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Bentley.STAAD.Pro.V8i.20.with.crack.rar
Bentley STAAD.Pro V8i (SELECTSeries 6) Full Crack to perform robust concrete design and analysis of columns, beams, walls, and floors in compliance with
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عنوان خریدن ‘Bentley Staad Pro V8i SELECTseries 6’
Bentley.STAAD.Pro.V8i.20.with.crack.rar
Bentley STAAD.Pro V8i (SELECTSeries 6) Full Crack to perform robust concrete design and analysis of columns, beams, walls, and floors in compliance withQ:
Why is the image of L^2 in L^1 still in L^1?
The following is from the book “Introduction to the Analysis of the Heart Disease”
I know we can prove that if $f \in L^p, p \in (1, \infty)$ and $g \in L^q$, where $1/p + 1/q =1$, the image $g(f)$ has to be in $L^1$ by the $L^1$ Domination Theorem. This is to prove that if $g \in L^1$ and $f = g$ a.e, then $g \in L^2$.
But I’m wondering why if $g \in L^1$ and $f \in L^2$, we still have $g(f) \in L^1$? In the book it says “It is known that L^2 is continuously embeddable in L^1”. So my question is, when we say the image of a function is in a space, is it continuous embedding or we just mean we have a injection?
A:
$f$ is certainly in $L^2$,
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