator $\Omega$ on $C(X)$ is closed if its graph is a closed subset of $C(X)\times C(X)$. A linear operator $\bar{\Omega}$ is called the closure of $\Omega$ if $\bar{\Omega}$ is the smallest closed extension of $\Omega$. Not every linear operator has a closure. The difficulty which may arise is that the closure of the graph of the operator may not be the graph of a linear operator. It may instead correspond to a "multi-valued" operator. An example of a linear operator with no closure is: $$D(\Omega)=\{f\in C[0,1],f'(0) \text{ exists}\}\text{ and } \Omega f(\eta)\equiv f'(0) \text{ for } f\in D(\Omega) $$ where $D(\Omega)$ is the domain of $\Omega$.
I don't understand how the closure of the graph of the operator defined above is not the graph of a linear operator?
The graph consists of pairs $(f,a)$ where $f$ is continuous, $f'(0)$ exist and $a=f'(0)$ (a constant function). Clearly $(0,0)$ belongs to this graph and hence to its closure. Now let $f_n(x)=\frac {(1-x)^{n}} n$. Then $f_n \to 0$ uniformly and $f_n'(0) \to -1$. Hence $(0,-1)$ belongs to the closure of the graph. If this closure is the graph of some operator then the second coordinate is uniquely determined by the first coordinate. Since $(0,0)$ and $(0,-1)$ are both in this set it cannopt be a graph of any operator.