3
$\begingroup$

For example, let's suppose $H=L^{2}[0,1]$ and our operator $T:D(T)\subset H\to H$ is defined by

$$T:=i\frac{d}{dx},\qquad D(T):=C^{1}_{0}(0,1).$$

It's easy enough to show analytically that this operator is symmetric (i.e. $T\subset T^{\ast}$). But the rest of the arguments; even proving that $T$ is unbounded and densely defined is very heuristic.

Certainly, to show that $T$ is unbounded we can define

$$u_{n}(x):=\begin{cases}1-nx,&\qquad 0\le x\le 1/n,\\ 0,&\qquad 1/n<x\le 1,\end{cases}$$ and this does the job, since $\|T\|=\sup\|Tu_{n}\|/\|u_{n}\|=\sqrt{3}n>n\to\infty$ as $n\to\infty$.

To prove that $T$ is densely defined we can appeal to an argument whereby $D(T)$ contains an orthonormal sequence (involving Hermite polynomials) and this sequence is total in $H$ so that $D(T)$ is dense.

As if this wasn't bad enough, in order to prove that $T^{\ast}$ (which obviously exists since $T$ is densely defined) is given by

$$T^{\ast}=i\frac{d}{dx},\qquad D(T^{\ast})=\{u\in AC[0,1]:u'\in L^{2}[0,1]\},$$

we have to again consider an arbitrary subinterval of $[0,1]$ on which we construct a sequence $(u_{n})_{n\in\mathbb{N}}\subset D(T)$ which converges to the indicator function of our arbitrary interval and et cetera...et cetera...

So my question is if someone can provide me with some analytical intuition for the aforementioned steps (proving unboundedness, densely definedness and even determining the adjoint...although that's a lot of work)? At the moment it just seems very heuristic.

$\endgroup$
  • $\begingroup$ I wonder what meaning you attribute to the word heuristic. $\endgroup$ – Mariano Suárez-Álvarez Jan 7 '17 at 21:01
  • $\begingroup$ @MarianoSuárez-Álvarez By heuristic I mean a "guessing" type approach (although in this case it is more constructive), rather than analytical deduction. $\endgroup$ – Jason Born Jan 7 '17 at 21:05
  • $\begingroup$ "Analytical deduction" does not exist. Everything we do is guided by our familiarity with the objects with which we deal. What you call "guessing" is the result of having become familiar with something to the point that you sort of expect it to behave in some way, much like you learn to know a sibling or your car. $\endgroup$ – Mariano Suárez-Álvarez Jan 7 '17 at 21:10
  • $\begingroup$ @MarianoSuárez-Álvarez I disagree. There is clearly a difference between determining the symmetry of the operator via a runover of the definition (and which remains fairly consistent for all symmetric operators), rather than determining its adjoint which is more heuristic in the way that I mean it. In any case, this seems a matter of semantics. $\endgroup$ – Jason Born Jan 7 '17 at 21:14
  • $\begingroup$ One does not 'analytically deduce" anything. One observes how things work —your question is essentially about integration by parts, for example— sees patterns, builds expectations about how other things work and if one is lucky, you get to formulate a theorem. A theorem is nothing but a very neat collection of examples, the subsumption of observations one has made. $\endgroup$ – Mariano Suárez-Álvarez Jan 7 '17 at 21:18
4
$\begingroup$

The density of $C_0^1(0,1)$ in $L^2[0,1]$ can be seen directly if you use measure theory for such an argument. This is because the space is defined in terms of Lebesgue measure. For example, consider the sets $S\subset(0,1)$ whose characteristic functions $\chi_{S}$ are $L^2$ limits of a sequence $\{f_n \}\subset C_0^1(0,1)$. This collection of sets is easily seen to include subintervals of $(0,1)$. Then show that this is a sigma algebra, which gets you up to the Borel subsets. You don't need to go beyond that because the equivalence classes of $L^2[0,1]$ don't distinguish beyond that.

Symmetry gives you a lot. For example, the symmetry of $T$ forces $T$ to be closable, which means that the closure of the graph of $T$ in $L^2\times L^2$ is the graph of an operator $\overline{T}$, which is its closure. And the closure of $T$ remains symmetric. So the adjoint $T^*$ of $T$ is closed and densely-defined, with $\mathcal{D}(\overline{T})\subseteq\mathcal{D}(T^*)$. And $g\in\mathcal{D}(T^*)$ iff there exists $h$ such that $$ \langle Tf,g\rangle =\langle f,h\rangle,\;\;\; f\in C_0^1(0,1). $$ You can figure out what $h=T^*g$ has to be by carefully choosing the test functions $f$. For example, you can use limits of $C_0^1$ functions to work up to allowing the use of a function $f_{\epsilon,\epsilon'}$ that is $1$ on $[a,b]\subset(0,1)$, is $0$ on $(0,a-\epsilon)\cup(b+\epsilon',1)$ and is linear and continuous on $[a-\epsilon,a]$, $[b,b+\epsilon']$. When you plug this $f_{\epsilon,\epsilon'}$ into the above equation you get $$ i\frac{1}{\epsilon}\int_{a-\epsilon}^{a}\overline{g(t)}dt-i\frac{1}{\epsilon'}\int_{b}^{b+\epsilon'}\overline{g(t)}dt = \int_{a-\epsilon}^{b+\epsilon}f_{\epsilon,\epsilon'}(t)\overline{h(t)}dt. $$ As $\epsilon,\epsilon' \downarrow 0$, the right side has a limit $\int_{a}^{b}\overline{h(t)}dt$. The limit definitely exists. So the limits of the left exist as well, and this is true of any $a,b \in (0,1)$. By the Lebesgue differentiation theorem, the limits on the left give values of $g$ a.e.. So, for almost every $a$, $b$, $$ i\overline{g(a)}-i\overline{g(b)}=\int_{a}^{b}\overline{h(t)}dt, \\ ig(b)-ig(a) = \int_{a}^{b}h(t)dt $$ Threfore $g$ may be modified on a set of measure $0$ to be absolutely continuous, as it is the integral of an $L^2$ function. And $h=ig'$ a.e.. Therefore it is necessary that every $g \in \mathcal{D}(T^*)$ be equal a.e. to an absolutely continuous with $g'\in L^2$ and $h=T^*g = ig'$. Then, you can integrate by parts to see that all such $h$ are valid in the adjoint equation. So that characterizes $T^*h=ih'$ on the domain $\mathcal{D}(T^*)$ consisting of all absolutely continuous $h$ such that $h \in L^2$ and $h'\in L^2$.

Working up to partial differential operators is very tedious, no matter how you handle it. The subject is inherently difficult. Already the notion of absolute continuity comes up, which is not an easy measure theory topic. You can see, however, the absolute continuous functions with derivatives in $L^2$ is the correct domain for the derivative operator because of the adjoint relation. The ideas used in this case serve as a general template for other cases: extract as much information as you can from the adjoint relation by careful use of test functions in the adjoint relation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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