Proving that an operator is unbounded and not closed.

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.

-

a) With the hint this is just a calculus exercise: plug in, evaluate and estimate. For example plug the appropriate formula for $f_n'$ and you should get something like (put $y=n(1-x)$) $$\int_0^1 \left(xf_n'(x)\right)^2dx=\int_0^n(n-y)^2f'(y)^2dy = \int_0^1 (n-y)^2f'(y)^2dy \geq (n-1)\| f'\|_{L^2}^2$$ which is clearly unbounded, while the $L^2$ norm of the $f_n$ is bounded (it's constant!).

b) This is easy if you know a little of Sobolev spaces: Take some $f\in W^{1,2}_0\setminus C^1$ (think of the first space as the absolutely continuous functions vanishing at $\{ 0,1\}$ with $f,f'$ square integrable), with $supp(f) \subset (0+r,1-r)$ for some $0<r<<1$, and take the regularizations $f_\epsilon = \eta_\epsilon * f$ (where $\eta_\epsilon$ are the usual mollifiers), then, for $\epsilon >0$ small enough $f_\epsilon\in Dom(T)$, $f_\epsilon \to f$ in $L^2$ and $Tf_\epsilon \to xf'$ in $L^2$, but $f\notin Dom(T)$.

For a reference on this approximation method you could take a look at DiBenedetto's book on real analysis.

Edit: Here's another approach I think works: Integration by parts, and noting that $Dom (T)$ is dense we calculate $$(g,Tf)_{L^2}=\int_0^1 g\overline{xf'} = -\int(xg'+g)\bar{f} =-(Tg+g,f)_{L^2}$$ and we notice that this calculation makes sense for $g$ an absolutely continuous function with square integrable derivative that vanishes at the endpoints so $(W_0^{1,2}=)AC_{0,2}[0,1] \subset Dom(T^*)$ and $T^*=-(T+I)$, now we note that $Dom(T^*)$ is dense, so $T^{**}$ is defined and an analogous calculation gives $AC_{0,2}[0,1] \subset Dom(T^{**})$, moreover we have $T^{**}=-(T^*+I)=T$ in $Dom(T)$ so $T$ has a closed extension, so it's closable, but then $\bar{T}=T^{**}$ which is a proper extension of $T$, so $T$ is not closed.

-
For part b) I don't actually know what a Sobolev space is. Can you think of another way to prove it? – Alex Kite May 6 '12 at 2:40
@Alex I have edited the question to add another, more 'functional-analytic' argument. – Jose27 May 6 '12 at 4:09
Sorry for the late reply and thanks very much for the answer! – Alex Kite May 10 '12 at 11:47