# Lp spaces set inclusion

• May 11th 2010, 10:49 PM
willy0625
Lp spaces set inclusion
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!
• May 12th 2010, 01:55 PM
willy0625
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

$\displaystyle \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\inf ty.$

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

$\displaystyle \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\tex t{almost everywhere}$

from which it follows that

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

Similarly for $\displaystyle h$, we obtain

$\displaystyle \Vert h\Vert_{s} := \Bigg( \int \vert h\vert^{s}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{s}}<\i nfty_{.}$
• May 14th 2010, 07:44 PM
Jose27
Quote:

Originally Posted by willy0625
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

$\displaystyle \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\inf ty.$

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

$\displaystyle \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\tex t{almost everywhere}$

from which it follows that

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

Similarly for $\displaystyle h$, we obtain

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

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.
• May 15th 2010, 12:10 AM
Opalg
Quote:

Originally Posted by Jose27
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.