I am revising for my Functional Analysis exam and am stuck on the following question.
Consider the Hilbert Space $L^2([0,1]),$ and define the operator $T:D(T) \rightarrow L^2([0,1])$ by \begin{equation*} (Tf)(x) = xf'(x) \end{equation*} We take the domain to be $D(T) = \{ f \in C^1([0,1]) \, : \, f(0) = f(1) = 0 \}$
a) Show that T is unbounded. (Hint: Consider $f_n(x) = \sqrt{n}f(n[1-x]),$ where the function f is squeezed around $1$.)
b) Show that T is not closed.
I am really not sure how to do either of these. For b) I know that in order to show $T$ is not closed we have to show that the graph of $T$ given by $G(T) = \{ (f, Tf) \in D(T) \times L^2[0,1] : f \in D(T) \} $ is not closed. To show $G(T)$ is not closed we need to give a sequence $(f_n , Tf_n) \in G(T)$ such that $(f_n, Tf_n)$ converges to $(f,g)$, where $Tf = g$ but $f \notin D(T)$. I cannot think of any sequence though that satisfies this.