The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)\int_{0}^{x}f(t)tdt+x\int_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f \in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
\begin{align}
(Sf)' &= (1-x)f(x)x-\int_0^xf(t)tdt-xf(x)(1-x)+\int_x^1f(t)(1-t)dt \\
&= -\int_0^x f(t)tdt+\int_x^1f(t)(1-t)dt.
\end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'\in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) \;\;\; a.e..
$$
Therefore $Sf\in\mathcal{D}(T)$ for all $f\in L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $f\in \mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]\rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
\begin{align}
\langle T_e f,g\rangle -\langle f,T_e g\rangle &= -\int_0^1 (f''g-fg'')dx \\
& = -(f'g-fg')|_{0}^{1}.
\end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $f\in\mathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) \in\mathcal{D}(T)
$$
and
\begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \\
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \\
+&E_0(f)T_e x+E_1(f)T_e(1-x)
\end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
In the case of power series, most of the results in $\mathbb{R}$ or $\mathbb{C}$ readily extends with due modifications. For example, let
$$ f(z) = \sum_{n=0}^{\infty} A_n z^n $$
be a power series in $z \in \mathbb{C}$, where $A_n$'s lie in a Banach space $(\mathscr{X}, \|\cdot\|)$. If we let
$$ R = \frac{1}{\limsup_{n\to\infty} \|A_n\|^{1/n}} \in [0, \infty], $$
then it is clear that the series converges absolutely if $|z| < R$ and diverges if $|z| > R$. That is, this $R$ is precisely the radius of convergence for $f(z)$. It is also easy to check that
$$ g(z) = \sum_{n=1}^{\infty} n A_n z^{n-1} $$
has the same radius of convergence as $f(z)$.
Now suppose $|z| < R$, and fix $r$ so that $|z| < r < R$. Also, consider any $h$ with $0 < |h| < r-|z|$. Then by using the identity
\begin{align*}
\frac{(z+h)^n - z^n}{h} - n z^{n-1}
&= n \int_{0}^{1} \bigl[ (z + ht)^{n-1} - z^{n-1} \bigr] \, \mathrm{d}t \\
&= n(n-1)h \int_{0}^{1} \int_{0}^{t} (z + hu)^{n-2} \, \mathrm{d}u\mathrm{d}t,
\end{align*}
we find that
\begin{align*}
\left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right|
&\leq n(n-1)h \int_{0}^{1} \int_{0}^{t} (|z| + |h|)^{n-2} \, \mathrm{d}u\mathrm{d}t \\
&= |h| \frac{n(n-1)}{2}(|z| + |h|)^{n-2}.
\end{align*}
Now by this estimate and the inequality $|z| + |h| < r$ altogether, it follows that
\begin{align*}
\left| \frac{f(z+h) - f(z)}{h} - g(z) \right|
&\leq \sum_{n=1}^{\infty} \|A_n\| \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| \\
&\leq \left( \sum_{n=1}^{\infty} \frac{n(n-1)}{2} \|A_n\| r^{n-2} \right) |h|
\end{align*}
and therefore $f'(z)$ exists and is equal to $g(z)$.
Best Answer
Let $ D_0 = \{ f \in L^2[0,1] | f \text{ is absolutely continous and } f'\in L^2[0,1] \} $ and $ T : D_0 \rightarrow L²[0,1], Tf=i\frac{d}{dt}f = i \cdot f'$.
Take a sequence $ \{ u_n \} $ which convergences to $ u $ in $ L^2 $- Norm and with $ { Tu_n } $ beeing convergent in $ L^2 $ with limit $ x $. We have to show that $ Tu = x $.
First we show that $ \{ u_n \} $ is uniform convergent. By Hölder's inequality
$$ \int_{0}^{t} {| i \cdot u_n'(s) - x(s)| ds} \le \int_{0}^{1} {|i \cdot u_n'(s) - x(s)| ds} \le \left( \int_{0}^{1} {ds} \right)^{\frac{1}{2}} \cdot \left( \int_{0}^{1} {|i \cdot u_n'(s) - x(s)|^2 ds} \right)^{\frac{1}{2}} = \| Tu_n - x \|_{L^2} $$
From this inequality we conclude that
$$ \| t \mapsto \int_{0}^{t} {i \cdot u_n(s)ds} - \int_0^{t}{x(s)ds} \|_{\infty} \le \| Tu_n - x \|_{L^2} $$
which gives uniform convergence of $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(t) \} $ to $ t \mapsto \int_0^{t}{x(s)ds} $. Especially $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(s)ds \} $ is a uniform Cauchy Sequence. Further we have
$$ | i \cdot u_n(0) - i \cdot u_m(0) | = \left( \int_{0}^{1} | i \cdot u_n(0) - i \cdot u_m(0) |^2 dt \right)^{\frac{1}{2}} = \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) - \int_{0}^{t} {(i \cdot u_n'(s)- i \cdot u_m'(s))ds} |^2 dt\right)^{\frac{1}{2}} \\ \le \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) |^2 dt \right)^{\frac{1}{2}} + \left( \int_{0}^{1} | \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} |^2 dt \right)^{\frac{1}{2}} $$
which implies
$$ | i \cdot u_n(0) -i\cdot u_m(0) | \le \| i \cdot u_n - i \cdot u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{L^2} \\ \le \| u_n - u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{\infty} $$
Hence $ \{ i \cdot u_n(0) \} $ is a Cauchy Sequence. Because each $ u_n $ is absolutly continous we have
$$ i \cdot u_n (t) = i \cdot u_n(0) + \int_{0}^t{i \cdot u_n'(s)ds} $$
which shows that $ \{i \cdot u_n\} $ and $ \{u_n\} $ are uniform convergent.
Now we show $ Tu = x $. Define $ g(t) = \lim_{n \rightarrow \infty} \frac{1}{i} u_n(0) + \frac{1}{i} \int_{0}^{t}{x(s)ds} $. Then we have $ Tg = x $ and $ i \cdot g' = x $ a.e. in $ [0,1] $. But $ \{ u_n \} $ convergences to g uniformly. Hence $ u = g $ a.e. in $ [0,1] $ which gives $Tu = x $.
Note: A similar reasoning can be found in Werner's Funktionalanalysis.