Closed operators integral property

Question

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 $x \in [a,b]$.

Answer 1

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 $$ in $D(T)$ be such that both $$ and $$ converge, say, $$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.