# Thread: Lp spaces set inclusion

1. ## Lp spaces set inclusion

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

Cheers!

2. I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
Here it goes:

Let $f\in L_{p}$ which means we have

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

Write $f:=g+h$ where the functions $g$ and $h$ are measurable functions.

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

Then WLOG we have

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

from which it follows that

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

Similarly for $h$, we obtain

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

3. 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 $f\in L_{p}$ which means we have

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

Write $f:=g+h$ where the functions $g$and $h$ are measurable functions.

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

Then WLOG we have

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

from which it follows that

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

Similarly for $h$, we obtain

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

Some cases are simple enough. For example if $\mu (X)<\infty$ and $1\leq p \leq q \leq \infty$ then $L^q \subset L^p$ (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, $X$ 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.

4. Originally Posted by Jose27
Some cases are simple enough. For example if $\mu (X)<\infty$ and $1\leq p \leq q \leq \infty$ then $L^q \subset L^p$ (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, $X$ 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 $f\in L^p(X)$, and let $Y = \{x\in X:|f(x)|\geqslant1\}$. Let $g = f\big|_Y$ (that is, g equals f on Y and g is zero outside Y) and similarly $h = f\big|_{X\setminus Y}$. Clearly $f=g+h$. Then $\mu(Y)<\infty$ and so (as Jose27 points out) $g\in L^r(X)$. Also, it's easy to see that $h\in L^s(X)$, because $\int\!\!|h|^s = \int\!\!|h|^p|h|^{s-p}\leqslant\int\!\!|h|^p<\infty$. Thus $f\in L^r(X) + L^s(X)$, as required.