Let $1\leq p<q\leq\infty$ and $-\infty\leq a<b\leq\infty$ be arbitrary. What specific functions satisfy $f\in L^p(a,b)$ and $f\notin L^q(a,b)$?
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I suppose a function $\psi(x)$ is in $L^n(a,b)$ if $\int_a^b|\psi(x)|^ndx$ converges. I also assume that $-\infty<a<b<+\infty$ then i guess $$\psi(x)=\frac{1}{(x-a)^\alpha}$$ will do for all $\alpha\in(q^{-1}, p^{-1})$

K. Sadri
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Why is it not in $L^q(a,b)$ ? – John Cataldo Apr 02 '18 at 07:23
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1$\alpha q>1$ and hence the integral diverges. (The singularity will be huge) – K. Sadri Apr 02 '18 at 07:36
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$L^{q}(X)\subseteq L^{p}(X)$ for $1\leq p\leq q\leq\infty$ for finite measure space $X$, this is by Holder's inequality.
For infinite measure, say, $L^{1}(1,\infty)$ and $L^{2}(1,\infty)$, consider $1/x\in L^{2}(1,\infty)$ and $1/x\notin L^{1}(1,\infty)$.
On the other hand, say, $L^{1}(0,1)$ and $L^{2}(0,1)$, consider $1/x^{1/2}\in L^{1}(0,1)$ and $1/x^{1/2}\notin L^{2}(0,1)$.

user284331
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1And though your example is a good one, I was wondering about general $p<q$ (the function would have to depend on p and q I guess) – John Cataldo Mar 31 '18 at 19:11
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1@StanislasHildebrandt Well, I guess you can generalize these examples for general $p,q$ quite easily, let $x^{\alpha}$ with a correctly chosen $\alpha$. – Surb Mar 31 '18 at 19:22