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There is a corollary in my textbook that says,

Let E be measurable, $m(E) < \infty$ , and $1 \leq p_1 < p_2 \leq 1$. Then $L_{p_2}(E) \subseteq L_{p_1}(E)$. Furthermore, $||f||_{p_1} \leq c||f||_{p_2}$ for all f in $L^{p_2}$ where $c = (m(E))^{(p_2−p_1)/(p_1p_2)}$ if $p_2 < \infty$ and $c = (m(E))^{1/p_1}$ if $p_2 = \infty$.

The textbook gives us two examples which I was a bit confused about...

Example 1 When $m(E) <\infty$, the inclusion is strict. For example, with E = (0, 1] and such that $−1/p_1 < \alpha < −1/p_2$. Define $f(x) = x^{\alpha}$. Then $f \in L^{p_1}(E)$ and $f \not\in L^{p_2}(E)$ (where, as above, $1 \leq p_1 < p_2 < 1$).

I don't understand why/how f is in $L^{p_1}$ but not $L^{p_2}$. I know that we have to check whether or not $[\int_E |x^{\alpha}|^{p_1}]^{1/(p_1)}$ and $[\int_E |x^{\alpha}|^{p_2}]^{1/(p_2)}$ are finite. But I'm not sure how that's done in this particular case.

Example 2 If $m(E) = \infty$, there is no inclusion relationship between the $L^p$ spaces. For example, with $E = (0,\infty)$ and $f(x) = \frac{x^{−1/2}}{1 + ln x}$ for x>1, , then $f \in L^p(0,1)$ if and only if p = 2.

I don't see how they can conclude that "$f \in L^p(0,1)$ if and only if p = 2".

Thanks in advance

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  • $\begingroup$ $1≤p_1<p_2≤1$? that trivialise the problem significantly :D $\endgroup$ – Lost1 Apr 11 '13 at 10:38
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Let's plug in some numbers :D

Let $p_1 = 1$ and $p_2 =2$. I claim $L^2 \subset L^1$ Moreover, the inclusion is strict.

So he said take $-1<\alpha < -1/2$, so take $\alpha=-3/2$ and $f=x^{-2/3 x}$

$||f||_1=\int^1_0 x^{-2/3 x} \text{dx}<\infty$

$||f||_2^2=\int^1_0 x^{-4/3 x} \text{dx} = \infty$

so $f$ is in $L^1$, not $L^2$, hence the strict inclusion.

For the 2nd one, if you try to integate $f(x)=\dfrac{x^{-1/2}}{1+\log x}$

$||f||_p^p= \int^1_0\dfrac{x^{-p/2}}{(1+\log x)^p} \text{dx}$. He is claiming that if $p=2$, this finite, if not, it is not. I think you do this substitution $y = \log x$ and convince yourself why it is infinite for $p\neq 2$

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