I am trying to understand to what extent the following result of Hille is an extension of the usual theorems on differentiation under the integral sign.

Theorem (Hille). Let $(\Omega,\Sigma,\mu)$ be a measure space and $X$ and $Y$ be Banach spaces. Let $f:\Omega \to X$ be a Bochner-integrable function and $T:X \to Y$ a closed linear operator defined on the domain $\mathscr{D}(T) \subseteq X$. If the image of $f$ is (almost-everywhere) contained in $\mathscr{D}(T)$ and if $Tf:\Omega \to Y$ is Bochner integrable, then $\int_\Omega f \, d\mu \in \mathscr{D}(T)$ and \begin{equation} (1) \, \, \, \, \, \, \, \, \, T \int_\Omega f \, d\mu = \int_\Omega Tf \,d\mu. \end{equation}

While (1) is trivial to show for bounded linear operators, the differential operator $Tf = f'$ defined on $\mathscr{D}(T) = \mathscr{C}^1([0,1]) \subseteq \mathscr{C}([0,1])$ into $\mathscr{C}([0,1])$ is unbounded. As such, an extension to unbounded operators, such as above, is necessary to formulate a differentiation-under-the-integral-sign theorem for the Bochner integral. On the other hand, all linear partial differential operators \begin{equation} D = \sum_{|\alpha| \leq a} c_\alpha(x) D^\alpha \end{equation} with smooth coefficients $c_\alpha \in \mathscr{C}^a(\Omega)$ are closable as operators defined on $\mathscr{D}(D) = \{f \in \mathscr{C}^a(\Omega) \cap L^2(\Omega) : Df \in L^2(\Omega)\}$ into $L^2(\Omega)$ (as Liviu points out in the comment section), and so the above theorem of Hille is a reasonable instance of differentiation-under-the-integral-sign theorems in the finite-dimensional case.

Question. Is there an analogue of the above result on linear differential operators with smooth coefficients for infinite-dimensional differential operators in the sense of Gâteaux derivative / Fréchet derivative? In other words, is (1) applicable for enough differential operators on Banach-valued functions to consider it a differentiation-under-the-integral-sign-type result?


The question needs to be formulated more precisely. First a comment.

Any unbounded operator

$$ A: D(A)\subset E\to F $$

$E,F$ Banach spaces, can also be viewed as a bounded operator

$$ D(A)\to F $$

where $D(A)$ is equipped with the graph-norm $\vert-\vert_G$

$$\vert x\vert_G = \vert x\vert_E+\vert Ax\vert_F. $$

The operator is closed iff $(D(A),\vert-\vert_G)$ is Banach.

The above example illustrates the important role played by the domain of the operator.

Recall what closeness entails: if $(x_n)$ is a sequence in $D(A)$ such that

$$ x_n\to x_\infty\;\;\mbox{in}\;\;\vert-\vert_E, $$

$$ Ax_n \to y_\infty\;\;\mbox{in}\;\;\vert-\vert_F, $$

then $x_\infty \in D(A)$ and $y_\infty=A x_\infty$. Thus, the choice of the norms $\vert-\vert_{E,F}$ is also relevant to your question.

You mention that a differential operator with constant coefficients is unbounded and closed with respect to the uniform norm, but you give no suggestion of what its domain ought to be. To see that this issue is more subtle take a look at the following simple example.

Consider the operator $\newcommand{\pa}{\partial}$ $\newcommand{\bR}{\mathbb{R}}$ $A=\pa^2_x-\pa^2_y-\pa_z^2$ whose domain is a subspace $D(A)\subset C^0(\bR^3)$. Here I think that $E=F=C^0(\bR^3)$ with the usual $\sup$-norm. Can you think of a choice of $D(A)$ that will make this a closed operator closed?

Note that if you think of $A$ as an operator $C^2(\bR^3)\to C^0(\bR^3)$ it is obviously bounded. However, if you choose $D(A)= C^2(\bR^3)\subset C^0(\bR^3)$, it is not clear to me that the resulting operator

$$ A: D(A)\subset C^0(\bR^3)\to C^0(\bR^3), $$

is closed.

The above concrete operator was not chosen at random. The operator $A$ is a hyperbolic operator, and for such operators strange things could happen.

  • $\begingroup$ Although the domain of a hyperbolic differential operator is difficult to characterize, such operators are always closable. The reason is quite simple: If $x_n$ converges in $E$, then $Ax_n$ converges to $Ax$ in some weaker sense, e.g. in the sense of distributions. But if $Ax_n$ converges to $Ax$ in the sense of distributions, it cannot converge to something else in $E$. $\endgroup$ Nov 17 '13 at 16:22
  • $\begingroup$ It is certainly true that any partial differential operator with smooth coefficients is closable. $\endgroup$ Nov 17 '13 at 16:46
  • $\begingroup$ I'll try to make the question more precise. In the meantime: is it easy to show that any PDO with smooth coefficients is closable? $\endgroup$
    – Mark Kim
    Nov 17 '13 at 17:13
  • 1
    $\begingroup$ Have a look at Section II.6 of Yosida's book on functional analysis. There he proves that any pdo with sufficiently differentiable coefficients is closable. $\endgroup$ Nov 17 '13 at 17:19

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.