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Let $X = (0,1)\subset IR $ and consider the spaces $L_{5}(0, 1)$ and $L_{5/4} (0,1)$.

First question: Show that function $v(x) = ax^{- 3/5}$ is element of $L_{5/4} (0,1)$

Second question: Is function $u(x) = bx^{- 2/15}$ element of $L_{5} (0,1)$ ? Show that.

If I take $\|v\|_{5/4}=4^{4/5}|a|$, how can I show that. I am very new on this topic. Can you help me?

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    Try to apply the definition of the norm in $L^5$ and $L_{5/4}$ to your functions $v$ and $u$; the intergrals that you will obtain can be found explicitly. – 2017-01-10
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    @TZakrevskiy , $\|v\|_{5/4}=(\int v(x)^{5/4}d\mu)^{4/5} = (\int (ax^{- 3/5})^{5/4}d\mu)^{4/5}=(\int (ax)^{- 3/4}d\mu)^{4/5}=((ax)^{- 3/4}\mu)^{4/5}$ it is right? Excuse me I am so new on Lebesque spaces. – 2017-01-10
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    Does $\int_0^1 x^{-3/4} dx$ converge? What about $\int_0^1 x^{-10/15} dx$? – 2017-01-10
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    Yes they converge, so we can directly say that they are in L5 and L5/4,can't we ? – 2017-01-11

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