For any two different numbers $p,q\in[1,\infty)$ find functions $f\in L^p \setminus L^q$ and $g\in L^q \setminus L^p$.
Solution: let $$f(x)=x^{-1/p}(1+|\log x|)^{-2/p}$$ Then $$\int|f|^p = \int_0^\infty x^{-1}(1+|\log x|)^{-2} = \int_0^1 x^{-1}(1-\log x)^{-2} + \int_1^\infty x^{-1}(1+\log x)^{-2} \\ =\int_{-\infty}^0 \frac {du}{(1-u)^2} = \int_1^\infty \frac {2du}{u^2}<\infty$$ Let $$g(x)=x^{-1/q}(1+|\log x|)^{-2/q}$$ Then $$\int|g|^q = \int_0^\infty x^{-1}(1+|\log x|)^{-2} = \int_0^1 x^{-1}(1-\log x)^{-2} + \int_1^\infty x^{-1}(1+\log x)^{-2} \\ = \int_{-\infty}^0 \frac {du}{(1-u)^2} = \int_1^\infty \frac {2du}{u^2}<\infty$$
Is this a correct solution, thanks.
It's a reasonable solution. For extra clarity, you should mention that you work on $(0,\infty)$ with the Lebesgue measure, and what you actually construct is a function that belongs to $L^p$ only for one specific $p$. (This is something you definitely should have checked). This of course accomplishes the task. The second calculation is redundant: you could have just said that $g$ is like $f$, only with $q$ instead of $p$. No need to repeat the same integration steps.