1
$\begingroup$

I am reading a paper.

They define $ L_{p,q}(Q) = \{ u \in L_p((0,T); L_q(Y)) : u(t, \cdot) = 0 \text{ on } Y \backslash Y_t \text{ for a.e. $t \in (0,T)$}\}$ with norm $\lVert u \rVert_{L_{p,q}(Q)} = \left(\int_0^T \lVert u(x) \rVert^p_{L_q(Y_t)} dx\right)^{1\over p}$ for $p < \infty.$

They write

Since $q < \infty$, the function $x \mapsto \lVert u(x, \cdot) \rVert_{L_q(Y_t)}$ is measurable by Fubini's theorem. Thus, the space $L_{p,q}(Q)$ is well-defined.

Can someone explain this to me? By well-defined I guess they want to show that that norm inside the integral in the norm of the space $L_{p,q}(Q)$ exists. Is that right? I don't see how Fubini's theorem tells us that that function is measureable. And I guess measurability implies that it can be integrated over $(0,T)$ like in the norm?

  • 0
    @LeonidKovalev $u \in L_p((0,T); L_q(Y))$ if \int_0^T{\lVert u \rVert_{L_q(Y)}^p} < \infty. It's just the usual Bochner space.2012-07-05

1 Answers 1

1

The issue they deal with is making sense of the condition "$u(\cdot, t)=0$ on $Y\setminus Y_t$ for a.e. $t$". Given that the elements of the Bochner space are equivalence classes of functions from an uncountable set into a set of equivalence classes of functions, it's reasonable to exercise some caution. One way to make this condition precise is to introduce the characteristic function $\chi(x,t)=1$ if $x\in Y_t$ and $=0$ if $x\notin Y_t$. Then we can say that $u\in L_{p,q}$ iff $u=u\chi$ in the Bochner space, and the norm is the usual Bochner norm. However, we need $\chi$ to be measurable in the product $\sigma$-algebra, and this is where we need to know something about $Y_t$ (and apply Fubini).

You did not tell us anything about $Y_t$.

  • 0
    Ah, great. Thanks!2012-07-05