Let $V$ be separable and Hilbert. Let $\mathcal V = L^2(0,T;V)$. Assume for each $t \in [0,T]$, $$a(t;\cdot,\cdot):V \times V \to \mathbb{R}$$ is continuous and bilinear.

Or equivalently, we have $A(t) \in \mathcal L(V,V')$ with $$\langle A(t)u,v \rangle = a(t;u,v)$$ such that $a(\cdot,u,v) \in L^\infty(0,T).$ We also have that $a(t;\cdot,\cdot)$ is bounded in $V \times V$.

I wish to show that $$t \mapsto A(t)u(t)$$ is measureable for every $u \in \mathcal V$.

Proof: It suffices to show, by Pettis thoerem, that $t \mapsto A(t)u(t)$ is weakly measurable since $V'$ is separable and Banach. So let $g \in V''=V$. Then $$\langle g, A(t)u(t) \rangle_{V'',V'} = \langle A(t)u(t), g \rangle_{V',V} = a(t;u(t),g)$$ which by assumption is measurable since it is in $L^\infty(0,T)$.

Hence $$t \mapsto A(t)u(t)$$ is measurable.

What is wrong with this proof? Because in Showalter, the author proves it like this: enter image description here "The uniform bound above" refers to the bound on $a(t;\cdot,\cdot).$ Theorem 1.11 refers to Pettis' theorem.


1 Answer 1


The assumption is that for any fixed $u,v\in V$, the map $t\mapsto a(t,u,v)$ is in $L^\infty (0,T)$ (and hence measurable). In your proof, you have $a(t,u(t),v)$, so (a priori) you cannot directly conclude that this is a measurable map.

On the other hand, the proof in Showalter works because he writes the thing as $B(A(t)^*v, u(t))$ where $B$ is the scalar product. Both maps $A(t)v^*$ and $u(t)$ are measurable and $B$ is jointly continuous, so the whole thing is measurable.

If you want to show directly that $a(t,u(t),v)$ is measurable, you can certainly do it by choosing a sequence of simple functions converging to $u$.

  • $\begingroup$ You really shouldn't write entire words in mathmode, it makes the spacing all messed up and makes it harder to tell math stuff apart. If you want to get italics for emphasis, enclose the word (or phrase) in asterisks. $\endgroup$
    – tomasz
    Jul 26, 2013 at 12:29
  • $\begingroup$ That works fine, thanks. $\endgroup$
    – Etienne
    Jul 26, 2013 at 12:40

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