Proposition. Let $\mathcal{H}=L^2[0,1]$ with a differentiation operator $T=i\frac{d}{dt}$ on it, whose domain $D(T)$ consists of all the absolutely continuous functions in $\mathcal{H}$, and the derivatives of those functions is still $L^2$. Then $T$ is a closed operator. (which means its graph is closed in $\mathcal{H}\times\mathcal{H}$)

Here is my try so far. Let $\{u_n\}$ be a sequence of $L^2$ functions in $\mathcal{H}$ converging to $u\in\mathcal{H}$. Also, $Tu_n$ converge to another $v\in\mathcal{H}$. It suffices to show that $Tu=v$, which is to say, $\| Tu-v\|=0$. Since $$\|Tu-v\|\le\|Tu-Tu_n\|+\|Tu_n-v\|$$ and $\|Tu_n-v\|\rightarrow 0$, we only need to prove that $\|Tu-Tu_n\|\rightarrow 0$. $$\lim_{n\rightarrow\infty}\|Tu-Tu_n\|^2=\lim_{n\rightarrow\infty}\int \left|i\frac{d}{dt}(u-u_n)\right|^2dx.$$ Ideally, we “may” swap the “limits” with the “integral” and “derivative” and the “proof” is finished. However, it seems to me not very clear that theorems from real analysis allow me to do so.

I’ve been stuck here for a long time. Any hints?

  • 2
    $\begingroup$ This operator does not seem well defined to me. The derivatives of absolutely continuous functions are integrable but not square integrable in general. E.g. $u(x)=\sqrt{x}$ is such that $u\in D(T)$ but $Tu\notin L^2[0,1]$. $\endgroup$
    – Lorenzo Q
    Sep 22, 2018 at 9:15
  • $\begingroup$ @LorenzoQuarisa You are right. Sorry for my mistake. What if we only consider those functions whose images is still in $\mathcal{H}$ ? $\endgroup$
    – J. Lizy
    Sep 22, 2018 at 9:25
  • $\begingroup$ You also need boundary conditions I guess (but maybe that's only for self-adjointness, I should check) $\endgroup$ Sep 22, 2018 at 9:26

2 Answers 2


Let $ D_0 = \{ f \in L^2[0,1] | f \text{ is absolutely continous and } f'\in L^2[0,1] \} $ and $ T : D_0 \rightarrow L²[0,1], Tf=i\frac{d}{dt}f = i \cdot f'$.

Take a sequence $ \{ u_n \} $ which convergences to $ u $ in $ L^2 $- Norm and with $ { Tu_n } $ beeing convergent in $ L^2 $ with limit $ x $. We have to show that $ Tu = x $.

First we show that $ \{ u_n \} $ is uniform convergent. By Hölder's inequality

$$ \int_{0}^{t} {| i \cdot u_n'(s) - x(s)| ds} \le \int_{0}^{1} {|i \cdot u_n'(s) - x(s)| ds} \le \left( \int_{0}^{1} {ds} \right)^{\frac{1}{2}} \cdot \left( \int_{0}^{1} {|i \cdot u_n'(s) - x(s)|^2 ds} \right)^{\frac{1}{2}} = \| Tu_n - x \|_{L^2} $$

From this inequality we conclude that

$$ \| t \mapsto \int_{0}^{t} {i \cdot u_n(s)ds} - \int_0^{t}{x(s)ds} \|_{\infty} \le \| Tu_n - x \|_{L^2} $$

which gives uniform convergence of $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(t) \} $ to $ t \mapsto \int_0^{t}{x(s)ds} $. Especially $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(s)ds \} $ is a uniform Cauchy Sequence. Further we have

$$ | i \cdot u_n(0) - i \cdot u_m(0) | = \left( \int_{0}^{1} | i \cdot u_n(0) - i \cdot u_m(0) |^2 dt \right)^{\frac{1}{2}} = \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) - \int_{0}^{t} {(i \cdot u_n'(s)- i \cdot u_m'(s))ds} |^2 dt\right)^{\frac{1}{2}} \\ \le \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) |^2 dt \right)^{\frac{1}{2}} + \left( \int_{0}^{1} | \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} |^2 dt \right)^{\frac{1}{2}} $$

which implies

$$ | i \cdot u_n(0) -i\cdot u_m(0) | \le \| i \cdot u_n - i \cdot u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{L^2} \\ \le \| u_n - u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{\infty} $$

Hence $ \{ i \cdot u_n(0) \} $ is a Cauchy Sequence. Because each $ u_n $ is absolutly continous we have

$$ i \cdot u_n (t) = i \cdot u_n(0) + \int_{0}^t{i \cdot u_n'(s)ds} $$

which shows that $ \{i \cdot u_n\} $ and $ \{u_n\} $ are uniform convergent.

Now we show $ Tu = x $. Define $ g(t) = \lim_{n \rightarrow \infty} \frac{1}{i} u_n(0) + \frac{1}{i} \int_{0}^{t}{x(s)ds} $. Then we have $ Tg = x $ and $ i \cdot g' = x $ a.e. in $ [0,1] $. But $ \{ u_n \} $ convergences to g uniformly. Hence $ u = g $ a.e. in $ [0,1] $ which gives $Tu = x $.

Note: A similar reasoning can be found in Werner's Funktionalanalysis.


There is a simple solution provided that you are vaguely familiar with the notion of Sobolev space and weak derivative.

Here $D(T)$ is nothing but the Sobolev space $W^{1,2}(0,1)$, which embeds into the space of absolutely continuous functions on $[0,1]$. So, your problem (after eliminating the irrelevant $i$ factor) is equivalent to proving, for a sequence $\left\{u_n\right\}\subset W^{1,2}(0,1)$, $$ (\|u_n-u\|_{L^2}\to 0\;\land\;\|u_n'-v\|_{L^2}\to 0)\Rightarrow v=u'$$ Solution: The relations $\|u_n-u\|_{L^2}\to 0$ and $\|u_n'-v\|_{L^2}\to 0$ yield respectively \begin{align*}\int_0^1 u_n\psi&\to \int_0^1u\psi,\qquad \forall \psi\in C_c^{\infty}([0,1])\\ \int_0^1u_n'\varphi&\to \int_0^1v\varphi,\qquad \forall \varphi \in C_c^{\infty}([0,1]) \end{align*} On the other hand, integrating by parts and using the first relation we obtain $$\int_0^1u_n'\varphi=-\int_0^1u_n\varphi'\to -\int_0^1u\varphi',\qquad \forall \varphi \in C_c^{\infty}([0,1])$$ and hence $$\int_0^1u\varphi'=-\int_0^1v\varphi,\qquad \forall \varphi \in C_c^{\infty}([0,1]) $$ which tells us precisely that $v$ is the weak derivative of $u$, i.e. $u'=v$ (and since $u$ is a.e. differentiable, the weak derivative is equal to the classical derivative wherever the latter is defined).


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