I'm helping some students study for their qualifying exam, and I wanted to double check my interpretation of a question.

Suppose we define the operator $L$ on $H=L^2\left([0,\infty)\right)$ so that

$$ Lu: = i\frac{du}{dx},\quad u\in D(L) = \left\{u\in H\big\vert u \text{ abs. cont. }\&\;u^\prime\in H \right\} $$ Note that the domain of $L$ is dense in $H$ but includes no boundary conditions!

Now, upon computing the formal adjoint $L^*$, we find that $L^* = L$. However, by writing out $\langle Lu,v\rangle$ and integrating by parts, we find that

$$ D(L^*) = \{u\in H\big\vert u \text{ abs. cont.},\; u^\prime\in H,\;\&\;u(0) = 0\} $$ So $D(L^*)\neq D(L)$, and hence $L$ is not self-adjoint. OK, fine - now, the second part of the question asks if $L^*$ is self adjoint, formally or otherwise. My answer would be that $L^*$ is again formally self-adjoint, but now $L^*$ is not self-adjoint since $D(L^*)$ is not dense in $H = L^2([0,\infty))$. Is this the correct interpretation, or could we force $L^*$ to be self adjoint by looking at a new Hilbert space $H^\prime = \overline{D(L^*)}$?

up vote 2 down vote accepted

There are a few subtle points. Typically, the domain for $L$ would consist of all absolutely continuous $f \in L^2$ such that $f' \in L^2$. Absolute continuity is required in order to integrate by parts, for example. This operator $L$ is closed, and that's useful to know in cases like this. And $\mathcal{D}(L)$ is dense in $L^2$.

The adjoint $L^*$ consists of all $f \in L^2$ for which there exists $h\in L^2$ such that $$ (Lg,f) = (g,h),\;\;\; g\in\mathcal{D}(L). $$ Because this holds for all test functions $g \in \mathcal{C}_c^{\infty}(0,\infty)$, then any such $f$ must also be absolutely continuous with $f'=h$ a.e.. You correctly deduced that $f(0)=0$ must hold. And it is the case that any such $f$ is in $\mathcal{D}(L^*)$. Because $L$ is closed and densely-defined, then so is $L^*$. (Your conclusion that $L^*$ is not densely-defined is inaccurate.)

With closed, densely-defined operators $L$, $L^*$, you also have $L^{**}=L$. As you noted, $L^*$ is formally selfadjoint, but it cannot be selfadjoint because $L^{**}=L\ne L^*$.

The operator $L$ has a full half-plane of spectrum because, for $\Re\lambda < 0$, the function $e^{\lambda x} \in \mathcal{D}(L)$ and $$ i\frac{d}{dx} e^{\lambda x} = i\lambda e^{\lambda x} $$ So every $i\lambda$ is an eigenvalue if $\Re\lambda < 0$, which gives $$ \{ \mu : \Im\mu < 0 \} \subseteq \sigma(L) \\ \implies \{ \mu : \Im\lambda \le 0 \} \subseteq \sigma(L). $$ The implication follows because $\sigma(L)$ is closed. In fact, it is true that $\sigma(L) = \{ \mu : \Im\lambda \le 0 \}$. So you have an uncountable number of eigenvalues for this operator on a separable space, which also tells you that $L$ cannot be selfadjoint. The spectrum of $L^*$ is the closed upper half plane. $(L^*-\lambda I)$ has co-dimension 1 range for $\Im\lambda > 0$, corresponding to the one-dimensional eigenspaces of $(L^*-\lambda I)$ for $\Im\lambda < 0$.

  • 1
    I forgot to state that $D(L)$ does indeed require absolute continuity. And I see my error now - of course $D(L^*)$ is dense in $H$ with respect to the $L^2$ norm, I wasn't thinking. Thanks for the response! – icurays1 Jun 30 '16 at 20:00

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