# Why define a function space in this fashion? (PDE and functional analysis)

Suppose that for $t \in [0,1]$, $X(t)$ is a Hilbert space of functions, eg. $X(t) = L^2(\Omega_t)$ where $\Omega_t$ is a bounded domain.

Define a space $$H := \{\overline{v}:[0,1]\to \bigcup_{t \in [0,t]}X(t)\times\{t\}, t \mapsto (v_t,t)\}$$

Abusing notation, identify $\overline{v}(t) = (v_t, t)$ with $v(t) = v_t$.

Why is this space defined like this? Presumably the idea is that $v(t)$ should be in $X(t)$ for each $t \in [0,1]$. In that case, why the need for the $\times \{t\}$ in the definition? Can anyone explain this space and abusing notation part to me a bit more? Thanks.

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When the sets $\Omega_t$ have some overlap, the corresponding spaces $L^2(\Omega_t)$ overlap as well: $L^2(\Omega_t)\cap L^2(\Omega_s)$ is naturally identified with $L^2(\Omega_t\cap \Omega_s)$. This may be useful in some situations, but apparently not here. Here we want the elements of $X(t)=L^2(\Omega_t)$ to have nothing to do with $X(s)=L^2(\Omega_s)$, even if the sets $\Omega_t,\Omega_s$ overlap or happen to be the same set. This is accomplished by marking the functions with a label $t$ which indicates which space they belong to.
This way, when you add the function $f(x)=|x|^2$ on $\Omega_2$ to the function $g(x)=|x|^3$ on $\Omega_{5}$, the result is not something that is equal to $|x|^2+|x|^3$ on $\Omega_2\cap \Omega_5$, but rather the formal combination $(f,2)+(g,5)$.
Now now that we did this, writing $(f,2)$ instead of $f$ every time looks like a hassle, especially when we do computations with $f$. So we go back to writing $f$ but meaning $(f,2)$, trying to have a cake and also to eat it.
Thank you. So this new space $H$ is not a linear space technically, right? the second argument fails us. –  soup Feb 12 '13 at 18:36
No, it is a linear space. Multiplication by scalars multiplies the first coordinate only. It could also be written as $H=\prod_{t\in [0,T]}X(t)$, direct sum with uncountably many component. –  user53153 Feb 12 '13 at 20:09