Closed operator

Remarks

Equivalently, $T$ is closed when, for every sequence $(x_n)$ in $D(T)$ with $x_n \to x$ in $X$ and $T x_n \to y$ in $Y$, one has $x \in D(T)$ and $Tx = y$.

Example: on $X = Y = C[0, 1]$ with the supremum norm, let $D(T) = C^1[0, 1]$ and $Tf = f'$. If $f_n \to f$ and $f_n' \to g$ uniformly, then $f_n(x) = f_n(0) + \int_0^x f_n'$ gives $f(x) = f(0) + \int_0^x g$, so $f \in C^1[0, 1]$ and $f' = g$. Hence $T$ is closed, although it is not bounded.

Non-example: the same formula with $D(S)$ the space of polynomials is not closed. The Taylor polynomials $p_n$ of $e^x$ converge uniformly to $e^x$ and $p_n' = p_{n-1}$ converges to $e^x$, but $e^x$ is not a polynomial, so it is not in $D(S)$.

Every bounded operator defined on all of $X$ is closed. If $X$ and $Y$ are Banach spaces and $T$ is closed with $D(T) = X$, then $T$ is bounded, by the closed graph theorem.