Closed operators integral property Let $T$ be a closed operator from $X\to X$ with domain is dense in $X$. 
I want to show that
$$
T \int_a^b f(x)dx = \int_a^b Tf(x)dx
$$
if $\|f\|_X (x) , \|Tf\|_X(x)$ is continuos functions for $x \in [a,b]$.
 A: I think this is not true. Consider $C[0,1]$ the space of all real-valued continuous functions defined on $[0,1]$ with supremum norm. Take domain of $T$, denote by $D(T)$, as the subspace of functions $f\in C[0,1]$ which have a continuous derivatives. Then $P[0,1]$ (space of polynomials) is contained in $D(T)$, and by Weierstrass Approximation Theorem, $D(T)$ is dense in $C[0,1]$. Define $T : D(T)\longrightarrow C[0,1]$ as 
$$T(f) = f',$$
where $f'$ is the derivative of the function $f$. Observe that, for $f(x) = x^n$, we have
\begin{align*}
T\left(\int_{0}^{1} x^n dx\right) = T\left(\frac{1}{n+1}\right) = 0, \text{ and }\\
\left(\int_{0}^{1} T(x^n) dx\right) = \left(\int_{0}^{1} nx^{n-1}\right) = 1.
\end{align*}
To complete the proof, we show that $T$ is a closed operator. Let $<f_n>$ in $D(T)$ be such that both $<f_n>$ and $<Tf_n>$ converge, say,
$$f_n\longrightarrow f \text{ and } Tf_n={f'}_n\longrightarrow g.$$
Since the convergence ${f'}_n\longrightarrow g$ is uniform, 
\begin{align*}
\int_{0}^{x} g(t)dt &= \int_{0}^{x}\lim_{n\to \infty} {f'}_n(t) dt = \lim_{n\to \infty}\int_{0}^{x}{f'}_n(t) dt = f(x)-f(0),\\
&\implies f(x)= f(0)+\int_{0}^{x} g(t)dt
\end{align*}
So, $f\in D(T)$ and $f' = g$. Hence, $T$ is closed operator.
