Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.