About closed graph theorem I want to show that in the closed graph theorem, the completeness of $Y$ is essential. (a.e I want to find two norm space $X,Y$ which $Y$ isn't complete and linear function $T:X\to Y$ such that $T$ is closed but isn't bounded.) I know that when $X$ isn't complete, there exist such $T$. See here. I tried to find a counter example in $C^1[0,1]$ but I couldn't find closed linear function here. Please give me a hint to solve this question.
 A: The example mentioned in the question you quoted is what you are looking for. Let us take $X=C^\infty[0,1]$ with the supremum norm, so we don't have codomain issues, and let $D:X\to X$ be the derivative. This is an unbounded operator, and it's graph is closed. Indeed, if $(x_n,x_n')\to (x,y)$ in $X\times X$, this means that $x_n\to x$, $x_n'\to y$ uniformly. Now, for some $t\in X$, 
\begin{align}
\left|\frac{x(t+h)-x(t)}h-y(t)\right|
&\leq\left|\frac{x(t+h)-x_n(t+h)}h\right|+\left|\frac{x_n(t+h)-x_n(t)}h-y(t)\right|
+\left|\frac{x_n(t)-x(t)}h\right|\\ \ \\
&\leq\frac{2\|x_n-x\|}h+\left|\frac{x_n(t+h)-x_n(t)}h-y(t)\right|\\ \ \\
&=\frac{2\|x_n-x\|}h+\left|x_n'(\xi_{h,n})-y(t)\right|  \ \ \ \ \ \ \textit{(Mean Value Theorem)}\\ \ \\
&\leq\frac{2\|x_n-x\|}h+\left|x_n'(\xi_{h,n})-x_n'(t)\right|+\left|x_n'(t)-y(t)\right|\\ \ \\
&\leq\frac{2\|x_n-x\|}h+\left|x_n'(\xi_{h,n})-x_n'(t)\right|+\left\|x_n'-y\right\|\\ \ \\
\end{align}
Now fix $\varepsilon>0$. As $x_n'$ is uniformly continuous and $\xi_{h,n}$ lies between $t$ and $t+h$, there exists $\delta>0$ such that if we take $|h|<\delta$  we get $|x_n'(\xi_{h,n})-x_n'(t)|<\varepsilon/3$. So for $|h|<\delta$, we have 
\begin{align}
\left|\frac{x(t+h)-x(t)}h-y(t)\right|
&\leq\frac{2\|x_n-x\|}h+\frac\varepsilon3+\left\|x_n'-y\right\|,
\end{align}
for every $n$. Now we can take $n$ big enough so that $\|x_n-x\|<h/6\varepsilon$ and $\|x_n'-y\|<\varepsilon/3$, so 
\begin{align}
\left|\frac{x(t+h)-x(t)}h-y(t)\right|
&\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.
\end{align}
Thus, $x'=y$, and $D$ has closed graph. 
