Let $V$ be a Banach space and $(0,T)$ a time interval. Consider the space $C_0^\infty(0,T)$ of infinitely often differentiable functions with values in $\mathbb R$ and compact support in $(0,T)$ and the space $L^2(0,T;V)$ of square Bochner integrable functions.

It holds that $C_0^\infty(0,T)$ is dense in $L^2(0,T;\mathbb R)$, see, e.g., Cor. 4.23 in Brezi's book on Sobolev Spaces and PDEs .

Does a related result hold in the Bochner space setting, namely, is

$$ C_0^\infty(0,T)\cdot V := \left \{ f_v\in \bigl((0,T)\to V \bigr) :\quad f_v=\phi v, \quad v \in V,\quad \phi \in C_0^\infty(0,T)\right \} $$ dense in $L^2(0,T;V)$?

  • $\begingroup$ I know it's a bit late but you might want to check the book of Ben Schweizer "Partielle Differentialgleichungen". In Addition there is a nice pde script of John Hunter: math.ucdavis.edu/~hunter/pdes/pde_notes.pdf where also Bochner integrals are introduced $\endgroup$ – Quickbeam2k1 Apr 10 '14 at 10:20

No. First you should observe that $C_0^\infty(0,T) \cdot V$ is not a subspace. For two (not colinear) vectors $v_1, v_2$, the function $f(t) = \chi_{(0,1/2)}(t) \, v_1 + \chi_{(1/2,1)}(t) \, v_2$ is cannot be approximated.

The set of all finite linear combinations of functions from $C_0^\infty(0,T) \cdot V$ is dense in $L^2(0,T; V)$.

  • $\begingroup$ Yes, true. Fortunately, the linear hull of $C_0^\infty(0,T)\cdot V$ is still OK for my purposes. Do you know a reference to the latter claim? $\endgroup$ – Jan Oct 7 '13 at 8:17
  • $\begingroup$ Actually, I am trying to prove that the definitions of weak derivatives $\dot v \in L^2(0,T;V')$ of $v\in L^2(0,T;V)$ in the weak sense and in the 'pointwise' sense via $\langle \dot v(\cdot), w \rangle \colon (0,T) \to \mathbb R$, for all $w\in V$, are equivalent. Cf. this question $\endgroup$ – Jan Oct 7 '13 at 8:38
  • $\begingroup$ No, I do not know a reference. However, it should be easy to prove by using an argument involving mollifiers. $\endgroup$ – gerw Oct 7 '13 at 11:14

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.