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I am not sure about this one, because if I assume that $f \in L^p$ then it means $$({\int |f|^p du)}^{1/p} < \infty$$ but if this is true, then, since $1 < p < \infty$ I can remove it from the norm and get $$({\int |f|^p du)} < \infty$$ and if this is true $$({\int |f|^r du)} < \infty$$ is surely true, because $r \leq p$ and if so, then adding the 1/r to the power of the integral should only make it smaller (since $1 \leq r)$ which means $f \in L^r$

This looks a bit too simplified to me, am I wrong here?

thanks in advance, I really appreciate this forum and all of your help!!

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    I think Hölder's inequality does the trick somewhere. I did this as an exercise once but I can't remember how. – 2017-01-24
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    For $0 < \lvert f(x)\rvert < 1$ and $r < p$, we have $\lvert f(x)\rvert^p < \lvert f(x)\rvert^r$. – 2017-01-24
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    @DanielFischer that is true, but if $|f(x)| <$ 1 then since $r \geq 1$ I will still get a finite number – 2017-01-24
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    But who says that the integral will remain finite? – 2017-01-24
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    @DanielFischer I see. so would I need to handle this case in a different way or is my whole solution wrong? – 2017-01-24
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    The result requires a condition on the measure (it's wrong in general, e.g. for the Lebesgue measure $\lambda$ on $\mathbb{R}$ we have $L^r(\lambda) \subset L^p(\lambda) \iff r = p$). You need to use that condition. – 2017-01-24
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    Ok. I forgot about the issue that the measure is finite, but I can't see how to use that – 2017-01-24

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This is incorrect in general. The added assumption that your measure space is finite would imply the containment (using Holder's inequality proves this).

As for a counter example to your case, consider

$$f(x)= \frac{1}{x^{1/p}}$$ for $|x| \le 1$ and

$$f(x) = \text{something in $L^p \cap L^r$,}$$ for $|x| \ge 1.$ We are using $\mathbb{R}$ with the Lebesgue measure as our measure space.

Hint for applying Holder's: Holder's inequality states that

$$\|hg\|_{L^1} \le \|h\|_{L^p} \|g\|_{L^q},$$

if $\frac{1}{p} + \frac{1}{q}=1$. If $r0$ such that $$ \frac{r}{q} + \frac{1}{s} = \frac{1}{q/r} + \frac{1}{s} =1.$$

Take $h=|f|^r$. Now what could you take $g$ as in Holder's inequality (as I've written above)?

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    I really cant see how finite measure space is helping me here :( – 2017-01-24
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    Starting with $$ \int |f|^r \cdot 1 d\mu$$ try applying Holder's with appropriate $p,q$. – 2017-01-24
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    What the indicator function has to do with all of this? I'm really at a loss – 2017-01-24
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    Sub in $h=|f|^r$, $g=1$, $p=q/r$ and $q=s$ into Holder's inequality (the one I wrote after 'hint:') I've abused notation by using $q$ twice, but I'm sure you can work it out. – 2017-01-24
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    For finite measure spaces, the inclusion is in the other direction, $1 \leqslant r \leqslant p \leqslant \infty \implies L^p \subset L^r$. – 2017-01-24
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    Yup, thanks. The method I've suggested proves exactly that. – 2017-01-25