# Need explanation of passage about Lebesgue/Bochner space

From a book:

Let $V$ be Banach and $g \in L^2(0,T;V')$. For every $v \in V$, it holds that $$\langle g(t), v \rangle_{V',V} = 0\tag{1}$$ for almost every $t \in [0,T]$.

What I don't understand is the following:

This is equivalent to $$\langle g(t), v(t) \rangle_{V',V} = 0\tag{2}$$ for all $v \in L^2(0,T;V)$ and for almost every $t \in [0,T]$.

OK, so if $v \in L^2(0,T;V)$, $v(t) \in V$, so (2) follows from (1). How about the reverse? Also is my reasoning really right? I am worried about the "for almost every $t$ part of these statements, it confuses me whether I am thinking correctly.

Edit for the bounty: as Tomas' comment below, is the null set where (1) and (2) are non-zero the same for every $v$? If not, is this a problem? More details would be appreciated.

For $(2)$ implies $(1)$, consider the function $v\in L^2(0,T;V)$ defined by $$v(t)=w, \forall\ t\in [0,T]$$

where $w\in V$ is fixed. Hence, you have by $(2)$ that $$\langle g(t),v(t)\rangle=\langle g(t),w\rangle=0$$

for almost $t$. By varying $w$, you can conclude.

• Thank you. Is my proof of (1) implies (2) right? Maybe i am overthinking the issue a bit. – michael_faber Jun 12 '13 at 19:36
• I hhave a doubt: The null set in $[0,T]$ where $\langle g(t),v\rangle \neq 0$ is the same for all $v$? – Tomás Jun 12 '13 at 20:17
• I am not sure. If they're not the same for all $v$ is there a problem? – michael_faber Jun 12 '13 at 20:27
• I think that if they are not the same for all $v$, then $(1)$ does not implies $(2)$, but I have to think more. Nevertheless, if they are the same, then your argument is right and it is simple like this. – Tomás Jun 12 '13 at 20:32

Your reasoning for (1) implies (2) is almost correct: $v\in L^2(0,T;V)$ implies $v(t)\in V$ only for almost every $t$. The function $v(t) = (t-a)^{-1/3} \in L^2(0,T;\mathbb R)$ for some number $a\in [0,T]$ serves as an example. Consequently, for a given $v\in L^2(0,T;V)$, we also have to exclude the points where $v(t)\not\in V$.

Nevertheless, (2) requires the statement for all $v$, such that the remark above is just a technicality.