Lp spaces set inclusion

Mar 2010
32
3
Melbourne
I am stuck on trying to show for that a fixed measure space \(\displaystyle (X,\cal{S},\mu)\) with \(\displaystyle 1 \leq r \leq p \leq s < \infty\) we have
\(\displaystyle L_{p} \subset L_{r} + L_{s} \).

Cheers!
 
Mar 2010
32
3
Melbourne
I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
Here it goes:


Let \(\displaystyle f\in L_{p}\) which means we have

\(\displaystyle \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\infty.\)

Write \(\displaystyle f:=g+h\) where the functions \(\displaystyle g\) and \(\displaystyle h\) are measurable functions.

Since the measuable function \(\displaystyle \vert\hspace{0.5mm}g+h\vert\) is integrable, it is almost everywhere finite.

Then WLOG we have

\(\displaystyle \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\text{almost everywhere}\)

from which it follows that

\(\displaystyle \Vert g\Vert_{r} := \Bigg( \int \vert g\vert^{r}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{r}}<\infty_{.}\)

Similarly for \(\displaystyle h\), we obtain

\(\displaystyle \Vert h\Vert_{s} := \Bigg( \int \vert h\vert^{s}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{s}}<\infty_{.}\)
 
Last edited:

Jose27

MHF Hall of Honor
Apr 2009
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México
I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
Here it goes:


Let \(\displaystyle f\in L_{p}\) which means we have

\(\displaystyle \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\infty.\)

Write \(\displaystyle f:=g+h\) where the functions \(\displaystyle g\)and \(\displaystyle h\) are measurable functions.

Since the measuable function \(\displaystyle \vert\hspace{0.5mm}g+h\vert\) is integrable, it is almost everywhere finite.

Then WLOG we have

\(\displaystyle \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\text{almost everywhere}\)

from which it follows that

\(\displaystyle \Vert g\Vert_{r} := \Bigg( \int \vert g\vert^{r}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{r}}<\infty_{.}\)

Similarly for \(\displaystyle h\), we obtain

\(\displaystyle \Vert h\Vert_{s} := \Bigg( \int \vert h\vert^{s}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{s}}<\infty_{.}\)
I don't like your argument, you can't say \(\displaystyle |g|\leq |g+h|\) (what if \(\displaystyle h\) is negative), and the part where you took integrals is fishy at best.

Some cases are simple enough. For example if \(\displaystyle \mu (X)<\infty\) and \(\displaystyle 1\leq p \leq q \leq \infty\) then \(\displaystyle L^q \subset L^p\) (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, \(\displaystyle X\) had no finite measure subset the inclusions would be reversed (I don't yet have a proof for this), and the general case would (?!) follow by decomposing your space in this (or some similar) fashion.
 
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Opalg

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Some cases are simple enough. For example if \(\displaystyle \mu (X)<\infty\) and \(\displaystyle 1\leq p \leq q \leq \infty\) then \(\displaystyle L^q \subset L^p\) (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, \(\displaystyle X\) had no finite measure subset the inclusions would be reversed (I don't yet have a proof for this), and the general case would (?!) follow by decomposing your space in this (or some similar) fashion.
Following up on that idea, let \(\displaystyle f\in L^p(X)\), and let \(\displaystyle Y = \{x\in X:|f(x)|\geqslant1\}\). Let \(\displaystyle g = f\big|_Y\) (that is, g equals f on Y and g is zero outside Y) and similarly \(\displaystyle h = f\big|_{X\setminus Y}\). Clearly \(\displaystyle f=g+h\). Then \(\displaystyle \mu(Y)<\infty\) and so (as Jose27 points out) \(\displaystyle g\in L^r(X)\). Also, it's easy to see that \(\displaystyle h\in L^s(X)\), because \(\displaystyle \int\!\!|h|^s = \int\!\!|h|^p|h|^{s-p}\leqslant\int\!\!|h|^p<\infty\). Thus \(\displaystyle f\in L^r(X) + L^s(X)\), as required.
 
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