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Let $E$ and $F$ be Banach spaces. Let $f(t)$ be a measurable map takes value in the space of bounded linear operator from $E$ to $F$, i.e. $f(t)\in L(E,F)$ for any $t\in Dom(f)\subset E$.

Now, I see that the range of $f(t)$ is the set of all elements in $L(E,F)$. This implies $f(t)$ may takes value in $F$? What is the domain of $f(t)$? $E$?

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    Your "i.e." is wrong, I think. What's true is that for any $x $ in the (unspecified) domain of $f$, we have $f(x) \in L(E, F)$. That means that if $e \in E$, then $f(x)(e) \in F$. – 2017-01-07
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    I edited my first sentence? Correct? In other word of what you said, $f$ does take value in $F$? – 2017-01-07
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    It is still wrong to say that $f$ takes values in $F$, for $f(x)$, a typical "value taken by $f$", is in $L(E, F)$, not in $F$. – 2017-01-07

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