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$$ f \in C^0 ([0,1] , W^{3,2} (K) ) $$

Here $W^{n,m}$ is a Sobolev Space, and K is a subset of $\mathbb R^3$ .

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@DavideGiraudo $f$ is a function defined on $[0, \infty) \times \mathbb R^3$. – Misaj May 27 '12 at 10:15
up vote 0 down vote accepted

$f$ is a continuous map from the unit interval into the function space $W^{3,2}(K)$.

Edit (after OP made his question more precise):

If $f:[0,\infty)\times \mathbb{R}^3 \rightarrow \mathbb{R}$

then $t\mapsto f(t, .)$ may be viewed as a map from the positive real axis to maps $\mathbb{R}^3 \rightarrow \mathbb{R}$

By restriction (on both arguments), such a map gives rise to a map $[0,1]\rightarrow \{\phi: K\rightarrow \mathbb{R}\}$. Then your notation means that for this restriction the first sentence of my answer applies.

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oh I'm sorry. $f$ was a function defined on $[0,∞) \times \mathbb R^3 $. – Misaj May 27 '12 at 10:17
Then the notation says that the corresponding restriction of the first argument of $f$ to $[0,1]$, when viewed as a map from $K$ to $\mathbb{R}$, has the property I wrote down. – user20266 May 27 '12 at 10:19

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