# Lp spaces set inclusion

#### willy0625

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!

#### 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

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

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

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

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#### Jose27

MHF Hall of Honor
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.

• willy0625 and Opalg

#### Opalg

MHF Hall of Honor
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.

• willy0625